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Contents
1 Prelude
1
1.1 What is music? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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1.2 What is math? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2
1.3 Solutions to exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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2 Math for Poets
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2.1 Meter as binary pattern . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6
2.2 The Hemachandra-Fibonacci numbers . . . . . . . . . . . . . . . . . . . . . . . .
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2.3 The expanding mountain of jewels . . . . . . . . . . . . . . . . . . . . . . . . . . 12
2.4 Naming meters and de Bruijn sequences . . . . . . . . . . . . . . . . . . . . . . . 13
2.5 Patterns in music and architecture . . . . . . . . . . . . . . . . . . . . . . . . . . 14
2.6 Solutions to exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
3 Rhythm
21
3.1 Measuring time . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21
3.2 Rhythm and groove . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23
3.3 Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24
3.4 The beat class circle and modulus . . . . . . . . . . . . . . . . . . . . . . . . . . . 25
3.5 Polyrhythm . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28
3.6 Additional applications of the modulus . . . . . . . . . . . . . . . . . . . . . . . . 29
3.7 Solutions to exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30
4 Pitched sounds
32
4.1 Periodicity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32
4.2 Frequency . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33
4.3 Amplitude and envelopes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36
4.4 Sinusoids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37
4.5 The Superposition Principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39
4.6 Beats . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42
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CONTENTS
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4.7 The wave equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46
4.8 Timbre and spectrum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51
5 Intervals, Pitch, and Pitch Class
53
5.1 Intervals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53
5.2 Measuring intervals using semitones . . . . . . . . . . . . . . . . . . . . . . . . . 55
5.3 Application: the major scale . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56
5.4 Absolute pitch . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57
5.5 Pitch classes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59
6 Transformation
62
6.1 Clapping math . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62
6.2 Rhythm class . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63
6.3 Transposition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65
6.4 Solutions to exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67
A Worksheets
68
Poetic meters of ixed duration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69
Musical rhythms with 2- and 3-beat notes . . . . . . . . . . . . . . . . . . . . . . . . . 70
The expanding mountain of jewels . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71
Patterns in the meruprastāra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72
Sinusoids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74
Intervals and pitch . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76
B Assignments
79
Chapter 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80
Chapter 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80
Chapter 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85
Chapter 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88
Chapter 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92
C Review Problems
94
Review Problems for Test 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95
Review Problems for Test 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99
Formula sheet . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105
Review for Final Exam (Part 1) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 106
Review for Final Exam (Part 2) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 108
CONTENTS
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D Quizzes
109
Index
115
Chapter 1
Prelude
Mathematics and music are just two of the many ways we respond to the world around
us. Mathematics describes quantity and space, while music involves the appreciation
of sound. Both math and music are intentional human creations. Like mathematicians,
musicians describe concepts like “chords” that cannot be touched or seen. Like musicians,
mathematicians speak of “beauty” or “elegance”—in their case, of a particular equation,
theorem, or proof. Both disciplines rely on precise notation.
1.1
What is music?
What is the difference between “sound” and “music?” We can probably agree that the
styles of popular and classical music that we make or consume are “music.” However, it is
dif icult to de ine “music” in general. Some composers, such as John Cage (1912-1992), have
deliberately challenged our notions of what music is. Watch the video of John Cage’s 4’33”
(1952).
Although there is no universally accepted de inition of music, I like the French composer
Edgar Varese’s description that “music is organized sound.” This organization can be seen
on many different levels, from the sounds that musical instruments make to the way a
composer structures a piece. I would add that music exists in time and that music is a form
of expression. For the purposes of this class, I’m going to use the de inition
Music is the art of organizing sound in time.
Discussion.
1. Watch The Triplets of Belleville perform. What is the difference between vacuuming
and playing a vacuum cleaner as a musical instrument? At what point does it become
clear that the triplets are performing music, rather than some other activity?
2. Can silence be music? What is the difference between sitting in silence and watching a
performance of John Cage’s 4’33”? Or performing the piece yourself? John Cage
focused on the listener as the creator of music (or, at least, the person who makes
something music rather than sound). He wrote, “If music is the “enjoyment” of
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“sound”, then it must center on not just the side making the sound, but the side
listening. In fact, really it is listening that is music. As we savor the sound of rain,
music is being created within us” (from “In this time”).
3. Is there a difference between poetry and music? Does rap count as music? What about
rap in sign language? Perhaps we should focus on the intent of the creator, or
musician. Taking this view, music is “the creation of a musician.”
4. Does my de inition of music as “the art of organizing sound in time” include things that
you do not consider “music?”
5. Should our aesthetic judgements—what we like—have any relevance in the de inition
of music?
Found sound. A found sound is a sound that is in the world around you but not intended
to be “music” by whoever or whatever made it. Some musicians record found sounds or
music produced by other musicians and incorporate them into their own compositions. This
process is called sampling. Do you think sampling is creating music?
When you record a sound, the picture of the sound that you see is called a waveform. It
is a visual representation of the vibration of air molecules that produced the sound. The
horizontal axis represents time, which is usually measured in seconds. The vertical axis
shows you how the air molecules are moving. The rough shape of the waveform when you
zoom out gives you information on how the volume of the sound changes. It often looks like
a series of blobs. You can also zoom in on the waveform. The patterns that you see when
you zoom in give information on the timbre (TAM-ber) of the sound. Timbre is dif icult to
describe, but, loosely, it’s the quality of the sound that allows you to distinguish between
different instruments, voices, and other sounds.
Exercise 1.1. Install software that allows you to record and analyze sound. I recommend Audacity
for a computer, TwistedWave for iPhone, and MixPad for Android. Audacity has been installed on
many computers on campus. Make ive-second recordings of yourself talking, humming, clapping,
and saying “shh.” Comment on the differences between the pattern (or lack of pattern) in their
waveforms, both zoomed out and zoomed in.
1.2
What is math?
The philosopher Michael Resnik (1981) called mathematics “a science of pattern.”
Mathematical science precisely describes structure, both in the physical world and in the
abstract. It has been part of a liberal arts education from the beginning. It trains us in
using abstraction and in forming logical arguments. Though few people are professional
mathematicians, all humans are mathematical thinkers: we engage informally in ideas of
quantity, pattern, space, and logic. Music theory, the study of structure in music, is a type of
mathematical thinking.
The distinction between mathematical science and mathematical thinking is in the use of
precision and rigor. Mathematical science is precise and logical. It’s what you learn in
math class and what mathematicians do. Proofs need to be so logically rigorous that they
CHAPTER 1. PRELUDE
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can withstand all potential challenges. Mathematical thinking is any activity that involves
number, geometry, informal logic, etc. It doesn’t have to be formal, and you don’t have to
have a precise answer. So reading a map or calculating a tip are examples of mathematical
thinking, but not mathematical science.
Here are some of the basic terms that mathematicians use to describe what they do.
De inition. A de inition is a precise description of a mathematical term. For example,
integers are de ined to be the numbers {. . . − 3, −2, −1, 0, 1, 2, 3, . . .} that have no fractional
part. (Some authors use the term “whole numbers” to refer to the nonnegative integers
{0, 1, 2, . . .}, while others say that whole numbers also include negative integers. We’ll use
the term “integers” because it is agreed on by everyone.) “A number is even if it equals 2n,
where n is an integer” is the de inition of the mathematical term even.
Statement. I de ine a statement to be a sentence that is either true or false. “Two is an even
number” and “3 + 2 = 5” are true statements. “Two is an odd number” is a false statement.
“Figure out whether two is even,” “Seven is the best number,” “5 + x = 2,” and “Is two an even
number?” are not statements. “This sentence is false” sounds like it might be a statement,
but it’s neither true nor false. It is an example of a paradox.
Axiom. In order to do any useful mathematics, we all have to agree that certain statements
are true without proof. For example, the Peano axioms are needed to establish the de inition
of integers and the rules of arithmetic. The Parallel postulate in geometry states that parallel
lines never intersect, where two lines are parallel if they are perpendicular to the same
line. Surprisingly, it is not possible to prove that they do not intersect. In fact, the parallel
postulate must be accepted as true in Euclidean geometry, which is the geometry of the
plane that you learned in high school. (Parallel lines do intersect on the globe, and this type
of geometry is called spherical geometry.)
Theorem. A theorem is a statement pertaining to mathematics or logic that has been
proven to be true. Theorems build on de initions, axioms, demonstrably true statements,
and sometimes other theorems. The statement “the sum of two even numbers is an even
number” is a theorem.
Proof. A proof is a rigorous, logical argument that demonstrates that a theorem is true.
The theorem “the sum of two even numbers is an even number” can be proven using a logical
argument, assuming that the Peano axioms are true and following from the de inition of an
even number. Here is the proof. The box at the end of the paragraph marks the end of the
proof.
Proof. Suppose a and b are even numbers. The de inition of “even number” means that there
are integers n and m where a = 2n and b = 2m. By the rules of addition and multiplication,
a + b = 2n + 2m = 2(n + m), which is an even number because n + m is an integer.
2
Notice that this proof implicitly relies on the Peano axioms, because we need the rules of
arithmetic and the de inition of integers to complete the proof.
CHAPTER 1. PRELUDE
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Example. An example is an illustration of a theorem in a particular instance. The equation
2 + 4 = 6 is an example of the theorem “the sum of two even numbers is an even number.”
This example doesn’t prove the theorem, because there are in initely many even numbers
to add, and you can’t test them all. Unless you are able to demonstrate that all possible
examples of a theorem are true, examples do NOT prove a theorem.
Conjecture. A conjecture is a statement pertaining to mathematics or logic that is made
without proof. Conjectures are usually “educated guesses” based on some pattern that has
been observed. “The sum of two integers is greater than either integer” is a conjecture.
There exist some famous conjectures that mathematicians have never been able to prove
true or false.
Counterexample. A counterexample is an example that proves that a conjecture is false. A
counterexample to “The sum of two integers is greater than either integer” is −1 + 2 = 1,
which is not larger than 2.
Algorithm. An algorithm is a set of mathematical instructions. An algorithm for
determining whether a number is even is the following: Divide the number by 2. If the
remainder is 0, the number is even.
Exercise 1.2. Prove that 0 is an even number.
Exercise 1.3. A number is odd if it equals 2n + 1, where n is an integer. Suppose that a student
wishes to prove that the sum of two odd numbers is an even number. Explain why the following is not
a proof.
Proof. We can test odd numbers: 1 + 1 = 2, 1 + 3 = 4, 3 + 1 = 4, 1 + 5 = 6, etc. This also works for
negative odd numbers. For example, −1 + 3 = 2, −1 + (−3) = −4, and −5 + 5 = 0. In each case, odd
numbers sum to an even number.
2
Exercise 1.4. Find a counterexample to the statement “the product of two odd numbers is an even
number.”
Exercise 1.5. Explain why the statement “All music is made by musical instruments or voices” is
false.
Exercise 1.6. A universal statement is one that is asserted for in initely many examples. Explain why
many examples don’t prove that a universal statement is true, but one counterexample disproves a
universal statement.
1.3
Solutions to exercises
Solution 1.1. Figure 1.1 shows a sample response. The top image shows the overall shape of the
waveforms and gives you information about the rhythm of the sound. Clapping has a repeating
pattern representing the rhythm of the claps. Zooming in on the waveform gives information about
the “quality” (timbre) of the sound. Only humming shows a repeating pattern. Notice that the speech
CHAPTER 1. PRELUDE
5
transitions between a random pattern coinciding with the “s” in music and a repeating pattern
coinciding with the “i.” In speech, vowels look more like humming than consonants such as “s” do.
Figure 1.1: In the top image, the irst waveform is me speaking the sentence “Music is the art of
organizing sound in time.” The others are, in order, humming, clapping, and saying “shh.” The
bottom image shows what they look like around the 0.7 second mark.
Solution 1.2. Proof. Since 0 = 2 · 0 and 0 is an integer, 0 is even by the de inition of even.
2
Solution 1.3. The student has not tested all odd numbers. That would be impossible because there
are in initely many odd numbers. A logical argument must be used, such as the proof on page 3.
Solution 1.4. A counterexample is 5 · 3 = 15, because 5 and 3 are odd and 15 is also odd.
Solution 1.5. You need to ind a counterexample—that is, some music that is not made by musical
instruments or voices. Two counterexamples are the Triplets of Belleville’s piece that uses household
objects and algorithmic music made by a computer.
Solution 1.6. In the case of a true universal statement, there are in initely many examples to test, and
testing some of them does not prove that they are all true. Therefore, the rules of logic must be used.
In the case of a false universal statement, demonstrating that it is false in one instance means that the
statement is not universal and therefore is false.
Chapter 2
Math for Poets
...
But most by Numbers judge a Poet’s Song,
And smooth or rough, with them, is right or wrong;
...
These Equal Syllables alone require,
Tho’ oft the Ear the open Vowels tire,
While Expletives their feeble Aid do join,
And ten low Words oft creep in one dull Line,
...
—Alexander Pope, An Essay on Criticism (1709)
Numerical patterns have fascinated humans for millennia: numbers that are powers of
other numbers, squares that are sums of squares, numbers that form intriguing lists. This
is the story of one of the earliest studies of rhythm, an investigation that led ancient Indian
scholars to discover the mathematical patterns that Westerners know as the Fibonacci
numbers, Pascal’s triangle, and the binary counting system. Although our story initially
concerns rhythm in poetry, the Ancient Indians’ ability and fascination with exploring
rhythmic patterns also had a profound in luence on their music.
2.1
Meter as binary pattern
In English, a poetic rhythm, called a meter, is a pattern of stressed and unstressed syllables.
English poets use about a dozen different meters. Much poetry, including Shakespeare’s
plays, is written in iambic pentameter— ive pairs of alternating unstressed and stressed
syllables to a line. Alexander Pope’s 700+ line iambic pentameter poem An Essay on Criticism
(1709) is a good example. In the excerpt that begins this chapter, he ridicules critics who
judge poetry “by numbers”—that is, solely on how well a poet follows strict metrical rules.
While English poets use relatively few meters, there are hundreds of different meters
in Sanskrit, the classical language of India. Syllables in Sanskrit poetry are classi ied by
duration (short or long) rather than stress. Any Sanskrit meter can be written as a binary
pattern—a pattern of any length formed by two symbols. For example, there are eight binary
patterns of length 3 that are formed from the letters L and S: LLL, SLL, LSL, SSL, LLS, SLS, LSS,
and SSS. These correspond with the eight three-syllable meters, using S for a short syllable
and L for a long syllable.
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CHAPTER 2. MATH FOR POETS
7
Pingala is thought to be the irst Indian scholar to study meter mathematically. He probably
lived in the last few centuries BC. As is typical in ancient Indian literature, Pingala’s writings
took the form of short, cryptic verses, or sūtras, which served as memory aids for a larger set
of concepts passed on orally. We are dependent on medieval commentators for transmission
and interpretation of Pingala’s writings.
Here are two of the questions that Pingala solved:
1. What is a reliable way to list all the meters with a given number of syllables?
2. How many meters have a given number of syllables?
Problem 1: listing meters. There are a number of ways to solve this problem. Pingala’s
solution would result in the one-syllable meters listed as
L
S
the two-syllable meters being listed as
LL
SL
LS
SS
and the three-syllable meters like this:
LLL
SLL
LSL
SSL
LLS
SLS
LSS
SSS
Here is how the four-syllable patterns would be listed:
LLLL
SLLL
LSLL
SSLL
LLSL
SLSL
LSSL
SSSL
LLLS
SLLS
LSLS
SSLS
LLSS
SLSS
LSSS
SSSS
Several patterns are observable in these lists. The irst column alternates L and S, the second
alternates pairs of L’s and pairs of S’s, the third alternates four copies of the letters, and so on.
There is symmetry in the lists, in the sense that the irst pattern is equivalent to the last with
the letters exchanged, and this is true for each pair of patterns that are at the same distance
from the beginning and end. In addition, there is a relationship between successive lists:
for example, the list of four-syllable meters is formed from the list of three-syllable meters
by irst adding L’s to the end of the list, then adding S’s. This last observation is useful for
describing an algorithm that will produce all meters of length n in the order that Pingala did.
CHAPTER 2. MATH FOR POETS
8
Exercise 2.1. Write the list of ive-syllable meters.
Theorem 2.1 (Listing n-syllable meters). Suppose the list of one-syllable patterns is {L, S},
and the list of n-syllable patterns is formed from the list of (n − 1)-syllable patterns by adding
L’s to the end of each (n − 1)-syllable pattern, followed by the list resulting from adding S’s to
the end of each (n − 1)-syllable pattern. Then each n-syllable pattern will occur exactly once in
the list.
Proof. It is clear that each of the one-syllable meters is listed once. Suppose that the
algorithm results in each of the (n − 1)-syllable patterns being listed exactly once. Choose
any n-syllable pattern. Use the algorithm to form a list of n-syllable patterns. We want to
show that your chosen pattern occurs exactly once in the new list. If the pattern ends in an L,
then the algorithm shows that it appears exactly once in the irst half of the list, because the
pattern of its irst (n − 1) syllables appear exactly once in the list of (n − 1)-syllable patterns.
If it ends in an S, it appears exactly once in the second half of the list for the same reason. 2
The proof essentially says that if the one-syllable patterns are correct, then the two-syllable
patterns are correct, then the three-syllable patterns are correct, and so on, until your chosen
length is reached—sort of like a row of dominoes falling down. This type of reasoning is
called proof by induction. Although it may seem reasonable to argue this way, in fact, the
axiom of induction is required to allow it.
Problem 2: counting meters. How many meters have n syllables? Counting the patterns
on the lists I have typed, you see the numbers 2, 4, and 8, which are equal to 21 , 22 , and
23 . You might conjecture that there are 16 (or 24 ) four-syllable meters, and, in general,
there are 2n n-syllable meters. This is correct. It follows so closely from the theorem that
mathematicians would called it a corollary, which is a theorem that may be proven from
another theorem without much effort.
Corollary 2.1 (Counting n-syllable meters). The number of n-syllable meters is 2n .
Exercise 2.2. How long is the list on which LLSSLSL appears?
Exercise 2.3. Assuming that the theorem has been proven true, explain why the corollary is true. You
don’t need to write a formal proof.
The binary number system Since there’s nothing special about the letters L and S, the
previous theorem and its corollary generalize to any set of binary patterns. For example,
there are 25 patterns of length 5 that are formed from the letters a and b. In some ways,
Pingala anticipated the development of the binary number system. The binary number
system is a base-two positional number system (our number system is a base-ten positional
system). It has two digits, 0 and 1, and its place values are powers of two—therefore,
every number is also a binary pattern. The decimal numbers 1, 2, 8, and 11 have binary
representation 1, 10, 1000, and 1011, respectively. The binary number system was not fully
described until Gottfried Leibniz did so in the seventeenth century.
Exercise 2.4. Suppose you lip a coin three times and write down the pattern of heads and tails,
using H and T. The order of lips makes a difference—that is HHT is different from HTH. How many
CHAPTER 2. MATH FOR POETS
1 beat
S
9
2 beats
L
SS
3 beats
SL LS
SSS
4 beats
LL SLS
SSL LSS
SSSS
5 beats
SLL
LLS
LSL SSLS
SSSL SLSS
LSSS
SSSSS
Figure 2.1: Meters listed by duration
patterns of three coin lips are there? List them and use your list to compute the likelihood of getting
tails exactly once if you lip three times. Describe how you would list and count the patterns for any
number of coin lips.
Exercise 2.5. Here are the binary numbers from 8 (1000 in binary) to 15 (1111 in binary):
1000
1001
1010
1011
1100
1101
1110
1111
How is this sequence related to the list of three-syllable meters? Conjecture how many
binary numbers have ive digits and list them. (Extra Credit: Learn more about the binary
number system and determine whether your answer is correct.)
2.2
The Hemachandra-Fibonacci numbers
The 12th-century writer Ācārya Hemachandra also studied poetic meter. A mora is the
durational unit of Sanskrit poetry; short syllables count as one mora and long syllables two
morae, which we’ll call “beats.” Instead of counting meters with a ixed number of syllables,
Hemachandra counted meters having a ixed duration. For example, here are the three
meters of three beats: SL, LS, and SSS. More meters are listed in Figure 2.1.
Exercise 2.6. Before you go on, count the number of meters for duration one through ive and make
a conjecture about the number of meters with six beats and the formula for inding the number
of meters with any arbitrary number of beats. Try the worksheet on page 69 for a more thorough
exploration of the problem.
Hemachandra noticed that each number in the sequence is the sum of the two previous
numbers. Since the irst two numbers are 1 and 2, the numbers form the sequence
1, 2, 3, 5, 8, 13 . . .. In other words, he discovered the “Fibonacci” numbers—about ifty years
before Fibonacci did.¹ Indian poets and drummers know these numbers as “Hemachandra
¹Fibonacci may have learned the sequence from the Indians. Fibonacci was educated in North Africa and
was familiar with Eastern mathematics. His Liber Abaci (1202), in which the sequence appears, introduced
the Indian positional number system—the system we use today—to the West. However, his description of the
CHAPTER 2. MATH FOR POETS
10
Start with all the n × 1
patterns in one big pile
Separate them into two
smaller piles:
those
ending
in a
square
those
ending
in a
domino
Remove the inal blocks.
You are left with all the
(n − 1) × 1 patterns plus
all the (n − 2) × 1 patterns.
Figure 2.2: Hemachandra’s problem is equivalent to the “domino-square problem”: in how many ways can
1 × 2 dominoes and 1 × 1 squares tile a 1 × n rectangle? Here is a visual demonstration that the nth number
in the sequence is the sum of the two preceding numbers.
numbers.”
Theorem 2.2. The sequence of numbers of meters with n beats, beginning with n = 1, is the
Hemachandra sequence, 1, 2, 3, 5, 8, 13, . . .. When n > 2, each number in the sequence is the
sum of the two previous numbers.
Proof. Suppose Hn represents the nth number in the sequence, which equals the total
number of patterns of duration n. Since there is one pattern (S) of duration 1, H1 = 1, and
since there are two patterns (SS and L) of duration 2, H2 = 2. When n > 2, partition the
collection of n-beat patterns into two sets: patterns of duration n − 2 followed by a long
syllable and patterns of duration n − 1 followed by a short syllable. The number of patterns
in the irst set equals Hn−2 , since they are formed by adding L to the patterns with n − 2
beats, and the number of patterns in the second set equals Hn−1 , since they are formed by
adding S to the patterns with n − 1 beats. The partition shows that Hn = Hn−1 + Hn−2 when
n > 2. Therefore, the list of numbers forms the Hemachandra sequence.
2
Figure 2.2 gives a visual demonstration in which short and long syllables are represented by
squares and dominoes, respectively. In the picture, n = 5, but the same argument may be
made for any n greater than 2.
Exercise 2.7. The procedure for writing the patterns with duration n as a combination of patterns of
durations n − 1 and n − 2 that is explained in the proof suggests how use the information in Figure 2.1
to list the 13 patterns with six beats. Write them.
Recursion. Recursion is a process in which one structure is embedded inside another
similar structure, rather like nesting Russian dolls, or the Droste cocoa box. Recursion is the
lifeblood of computer programming and is crucial in mathematics as well. An algorithm is a
recursive if you start with some information (called a base case) and arrive at all subsequent
information by repeatedly applying the same rule, called a recursive rule.
In the example of the Hemachandra numbers, there are two meters of one syllable each, L
and S. This is the base case. If you know all the meters that have n syllables, you can list the
number sequence as counting the sizes of successive generations of rabbits is not found in India.
CHAPTER 2. MATH FOR POETS
11
Rhythms of one- and two-beat notes
merengue bell part (Dominican Rep.)
cumbia bell part (Columbia)
mambo bell part (Cuba)
bintin bell pattern (Ghana)
also bembe shango (Afro-Cuban)
Rhythms of two- and three-beat notes
lesnoto (Bulgaria)
bomba (Puerto Rico)
guajira (Spain)
12-beat clave (Cuba)
Figure 2.3: Musical rhythms.
meters that have n + 1 syllables by irst adding an L to the beginning of the meters with
n syllables, then adding an S to the beginning of the meters with n syllables. This is the
recursive rule that is expressed by the formula Hn = Hn−1 + Hn−2 .²
The Padovan sequence. The poetic meters that Pingala and Hemachandra studied have
an analogue in music. Music from India, the Middle East, and the Balkans is often written
in additive meter—that is, a rhythmic organization founded in grouping beats rather than
subdividing larger units of time called measures, which is the typical structure of Western
European music (see Chapter 3 for a fuller explanation).
Figure 2.3 shows a few examples. The Bulgarian dance called Daichovo horo has a nine-beat
measure, grouped 2+2+2+3. This means that the irst, third, ifth, and seventh beats normally
receive an accent; they are also the beats on which the dancers step. The jazz pianist and
composer Dave Brubeck (1920-2012) used the same rhythm in his “Blue Rondo à la Turk”
(1959). A Gankino horo has an eleven-beat measure, with beats grouped 2+2+3+2+2.
Many additive meters are binary patterns formed of two- and three-beat groupings. Pingala’s
algorithm will list all additive meters formed of n groups of beats for any n. However,
musical patterns are typically classi ied by their number of beats. In this situation, we need
something like Hemachandra’s sequence for counting meters of a given duration, as explored
in the following exercise:
Exercise 2.8. This problem is explored in the worksheet on page 70. The Hemachandra numbers
count meters formed from one- and two-beat groups. What sequence counts meters consisting of
two- and three-beat groups? Find the irst few numbers in the sequence—the base case—and a
recursive rule that generates the sequence. Explain why your rule is correct. This number sequence
is called the Padovan sequence and has a rich history. Discuss. (The On-Line Encyclopedia of Integer
Sequences is a wonderful research tool for this sort of problem.)
Exercise 2.9. (Nonretrogradable rhythms) A rhythm is nonretrogradable if its sequence of durations
²If the notation Hn , Hn−1 , etc. is unfamiliar, it’s worth taking time to understand it. Since Hn refers to the
nth number in the sequence, Hn−1 is the (n − 1)th number—that is, the number preceding Hn . The equation
Hn = Hn−1 + Hn−2 means “the nth number is the sum of the number that is one place before it and the number
that is two places before it.” For example, if n = 5, then H5 = H5−1 + H5−2 = H4 + H3 . In words, the ifth
number is the sum of the fourth number and the third number. Note that a small difference in typography makes
a big difference in meaning: H5−1 ̸= H5 − 1 because H5−1 = H4 = 5, while H5 − 1 = 8 − 1 = 7.
CHAPTER 2. MATH FOR POETS
12
Figure 2.4: The meruprastāra from North Africa (c.1150) to China (1303) to Germany (1527)
reads the same forwards as backwards. Examples include 1+2+1 and 2+2+1+1+2+2. Find the number
of nonretrogradable rhythms of length n beats and composed of notes of durations 1 and 2.
2.3
The expanding mountain of jewels
Before reading this section, I recommend trying Worksheet A.
Pingala is credited with the discovery of “Pascal’s” Triangle in India, which he called the
meruprastāra, or “the expanding mountain of jewels.” Mount Meru is a mythical mountain
made of gold and precious stones. The nth row in this triangle counts the number of
unordered combinations of n syllables: taking all n, taking (n − 1), taking (n − 2), and so on,
where each syllable is considered different, rather than just long or short.
Here is how to compute the third row in the meruprastāra. There is one way of choosing
all three syllables from the word “prastāra” (that is, pras+tā+ra), there are three ways of
choosing two syllables (pras+tā, pras+ra, tā+ra), and there are three ways of choosing one
syllable (pras, tā, ra). In order to complete the third row, note that there is one way to choose
no syllables. Therefore, the third row is 1 3 3 1.
When each list counting combinations of r syllables drawn from sets of n syllables is written
in a row, and the rows are stacked, the numbers form a triangular array that extends forever:
1
1
1
1
1
2
3
4
1
3
6
1
4
1
Pingala recognized that each interior number is the sum of the two numbers above it. This
array is known in the West as Pascal’s triangle—though, of course, it wasn’t yet named for
Pascal, who was born in France in 1623. This triangle has long been recognized all over the
world. Figure 2.4 shows three familiar images.
CHAPTER 2. MATH FOR POETS
13
Exercise 2.10. Bhatt discovered the meruprastāra in a different context: he found the number of
meters of n syllables having r short syllables. This is the problem that was solved in Worksheet A.
The fact that the two problems produce the same triangle is, of course, no coincidence. Find an exact
correspondence between the number of combinations of r objects drawn from a set of n different
objects and the number of meters of n syllables having r short syllables.
Recursion and the meruprastāra The irst row of the meruprastāra contains the
numbers 1, 1. This is the base case. If you know any row in the meruprastāra, each number
in the following row equals the number directly above it plus the number diagonally above
and to the left (if there is no number in these positions, add zero). This is the recursive rule.
The 12th-century writer Bhaskara gives another recursive algorithm for inding the numbers
in the meruprastāra in his work Lilavati. To ind the nth row in the meruprastāra, start by
writing the numbers 1, 2, . . ., n, and above them write the numbers n, n − 1, . . ., 2, 1, like so
(shown for n = 5):
5
1
4
2
3
3
2
4
1
5
The irst number in the row is 1 (this is true for every n). Obtain the other numbers in the
row by successively multiplying and dividing by the numbers you have written:
5
= 5;
1
5·
4
= 10;
2
10 ·
3
= 10;
3
5
10
10 ·
2
= 5;
4
5·
1
= 1.
5
This tells us that the ifth row is
1
10
5
1
The numbers in row n are built up recursively, one from the next, starting from 1 (the base
case).
Exercise 2.11. Find the sixth row of the meruprastāra using Bhaskara’s method. Check your work
using the addition algorithm, starting with the ifth row.
2.4
Naming meters and de Bruijn sequences
Since there are hundreds of Sanskrit meters, remembering the pattern for any particular
meter requires some effort. Pingala’s indexing procedure helps somewhat; for example, it
identi ies the pattern LLLSLS as “number forty-one in the catalog of six-syllable meters.”
However, Pingala’s best and most well-known solution to the problem of remembering
meters involves the following mapping of groups of three syllables to letters:
m LLL
y SLL
r
s
LSL
SSL
t
j
LLS
SLS
bh
n
LSS
SSS
To encode the meter LLLSLS, begin by breaking it into groups of threes (LLL-SLS). These
groups correspond to the letters mj. The letters mj can be embedded in a word—say,
CHAPTER 2. MATH FOR POETS
14
“mojo”—that is more memorable that “number forty-one.” Musicians also use this method
for remembering rhythm patterns.
Either in Pingala’s time or later, the nonsense word yamātārājabhānasalagām came to be
used as a way to remember the mapping of triplets of syllables to letters. The word de ines
a pattern of long and short syllables (in the English transliteration of Sanskrit, a is a short
vowel and ā is a long vowel):
ya mā
S
L
tā
L
rā ja
L S
bhā
L
na
S
sa
S
la
S
gām
L
The pattern SLLLSLSSSL has the curious property that each sequence of three syllables
occurs exactly once. For example, the irst three syllables form the pattern SLL, the second
through fourth syllables form LLL, and so on. These patterns are named using Pingala’s
table, so that ya represents SLL. Since the number of syllables in a meter does not have to
be a multiple of three, the last two syllables, la and gām, are used for the leftovers. It is not
known whether Pingala knew this mnemonic for the triplets, or if it was discovered by poets
and drummers that came after him.
The mathematician Sherman Stein recognized that the pattern SLLLSLSSSL is close to being
a de Bruijn sequence. A de Bruijn sequence is a binary pattern where, if the pattern were
wrapped on a circle, every “word” of n letters would appear exactly once. The pattern
SLLLSLSS is a binary de Bruijn sequence for three-letter words. Figure 2.5 depicts de Bruijn
sequences for three- and four-letter words arranged on circles.
S
S
L
S
S
L
L
L
S
L
L
L
S
S L L
S L S
L
S
L
S
S
Figure 2.5: De Bruijn sequences for binary patterns of three and four letters, arranged on circles.
Exercise 2.12. Although LLLSLSSS is also a de Bruijn sequence, it is not fundamentally different
from SLLLSLSS , as each sequence produces the three-letter words in the same order, though starting
at a different point in the cycle. Using this notion of equivalence, there are only two possible de Bruijn
sequences for three-letter words using the alphabet {L, S}. How can we ind the other one?
2.5
Patterns in music and architecture
Indian scholars had a great enthusiasm for solving mathematical problems related to poetry.
However, the situation of pattern in Indian music tells a more remarkable story. There,
the list of patterns has transcended its role as a “dictionary” of available patterns and has
CHAPTER 2. MATH FOR POETS
15
Theme
Dhi
ta
ta
ghi
(ta
Dhi
Dhi
Ta
|: (ta
Dhi
|: (ta
Ta
|: (ta
ta
|: ta
Ta
|: (ta
Dhi
(ta
Dhi
Ta
ta
Variation I.
6+8+6+6+6
Variation II.
6+6+4+6+6+4
Variation III.
6+4+6+6+4+6
Variation IV.
10+6+10+6
Variation V.
4+6+10+6+6
Variation VI.
6+10+6+10
Variation VII.
6+6+6+8+6
ne
ke
ge
de
ge
ne
ne
ke)
ge
ne
ge
ke)
ge
ke)
ke
ke)
ge
ne
ge
ne
ke)
ke)
Ta
ta
Dhi
na
Dhi
Ta
Ta
(ta
Dhi
Ta
Dhi
(ta
Dhi
(ta
ta
(ta
Dhi
Ta
Dhi
Ta
(ta
(ta
ge
ge
ne
ge
ne
ke
ke)
ge
ne
ge)
ne
ge
ne
ge
ke)
ge
ne
ke
ne
ke)
ge
ge
Dhi
Dhi
Ta
Tin
Ta
ta
(ta
Tin
Ta
(Dhi
Ta
Dhi
Ta
Dhi
(ta
Dhi
Ta
ta
Ta
(ta
Dhi
Dhi
ne
ne
ge
na
ke)
ke)
ge
ne
ke)
ne
ge)
ne
ke
ne
ge
ne
ke)
ke
ke)
ge
ne
ne
Ta
Ta
te
Ta
(ta
(ta
Dhi
Ta
(ta
ta
(Dhi
ta
ta
Ta
Dhi
Ta
(ta
ta
(ta
Dhi
Ta
Ta
ge
ke
te
ke
ge
ge
ne
ke)
ge
ke) :|
ne
ke) :|
ke
ke) :|
ne
ke :|
ge
ke) :|
ge
ne
ke
ke)
Figure 2.6: Theme and variations for tabla, as taught by Lenny Seidman. Syllables such as “Dhi” and “ne”
indicate particular ways of hitting the drums’ heads to produce sounds. Each syllable occupies the same
amount of time. The symbols |: and :| indicate repeats and parentheses enclose phrases.
become itself an interesting and valuable musical structure. In music, “prastāra”—meaning
systematic permutation of rhythmic elements—is commonly recognized as a principal part
of the process of rhythmic variation. Indian musicians typically use prastāra towards the
end of a piece, since progression through all the permutations of a rhythmic pattern is a
process that has a de inite ending—that is, when all the possibilities have been exhausted.
Lewis Rowell’s description of prastāra in early Indian music is also applicable today:
Once again we can draw an important formal conclusion from the popularity of
prastāra: endings are to be signaled well in advance by the onset of some
systematic musical process, a process of playful exploitation that can be followed
along a course of progressively narrowed and focused expectations and that
leads inexorably to a predictable conclusion. […] But prastāra has symbolic
overtones that transcend its local role as a simple tactic of closure: the device
mimics the series of transformations through which all substance must
eventually pass [?, p. 251].
Figure 2.6 demonstrates the use of permutation in a simple composition for tabla (Indian
drums). Each variation on the theme is a permutation of groups of 4, 6, 8, and 10 beats:
I.
68666
II.
664664
III.
646646
IV.
10 6 10 (6
V.
4) 6 10 6 6
VI.
6 10 6 10
VII.
66686
Variations II and III state two of the three permutations of {6, 6, 4}, with 4+6+6 missing.
Variations IV, V, and VI exhaust the permutations of the 10- and 6-beat phrases (note that
variation V begins with the last four beats of the 10-beat phrase combining with the last 6
beats of variation VI to form the 10-beat phrase). Finally, variation VII, a mirror image of I,
signals that the permutation process has come to a close.
As Rowell points out, we can also understand prastāra as manifesting a fascination with
recursive generation and transformation that appears in Indian art, architecture, and
religion from ancient times. The medieval Śekharī (“multi-spired”) temples of western
and central India gave form to the view that the cosmos was recursively generated. The
eleventh-century Kandāriyā Mahādeva temple (Figure 2.7) is a celebrated example of this
CHAPTER 2. MATH FOR POETS
Figure 2.7: Recursion in Indian architecture: the eleventh century Kandāriyā Mahādeva temple [?].
16
CHAPTER 2. MATH FOR POETS
17
Fig. I4 Rajaranitemple, Bhubaneshwar,
Orissa, ca. Io030
Fig. I5 Cumulative projection:
a. Latina
b. Type I (Latina as central projection)
c. Type I as central projection, as
at Rijarani temple, Bhubaneshwar
d. Late composition with "c"as central
projection, as in Figure 4I
L.
b.
d.
Figure 2.8: Śekharī temple construction. ©2002 Adam Hardy
1
1
1
1
1
1+1
2+1
3+1
4+1
1
1
1+2
3+3
6+4
1
1+3
4+6
1
1+4
1
Figure 2.9: The meruprastāra is made of copies of itself.
style; it is composed of miniature shrines (aedicules) emanating from a central shrine. Adam
Hardy writes,
As soon as the dynamic relationships between the aedicules are considered, the
vision of a theological hierarchy can be seen as a dynamic process of
manifestation: the emerging, expanding, proliferating, fragmenting, dissolving
patterns are so closely analogous to the concept, perennial in India, of a world of
multiplicity recurrently manifesting from unity and dissolving back into unity,
that the idea can be said to be embodied in the forms [?, p. 91-2].
It is no wonder that ancient and medieval Indian mathematicians developed an outstanding
facility with recursion.
2.6
Solutions to exercises
Solution 2.1. I’ve written the list in two columns to save space.
CHAPTER 2. MATH FOR POETS
18
LLLLL
SLLLL
LSLLL
SSLLL
LLSLL
SLSLL
LSSLL
SSSLL
LLLSL
SLLSL
LSLSL
SSLSL
LLSSL
SLSSL
LSSSL
SSSSL
LLLLS
SLLLS
LSLLS
SSLLS
LLSLS
SLSLS
LSSLS
SSSLS
LLLSS
SLLSS
LSLSS
SSLSS
LLSSS
SLSSS
LSSSS
SSSSS
Solution 2.2. Since LLSLSL has seven syllables, the list has 27 = 128 meters.
Solution 2.3. Pingala’s algorithm means that each list is twice the size of the previous one (formally, if
there are k (n − 1)-syllable meters, there are (2 × k) n-syllable meters. Since there are 2 one-syllable
patterns, the numbers of meters of each length follows the pattern 2, 4, 8, 16, and so on. A formal
proof would require induction.
Solution 2.4. Since patterns of H and T are binary, there are 23 = 8 patterns, and they are HHH THH
HTH TTH HHT THT HTT TTT
Each of these patterns is equally likely, and three of them have one T and two H’s, so the likelihood of
getting one T is 3/8 or 37.5%. In general, use the results from the study of meters, substituting H for
L and T for S.
Solution 2.5. Start with 1, then use the list of three-syllable patterns, replacing 0 with L, 1 with S,
and writing the patterns backwards. There are 24 = 16 ive-digit binary numbers. You write them
by following this same procedure with the list of four-syllable meters. Proving that this answer is
correct involves understanding how place-value number systems work in bases other than 10.
Solution 2.6. If your conjecture was something like “there are 13 meters with 6 beats, and you get
any number in the sequence by adding the two previous,” you would be correct.
Solution 2.7. The procedure is to add a L to all the 4-beat patterns, then add an S to all the 5-beat
patterns, which are listed in igure 2.1. The answer is: LLL SSLL SLSL LSSL SSSSL SLLS LSLS SSSLS LLSS
LLSLS SLSSS LSSSS SSSSSS
Solution 2.8. Here are the irst ten entries of the Padovan sequence:
duration
num. patterns
1
0
2
1
3
1
4
1
5
2
6
2
7
3
8
4
9
5
10
7
If Pn is the number of n-beat patterns, then a recursive rule is Pn = Pn−2 + Pn−3 when n > 3. The
proof of this statement is similar to the proof of Theorem 2.2 (the Hemachandra numbers). In this
case, partition the patterns of duration n into patterns of duration n − 2 followed by a two-beat
note and patterns of duration n − 3 followed by a three-beat note. Incidentally, a student noticed
that Pn = Pn−1 + Pn−5 (when n > 5) and conjectured that all Padovan numbers also follow this
rule. Is this correct? Hint: use the irst rule to rewrite Pn−1 . The Padovan sequence has some
beautiful properties—for example, it is related to a spiral of equilateral triangles in the way the
Hemachandra-Fibonacci sequence is related to a spiral of squares (see below), and it is closely
connected to the Perrin sequence, another extremely cool sequence. See Ian Stewart’s “Tales of a
Neglected Number” (in Math Hysteria) for more examples.
CHAPTER 2. MATH FOR POETS
19
16
12
13
9
2
8
2 3
2 1
11
3
1 1
21
5
4
5
21
7
Solution 2.9. Suppose N Rn stands for the number of nonretrogradable rhythms of length n. Then
the irst four numbers in the sequence are 1, 2, 1, 3. For n > 4, use N Rn = N Rn−4 + N Rn−2 .
Solution 2.10. Any way of choosing objects from a collection of n different objects can be represented
by a binary pattern of length n in this way: assign numbers from 1 to n to the objects, and write I
(in) if an object is selected and O (out) if it is not. For example, suppose you choose {2, 5, 7} from
the collection {1, 2, . . . , 8}. That choice corresponds to the binary pattern OIOOIOIO. Each choice of
r objects corresponds to a pattern of n letters with r I’s. Substituting S for I and L for O produces a
n-syllable meter with r short syllables.
Solution 2.11.
6
1
5
2
4
3
3
4
2
5
1
6
The irst number in the row is 1 (this is true for every n). Obtain the other numbers in the row by
succesively multiplying and dividing by the numbers you have written:
6
5
4
3
2
1
1 = 6; 6 · 2 = 15; 15 · 3 = 20; 20 · 4 = 15; 15 · 5 = 6; 6 · 6 = 1. This tells us that the sixth row is
1 6 15 20 15 6 1.
Solution 2.12. Consider the eight three-letter words. A de Bruijn sequence can be thought of as an
ordering of these words: the irst word is the irst three letters in the sequence; the second word is
the second through fourth letters, and so on. You probably have noticed that there are some rules
about which words can follow each other. For example, SSL can be followed by either SLS or SLL.
A powerful representation called a directed graph is useful in organizing these possibilities. The
vertices of the graph represent states (in this case, three-letter words). Arrows indicate that one state
can legally follow another. The graph is shown below. Any path that visits each vertex exactly once
de ines a de Bruijn sequence.
SLL
SSL
LLL
LSL
SLS
SSS
LLS
LSS
CHAPTER 2. MATH FOR POETS
The “yamātārājabhānasalagā” sequence comes from following the path in the upper right.
20
Chapter 3
Rhythm
The beat is the foundation of music. Even before birth, we hear the regular pulse of our
mothers’ heartbeats. We experience the beat as simultaneously linear (progressing through
time) and cyclic (repeating, as on a clock). Musical events—sounds and silences—occur
within a repeating framework called a meter. A rhythm is a pattern of note onsets that is
actually present in a piece. In traditional and popular music, repeated rhythmic patterns, or
time-lines, overlay the meter.
3.1
Measuring time
Watch the video “Tāla in Carnatic classical music.” In this video, the performer’s repeated
hand gestures indicate the tāla, or repeating rhythmic organization of time in the piece.
These same gestures accompany any piece using this tāla. Compare also the hand gestures
used in Western classical conducting (Figure 3.1).
The tactus is the basic pulse of a piece of music—it’s where you naturally tap your foot. In
music with a regular rhythm, each pulse, or beat, has the same length. When tapping along
with the tactus, we normally expect some musical “event” to happen at every beat we tap.
Normally this event is something we can hear (a drum hit, or the beginning of a note) but
sometimes there is a silence—a rest. The tempo, or speed, of the tactus is measured in
beats per minute. Tempos of about 120 beats per minute are typical for pop songs, while a
comfortable walking tempo is around 100 beats per minute.
···
|
|
| |
|
|
|
|
|
|
|
|
···
Beats in the tactus are grouped into units called measures, or bars, just as seconds are
grouped into minutes. Musicians normally count beats by their position in the measure.
Here is an example of counting in four beats per measure.
···
|
1
|
2
|
3
|
4
|
1
|
2
|
3
|
4
|
1
|
2
|
3
| ···
4
Beats may also be subdivided. In Western classical and popular music, divisions into 2, 3, or
4 equal parts are common. This diagram shows a four-count measure with beats subdivided
21
CHAPTER 3. RHYTHM
22
Figure 3.1: Curwen’s conducting diagrams.
into 2 or 4 parts. The “&” should be read as “and.” Practice clapping at a constant rate on
the beats marked “|” and speak each line of the drill along with your claps. Then practice
three-count measures, using the patterns 123, 1&2&3&, and 1e&a2e&a3e&a.
|
1
1
1
e
&
&
a
|
2
2
2
e
&
&
a
|
3
3
3
e
&
&
a
|
4
4
4
e
&
&
a
Figure 3.2: Duple and quadruple subdivisions of a four-count measure.
|
1
1
e
a
|
2
2
e
a
|
3
3
e
a
|
4
4
e
a
Figure 3.3: Triple subdivisions of a four-count measure.
In some pieces, there are intermediate levels of grouping between the measure and the
tactus and, in addition, higher-level groupings of measures. In Western music notation,
the time signature indicates the internal structure of the measure. For example, the time
signature 12/8 is normally played as four counts per measure, each subdivided into three
parts. H.W. Day’s “Tree of Time” (Figure 3.4) is a fanciful representation of time signatures.
Exercise 3.1. Do this exercise with a partner. Both of the patterns “123123” (that is, two three-count
measures) and “121212” (three two-count measures) occupy six pulses in the tactus. Each person
picks one of the patterns. Start by clapping six beats, then count the tactus using the two different
patterns simultaneously while keeping your claps synchronized. Continue counting until you are
comfortable working together. Now change the volume of your counting so that the “1” is loud and
the other numbers are almost silent. The pattern of “1s” you hear is called a two-against-three
pattern. It should sound like 1.111.1.111.1.111., etc., where the dots represent quiet beats.
CHAPTER 3. RHYTHM
23
Figure 3.4: A fanciful representation of meter in Western classical music.
The pattern repeats every six beats. Try the patterns 3-against-4, 2-against-4, and 6-against-4 in the
same way. If you start saying “1” together, how many beats, in each example, are needed before you
both say “1” at the same time again?
3.2
Rhythm and groove
Groove is a dif icult term to de ine. Most musicians agree that groove is essential to most
styles of popular music. It normally refers to characteristic repeating rhythm patterns
and accents that identify different styles of dance music, such as ska and reggae. Listen to
the audio, starting around 3:00, in which James Brown, “The Godfather of Soul,” discusses
grooves in his music in this 2005 interview with Terry Gross of NPR. He compares two
versions of the “I Got You” groove and describes how his groove changed with his 1965 song
“Papa’s Got A Brand New Bag.”
The irst count in every measure is called the downbeat and the last count is called the
upbeat, because it comes right before the downbeat of the next measure. The most basic
grooves in Western music use four-count measures. Counts one and three are called the
on-beats and counts two and four are the off-beats. “Backbeat” grooves, popular in blues and
R&B, emphasize the off-beats, while downbeat-oriented grooves have a primary accent on
the downbeat and a secondary accent on count three. In the NPR interview, James Brown
said that “Papa’s Got a Brand New Bag” was his irst song to give a strong emphasis to the
downbeat.
Unlike the basic grouping of pulses in the tactus into measures, a groove can be highly
complex, with several musicians playing interlocking rhythms, each with different accent
patterns. A hip-hop beat is an example of what I’m calling a groove. We can study both the
pattern that each instrument plays and the combination of all these patterns. In addition,
musicians sometimes intentionally play a little ahead or behind the tactus—this kind of
variation is called microtiming or swing. Although microtiming is an important feature of
dance music grooves, it is dif icult to write down in music notation. Musicians normally
learn it by ear. Just to keep things simple, we won’t be studying microtiming in this class, but
CHAPTER 3. RHYTHM
24
it is an important part of groove in a lot of music.
Exercise 3.2. Practice both on-beat and off-beat clapping with different kinds of dance music in
four counts per measure. Find a piece that clearly emphasizes the on-beats and another piece that
clearly emphasizes the off-beats. Do you ind it easier to clap to “Papa’s Got a Brand New Bag” on the
on-beats, or on the off-beats?
3.3
Notation
Drum tablature, or drum tab, is a common way to notate rhythm. Sequences of drum hits can
be written with x’s, indicating hits, and .’s (periods), or rests, indicating that the drum is
not sounded. Each symbol occupies the same amount of time, a pulse in either the tactus or
some regular subdivision of the tactus. For example, the notation x..x..x. means “hit rest
rest hit rest rest hit rest.” This is a common pattern in music with four counts to a measure,
subdivided once. Practice clapping the pattern while you speak the names of the counts:
Count: 1 & 2 & 3 & 4 &
Clap: x . . x . . x .
Exercise 3.3. Write the pattern for a four-count measure, with hits (a) on the downbeat only (b) on
the upbeat only (c) on the on-beats, and (d) on the off-beats. Each x or . occupies one count. How
would your patterns change if each hit or rest takes half a count?
Exercise 3.4. How many patterns of hits and rests are possible in an eight-beat measure? Of those,
how many have three hits? (Don’t try to do this from scratch—use the results of the last chapter!)
Steve Reich’s “Clapping Music.” We can use drum tab to visualize Steve Reich’s “Clapping
Music.” Player A claps the same rhythm over and over until the piece ends.
Player A:
x x x . x x . x . x x .
Player B progresses through a number of patterns, playing each one eight times. The irst
pattern is the same as Player A’s, then the pattern is repeatedly shifted by one beat.
(1)
(2)
(3)
(4)
x
x
x
.
x
x
.
x
x
.
x
x
.
x
x
.
x
x
.
x
x
.
x
.
.
x
.
x
x
.
x
x
.
x
x
.
x
x
.
x
x
.
x
x
.
x
x
x
When two drum patterns are played simultaneously, a third rhythm is heard, called the
resultant rhythm. It is the rhythm that has rests whenever both of the original drum patterns
had a rest and drum hits otherwise.
Drum tab is convenient for inding resultant rhythms. Any beat that has an x in either A
or B (or both) has an x in the resultant rhythm. Here are patterns A and B in the previous
example, plus the resultant rhythm. The seventh beat is the only rest in the resultant rhythm.
Player A:
x x x . x x . x . x x .
Player B:
x . x x . x . x x . x x
resultant:
x x x x x x . x x x x x
CHAPTER 3. RHYTHM
1
25
2
3
4
Figure 3.5: The irst four patterns in “Clapping Music” that are clapped by player B.
Led Zeppelin’s “Kashmir” Here is a simpli ied version of the Kashmir pattern. The drums
hit on the numbered counts.
Count:
1 & 2 & 3 & 4 &|1 & 2 & 3 & 4 &|1 & 2 & 3 & 4 &|1
Guitar:
x x . x x . x x|. x x . x x . x|x . x x . x x .|x
Exercise 3.5. Explain why the sequence repeats after three measures.
Circular notation.
Because grooves repeat, it makes sense to visualize patterns on a circle, just as we visualize
hours on a clock. The downbeat goes at the top and the other beats proceed clockwise. Dots
show when the drum is hit. Figure 3.5 shows some of the patterns from “Clapping Music.”
3.4
The beat class circle and modulus
In music, time is of the essence. How should we measure it? Rather than using minutes or
seconds, we’ll use a unit of time that is relevant to the piece itself. Since the tactus is often
subdivided, using the tactus as a “ruler” to measure time would mean using a lot of fractions.
To avoid this, let’s measure time using the smallest level of subdivision, called the tatum.
(In music notation, the tatum might be a pulse of eighth or sixteenth notes.) I will call the
duration of a pulse in the tatum a beat. Each pulse in the tactus lasts a whole number of
beats.
The fact that a beat is actually a duration of time—not a speci ic point in time—means that
we ought to measure musical time starting at zero, just like a ruler starts at zero, so that
“beat 0” refers to the beginning of the irst measure and “beat a” refers to the time a beats
later.
Any pattern of hits corresponds to a pattern of time intervals (durations) that
measure the time from the beginning of one hit to the next. For example, the pattern
x..x..x.x..x..x. is described as “3+3+2+3+3+2” and shown in Figure 3.6. The hits in
the pattern fall on beats 0, 3, 6, 8, 11, and 14. The multiple levels of counting in Figure 3.6
give some idea of the complexity of musical time. The tatum is, here, twice as fast as the
tactus.
CHAPTER 3. RHYTHM
beats
count
26
0
1
2
3
4
5
6
7
8
9
10 11 12 13 14 15 16
1 & 2 & 3 & 4 & 1 & 2 & 3 & 4 &
Figure 3.6: Two repeats of the pattern x..x..x.
When we translate between the tactus and the measure, certain points in time are associated
with one another. Just as a clock represents the organization of time within a day, the beat
class circle represents beats in a musical measure. Figure 3.7 shows a repeated rhythm
pattern, called a timeline, on a beat class circle with eight beats. The blobs indicate which
time points are hits.
6
7
5
0
4
1
3
2
Figure 3.7: The timeline x..x..x. represented on the beat class circle.
Exercise 3.6. Write the rhythms in Figure 2.3 (see page 11) in drum tab and circular notation.
De inition 3.1 (Beat class). The beat class (bc) of beat a is its position in the measure,
calculated from the beginning of the measure.
Let beat a be the time a beats after the beginning of the irst measure of a piece. A measure
with n beats has n beat classes,
0, 1, 2, . . . , n − 1
Every downbeat starts at bc 0. Notice that beat class and “count” is not the same thing: the
downbeats are counted “1” but start at bc 0. The situation is similar to the relationship
between “the 1900s” and “the twentieth century.”
Suppose you start at the irst downbeat of a piece. After 54 beats, what is the beat class?
Since moving backward 12 beats has no effect on the position in the measure, we start
by subtracting as many multiples of 12 as we can, then looking at what’s left. Since
54 − 12 − 12 − 12 − 12 = 6, beat 54 is in beat class 6. It occurs 6 beats after the beginning of
a measure. Of course, we can do this more ef iciently using remainders: 6 is the remainder
when 54 is divided by 12.
In general, suppose there are n beats in a measure. Beat a falls in beat class R, where R is
the remainder when a is divided by n. Mathematicians call this remainder the modulus.
De inition 3.2 (Modulus). If a is an integer and n is a positive integer, then the modulus of a
relative to n is the remainder when a is divided by n. If R is the remainder, we write
a mod n = R
CHAPTER 3. RHYTHM
The beat class formula.
27
If there are n beats per measure, beat a falls in beat class a mod n.
Two beats have the same beat class if and only if they are in the same position in the
measure, which is the same thing as being separated by a whole number of measures.
Therefore, beat a and beat b belong to the same beat class if and only if the distance between
them, a − b, is divisible by n. (In general, an integer m is divisible by an integer n if m/n is an
integer.) Mathematicians call this relationship modular congruence:
De inition 3.3 (Modular congruence). Two numbers a and b are congruent modulo a positive
integer n if their difference, a − b, is divisible by n. We write a ≡ b (mod n).
The triple equals sign ≡ is read as “is congruent to.” Note that “their difference is
divisible by n” means that (a − b)/n is an integer. For example, 23 ≡ 51 (mod 7) because
(51 − 23)/7 = 28/7 = 4, which is an integer. It is true that a ≡ b (mod n) if and only if
a mod n = b mod n.
Theorem 3.1 (The beat class theorem). Suppose there are n beats in a measure. Beats a and
b lie in the same beat class if and only if a ≡ b (mod n).
For example, suppose there are seven beats to a measure. Then
• Beat class 0 corresponds to the downbeats, which occur on beats 0, 7, 14, 21 ….
• Beat 100 falls on beat class 2 because 100 ÷ 7 = 14r.2.
• Beat 28 falls on beat class 0 because 28 is divisible by 7 (28 ÷ 7 = 4r.0).
• Beats 100 and 849 belong to the same beat class because they are separated by
849 − 100 = 749 beats, which equals exactly 107 measures. Equivalently, use modular
congruence and the beat class theorem: 100 ≡ 849 (mod 7) because (100 − 849)/7 is
an integer. Another way to solve the problem is to use modulus. 100 is in bc 2 (see
above) and, since 849 ÷ 7 = 121r.2, 849 is also in bc 2.
Exercise 3.7. Evaluate (a) 144 mod 13 and (b) 169 mod 13.
Determine whether the equations are true:
(c) 25 ≡ 61 (mod 6)
(d) −4 ≡ 4 (mod 3).
Exercise 3.8. Suppose there are six beats to the measure. Which beats are downbeats? To which
beat class does beat 50 belong? Beat 90? Are beats 36 and 63 in the same beat class?
Exercise 3.9. Suppose there are four beats per measure and you clap on every third beat, starting
on beat 0. Find the sequence of beat classes on which your claps fall. Practice counting in four and
clapping every third beat.
Exercise 3.10. In Kashmir, there are eight beats per measure and the guitar pattern repeats xx.
every three beats. Find the beat classes of the hits in the guitar part of Kashmir.
CHAPTER 3. RHYTHM
3.5
28
Polyrhythm
I’ve presented the m-against-n rhythm as one person counting in cycles of m beats and the
other person counting in cycles of n beats. We saw that a m-against-n pattern repeats after
lcm(m, n) beats. What more normally happens is that a musician plays a repeating pattern
whose cycle length is not a divisor or multiple of the number of beats in a measure, while, at
the same time, keeping track of the underlying meter. An example is the Kashmir pattern,
which has three beats, where there are eight beats per measure. Eight repeats equal 24 beats
and ill a whole number of measures. This is expressed mathematically as (3 · 8) mod 8 = 0.
Note that 24 = lcm(3, 8). This 3-against-8 rhythm is an example of a polyrhythm.
De inition 3.4. Suppose n and m are positive integers. A m-against-n polyrhythm is either the
result of repeating a pattern of length n beats simultaneously with a pattern or measure of
length m beats.
In general, assuming that n and p are positive, repeating a pattern whose duration is p beats
r times takes rp beats. This equals a whole number of n-beat patterns or measures if
rp mod n = 0
The smallest positive value of rp that solves this equation is rp = lcm(p, n).
Example: Taylor Swift’s “22.” Taylor Swift’s 2012 hit “22,” written by Swift, Max Martin,
and Shellback, embeds a three-against-four polyrhythm in the line “We’re happy, free,
confused, and lonely at the same time.” The tatum has sixteen beats per measure, so each
count equals four beats.
& |1
&
2
&
3
&
4
& |1
&
x .|x . x x . x x . x x . x x . x x|. . x
Swift squeezes four repeats of the pattern x.x into the irst three counts of the measure,
corresponding to the lyrics “happy, free, confused, and lonely” before breaking out of the
polyrhythm at the end of the line. This phrase takes lcm(3, 4) = 12 beats, which is three
counts in the measure. At 1:29 in her video, she counts to four on her ingers, as if to
highlight the polyrhythm.
Exercise 3.11. Explain why np mod n = 0 for any positive integer values of p and n. What musical
fact does this re lect?
Exercise 3.12. Suppose you choose a value of p and clap the p-beat pattern of one clap followed by
p − 1 rests (like x.... or x..). If there are n beats per measure, the beat classes on which you
clap are 0, p mod n, 2p mod n, 3p mod n, 4p mod n, etc. Some interesting effects happen when p is
either one more or one less than n. In these cases, p and n can have no common factors (p and n are
relatively prime). If p = n + 1, then p mod n = 1, and so the sequence of beat classes on which you
clap is 0, 1, 2, 3, . . . , n − 1, 0, etc. In a four-count measure, the rhythm might look something like this
(the x’s are replaced by the count).
1 . . .|. 2 . .|. . 3 .|. . . 4|. . . .|1
What is the corresponding sequence of beat classes when p = n − 1? Illustrate your answer in a
four-count measure.
CHAPTER 3. RHYTHM
29
Exercise 3.13. George and Ira Gershwin’s 1924 song “Fascinating Rhythm” features the catchy
rhythm shown below. The song is performed in “swing eighths,” meaning that the “&’s” happen later
than the midpoint between the numbered counts.
1 & 2 & 3 & 4 &|1 & 2 & 3 & 4 &|1 & 2 & 3 & 4 &|1 &
x x x x x x . x|x x x x x . x x|x x x x . x x x|x x
fascinating rhythm, you’ve got me on the go, fascinating rhythm, I’m all a-quiver
The irst three phrases (“Fascinating rhythm, you’ve got me on the go, fascinating rhythm”) are the
beginning of a polyrhythm, though the pattern is broken in the fourth phrase. Which polyrhythm?
How long would the pattern have to continue to complete a whole number of measures?
3.6
Additional applications of the modulus
In what other musical situations can the modulus be useful?
Tihai endings. In classical Indian Carnatic music, a tihai often ends a piece or a solo. A
tihai consists of a rhythmic pattern that starts and ends on a hit and repeats three times,
with some ixed number of beats between the repeats. A tihai starts on bc 0 and ends on
bc 1, so its total duration equals some number of whole measures plus one beat. Here is an
example where there are eight beats per measure: xxxxx.xxxxx.xxxxx. In this case, the
pattern is xxxxx and the gap is one beat, so the total number of beats is seventeen—that
is, one more than a multiple of eight. Here’s a different tihai with eight beats per measure,
using the pattern xxx.x and a gap of ive beats.
Count:
1 & 2 & 3 & 4 &|1 & 2 & 3 & 4 &|1 & 2 & 3 & 4 &|1
Play:
x x x . x . . .|. . x x x . x .|. . . . x x x .|x
If n is the number of beats in a measure, p is the length of the pattern, and g is the length of
the gap, then the tihai’s length is (p + g + p + g + p) and satis ies the tihai equation
(3p + 2g) mod n = 1
The tihai equation is an example of a Diophantine equation—that is, an equation whose
solutions are required to be integers (remember that we are measuring beats at the tatum
level, so p and q can’t be fractions). Finding integer values of p and g that work for a given
n is not always straightforward. For example, if p is an even number, (3p + 2g) is even and
therefore (3p + 2g mod 8) is even, so the tihai equation cannot be solved in this case.
Exercise 3.14. Verify that, in a six-beat measure, a 3-beat pattern and a 5-beat gap works for a tihai
ending. Write a tihai of this type in drum tab. Explain why it is impossible to use a 4-beat pattern
when there are six beats per measure. Is it possible to have a 3-beat gap in a 6-beat measure?
Maximally even rhythms. Consider the Cuban tresillo rhythm x..x..x., which can also
be thought of as a 3 + 3 + 2 division. It’s also the rhythm of the bass part in Elvis Presley’s
“Hound Dog” and is common in rockabilly. Along with its cousins x..x.x.. (3 + 2 + 3) and
x.x..x.. (2 + 3 + 3), this rhythm is a an approximation of a division of eight beats into
three parts. Rhythms that approximate a division of n beats into k parts are called maximally
even.
CHAPTER 3. RHYTHM
3.7
30
Solutions to exercises
Solution 3.1. The 2-against-3 pattern repeats after 6 beats. The 3-against-4 pattern is
1..11.1.11.. (12 beats), the 2-against-4 pattern is 1.1. (4 beats), and the 6-against-4 pattern
is 1...1.1.1... (12 beats). In general, the number of beats in the pattern is the least common
multiple (lcm) of the number of beats in each count.
Solution 3.3. (a) x... (b) ...x (c) x.x. (d) .x.x If each hit or rest takes half a count, the patterns
are (a) x....... (b) ......x. (c) x...x... (d) ..x...x.
Solution 3.4. Hits and rests are binary patterns, and the number of binary patterns of length n is 2n .
Therefore, there are 28 = 256 patterns. The eighth row of Pascal’s triangle tells us that there are 56
eight-beat patterns that have three hits and ive rests.
Solution 3.5. The drum pattern repeats every measure (eight beats) and the guitar pattern repeats
every three beats. The entire pattern repeats after lcm(8, 3) = 24 beats, which equal three measures.
Solution 3.6. In general, a duration of one is written “x”; two is written “x.”, and three is
written “x..”. merengue: xxx.xxx.xxx.x.x.; cumbia: xx.xx.x.xx.x.x.x; mambo:
x.x.xxxx.xxxx.xx; bintin: x.x.xx.x.x.x; lesnoto: x..x.x.; bomba: x..x..x.; guajira:
x..x..x.x.x. (this is the rhythm of “America” in West Side Story); clave: x.x..x.x.x..
Solution 3.7. (a) 144 mod 13 = 1 because 144 ÷ 13 = 11r.1 and (b) 169 mod 13 = 0 because
169 ÷ 13 = 13 (remainder is 0). Determine whether the equations are true:
(a) 25 ≡ 61 (mod 6) because (25 − 61)/6 = 6, which is an integer.
(b) −4 ̸≡ 4 (mod 3) because (−4 − 4)/3 = −8/3 is not an integer.
Solution 3.8. The downbeats are beats 0, 6, 12, 18, etc. Beat 50 is in beat class 2 because
50 ÷ 6 = 8r.2. Beat 90 is in beat class 0 because 90 ÷ 6 = 15r.0.
Solution 3.9. You clap on beat classes
0 mod 4 = 0
3 mod 4 = 3
(3 + 3) mod 4 = 2
(2 + 3) mod 4 = 1
(1 + 3) mod 4 = 0, etc.
So you clap on beats 0, 3, 2, 1, 0, 3, 2, 1, …
Solution 3.10. The sequence of beat classes is 0, 1, 3, 4, 6, 7, 1, 2, 4, 5, 7, 0, 2, 3, 5, 6.
Solution 3.11. Since np mod n is the remainder when np is divided by n, and p is an integer,
np mod n = 0. Musically, a pattern of length p that is repeated n times, where n is the length of a
measure, results in a duration equal to a whole number of measures.
Solution 3.12. If p = n − 1, then p mod n = n − 1 and so p ≡ −1 (mod n). The sequence of
beat classes on which you clap is 0, n − 1, n − 2, . . . , 1, 0, etc. In a four-count measure, that’s
1 . . 4|. . 3 .|. 2 . .|1
Solution 3.13. The repeating pattern xxxxxx. has seven beats, while the measure has eight beats.
CHAPTER 3. RHYTHM
A 7-against-8 polyrhythm repeats after lcm(7, 8) = 56 beats, which equals seven measures.
Solution 3.14. In this case,
(3p + 2g) mod n = (3 · 3 + 2 · 5) mod 6 = 19 mod 6 = 1
The possible patterns are x.x.....x.x.....x.x and xxx.....xxx.....xxx. Since
(3 · 4 + 2g) mod 6 = (12 + 2g) mod 6 = 2g mod 6 is even, it cannot equal 1. Since
(3p + 2 · 3) mod 6 = 3p mod 6, which equals either 0 or 3, a 3-beat gap is not possible.
31
Chapter 4
Pitched sounds
Sound is vibration that travels through air or water. When we refer to “sound,” we normally
mean “audible sound”—that is, vibration that our ears can detect. Pitched sounds, or tones,
are present in most music. Loosely, a sound is pitched if you can hum along with it. The
human voice and most musical instruments, with the exception of some drums, can produce
pitched sounds. Physics and mathematics can explain not only the characteristics of pitched
sounds and how musical instruments produce them but also which combinations of tones
will seem most harmonious.
4.1
Periodicity
What is special about pitched sounds? Listen to the sound on the web site of a cellist playing
the open strings of the instrument. We can open the cello recording in Audacity and zoom in
on the waveform. What we see is a graph, with time in seconds on the horizontal axis and
displacement on the vertical axis. The computer reads the displacement measurements and
tells the speakers how to vibrate so you can hear the sound. These vibrations travel through
the air and are interpreted by our ears.
Figure 4.1: A 0.1 second portion of the waveform of a cello’s sound.
Figure 4.1 shows a 0.1 second portion of the waveform of the cello sound. This waveform
appears to repeat the pattern in Figure 4.2 over and over.
De inition 4.1. A pattern occurring in time is periodic if it repeats the same pattern at
regular time intervals. Its period is the length, in seconds per cycle, of the smallest portion of
32
CHAPTER 4. PITCHED SOUNDS
33
.
Figure 4.2: One cycle of the cello’s waveform.
the overall pattern that can be repeated to form the overall pattern. A cycle is a portion of the
overall pattern whose length equals one period.
Example: a sawtooth wave. Figure 4.3 depicts a sawtooth wave, which you can hear on
the web site. It sounds arti icial but is still is a pitched sound. This computer-created sound
is perfectly periodic, while the cello’s waveform is only approximately periodic.
Figure 4.3: A 0.1 second portion of a sawtooth wave.
The length of the clip shown is exactly 0.1 second. However, a shorter pattern—a cycle—can
be repeated to make the entire clip. There is no unique cycle, as any portion of the waveform
that lasts one period is a cycle. Three examples of cycles are shown in Figure 4.4. Each cycle
lasts 0.01 second, so the period of the vibration is 0.01 seconds per cycle.
this
or this
or this
Figure 4.4: Three cycles of the sawtooth wave (Figure 4.3).
Example: white noise (not periodic). Of course, not every waveform is periodic. A sound
produced by random vibrations is called white noise, as shown in Figure 4.5. It sounds like
hissing or static and is not pitched. You can listen to this clip on the web site.
Exercise 4.1. Estimate the period of the vibrations of the cello sound. Use correct units in your
answer.
4.2
Frequency
Pitched sound is sound produced by rapid, regular vibration. Frequency measures the speed
of vibration.
CHAPTER 4. PITCHED SOUNDS
34
Figure 4.5: The waveform of a white noise.
De inition 4.2. The frequency of a periodic vibration is its number of cycles per second. The
units of frequency are Hz, or Hertz.
Humans can hear pitched sounds that range from about 20 Hz to 20,000 Hz. Other animals
can hear sounds with lower or higher frequencies.
Period-frequency conversion. The frequency and period of any periodic waveform are
reciprocals.¹ That is,
frequency =
1
period
and period =
1
.
frequency
where the frequency is measured in Hertz (cycles per second) and the period is measured in
seconds per cycle. Note that the units of measurement are also reciprocals.
Example: using frequency to generate a tone. Let’s make a tone like the sawtooth wave
that we heard. In order to use Audacity’s sound generator, I need to know the frequency of
the vibration. I calculated that the period of the sawtooth wave is 0.01 second. That is, the
vibration is
0.01 seconds per cycle,
or
0.01 seconds
.
cycle
However, frequency is measured in cycles per second, or
the frequency.
0.01 seconds
corresponds to
cycle
cycles
.
second
1 cycles
0.01 second
Take the reciprocal to ind
= 100 Hz.
Now go to Generate → Tone in Audacity and make a 100 Hz sawtooth wave.
Pitched sound. I informally described a “musical tone” as “a sound you can hum along
with.” Instruments such as a violin, piano, guitar, or trumpet produce such sounds, as
does a person singing or whistling. We looked at the waveforms of recordings of musical
instruments and saw that they were created by periodic, or repeating, vibrations. You might
¹If n ̸= 0, the numbers n and 1/n are reciprocals. For example, 1/3 and 3 are reciprocals.
CHAPTER 4. PITCHED SOUNDS
35
wonder if any periodic vibration can be described as “musical.” The answer is no, because
vibrations that are either very slow or very fast cannot be heard.
De inition 4.3. A pitched sound is an audible sound produced by a periodic vibration.
A few observations…
• If two pitched sounds have the same frequency, they are said to have the same pitch.
• If one sound has a higher frequency than another, it is said to have higher pitch, or to
be a higher sound.
• If one sound has a lower frequency than another, it is said to have lower pitch, or to be
a lower sound.
• Because frequency and period are reciprocals, high pitches correspond to short
periods, while low pitches correspond to long periods.
Exercise 4.2. Use your estimate of the cello wave’s period to estimate the frequency of the cello
sound. Make a sound with that frequency in Audacity. Which of the four cello pitches is it most like?
Exercise 4.3. What is the period of the lowest audible sound? Of the highest audible sound?
Exercise 4.4. Suppose a violin plays a sound whose period is 0.002 seconds and a clarinet plays a
sound whose period is 0.0025 seconds. Compute the frequency of each sound. Which sound is lower?
Exercise 4.5. Generate or record a few seconds of pitched sound. In Audacity, select half of the
waveform and use the Effects → Change Pitch option to double its frequency. Listen to the entire clip.
Now, generate a white noise and do the same thing. How does white noise behave differently from
pitched sound?
Answers to Exercises.
Solution 4.1. The period of the cello sound is obtained by zooming in on the waveform and
locating a cycle. The length of the cycle is the difference between its starting and ending times. For
example, the cycle that starts at 3.570 seconds ends at 3.580 seconds. The length of the cycle is
3.580 − 3.570 = 0.01 seconds, which is the period. (The correct units for this measurement are
seconds.)
Solution 4.2. In a previous exercise, we calculated that the period of the cello sound is approximately
0.0102 seconds. The frequency is approximately 1/0.0102 cycles per second ≈ 98.04 Hertz. It is
most like the second tone.
Solution 4.3. The period of the lowest audible sound is 1/20 = 0.05 seconds. The period of the
highest audible sound is 1/20000 = 0.00005 seconds.
Solution 4.4. The frequency of the violin’s sound is 1/0.002 = 500 Hz. The frequency of the clarinet’s
sound is 1/0.0025 = 400 Hz. The clarinet’s sound is lower because its frequency is smaller.
Solution 4.5. Using Change Pitch to double the frequency of the pitched sound results in a higher
sound. It doesn’t seem to have an effect on the white noise.
CHAPTER 4. PITCHED SOUNDS
4.3
36
Amplitude and envelopes
So far, we have seen that pitched sounds are periodic, at least over small intervals of time.
Another quantity that affects what you hear is volume. Louder sounds generally result from
more powerful vibrations, and therefore the waveform is wider. A waveform’s amplitude is
half its width (that is, half the difference between its maximum and minimum points), which
normally is equal to its height above the horizontal axis. Increasing a waveform’s amplitude
increases how loud it sounds. Here is a ive-second clip from ZZ Top’s song “Sharp-Dressed
Man.” The wave gets wider on the loud drum beats, which occur about twice a second.
Figure 4.6: Five seconds of ZZ Top’s “Sharp-Dressed Man.”
An envelope is a representation of how a sound’s amplitude changes over time. Its graph
has roughly horizontal symmetry, with the top curve tracing the maximum points in the
waveform and the bottom curve tracing the minimum points. If a sound fades out, its
envelope will pinch together like a “>” symbol. The envelope often gives information about
the rhythm, since hitting a drum produces a sudden change in volume. The outer curves in
Figure 4.7 trace the envelope of the ZZ Top song “Sharp-Dressed Man.”
Figure 4.7: Envelope of the beginning of “Sharp-Dressed Man.”
Is amplitude the same thing as loudness? The short answer is no. Volume, or perceived
loudness, is a complex psychological phenomenon determined by the frequency and quality
of a sound, as well as its amplitude. There is no widely-used method to measure volume.
However, if two sounds have the same frequency and are played on the same instrument, the
sound with a greater amplitude will be perceived as louder.
Exercise 4.6. The Envelope Tool on Audacity lets you modulate the volume of a sound wave (look
for the button on the top of the window). Clicking and dragging shapes the envelope of the sound
wave. I have produced a waveform with Audacity by generating ive seconds of a 440 Hz tone and
using the Envelope tool to modify it (Figure 4.8). Sketch the envelope of this waveform. Describe
what you would hear when the sound is played.
CHAPTER 4. PITCHED SOUNDS
37
Figure 4.8: Waveform produced using the Audacity Envelope tool.
Solution 4.6. Here is the envelope:
You would hear a 440 Hz tone that fades in for the irst second, fades out rapidly after second 2, is
silent from about second 2.2 to second 3.6, has a short burst of volume at second 4, then fades in after
second 5.
4.4
Sinusoids
When you strike a tuning fork, its tines vibrate at 440 Hz. If you attach a ine pen to the fork
and move it while it is vibrating, you can sketch a smooth wave, as in Figure 4.9. In class,
we used Audacity to record the sound of a tuning fork. Its waveform, shown in Figure 4.10,
is a sinusoid, which is a curve that has the same basic shape as the graph of y = sin(t)
(Figure 4.11). Sounds with sinusoidal waveforms are often described as “pure” tones,
like the sound of a tuning fork. Sinusoids also represent the oscillation (back and forth
movement) of a spring or pendulum.
Figure 4.9: A tuning fork’s vibration produces a sinusoidal pattern.
Not all sinusoids have the same frequency or amplitude. A sine wave with amplitude equal
to A and frequency equal to f Hz has the equation
y = A sin(2πf t)
where the sine is calculated in radians and t is time in seconds. The general formula for a
sinusoid is y = A sin(2πf t + φ), where φ is the wave’s phase, which is a number between 0
CHAPTER 4. PITCHED SOUNDS
38
hing Calculator
3/8/17, 9:37 PM
Figure 4.10: Waveform recorded from a tuning fork.
Figure 4.11: Graph of y = sin(t). It crosses the horizontal axis at t = 0, π, 2π, 3π, etc.
and 2π that indicates where in the wave’s cycle the time t = 0 falls. Changing the phase shifts
the wave horizontally (back and forth in time).
Example. Figure 4.12 shows the graph of a sinusoid with an amplitude of 0.8 and a period
of 0.01 seconds. Suppose I want to know its formula, in the form y = A sin(2πf t). The
value of A = 0.8 is given. I then calculate the frequency f = 1/P . Since P = 0.01 seconds,
f = 1/0.01 = 100 Hz. This wave is the graph of y = 0.8 sin(2π · 100 · t) = 0.8 sin(200πt).
Although it is true that π can be approximated numerically, it is preferred to write 200π
rather than something like 628.3185, which approximates 200π to four decimal places, while
200π is an exact answer.
Desmos Graphing Calculator
3/8/17, 10:58 PM
desmos.com/calculator
powered by
by
powered
Page 1 of 1
Figure 4.12: A sinusoid with amplitude = 0.8 and period = 0.01 seconds.
powered by
Exercise 4.7. Find the amplitude, frequency, and period of the waveform represented by
0.4 sin(100πt). Use correct units in your answer.
Exercise 4.8. Find the formula for this wave. The numbers on the horizontal axis are seconds.
by
powered
powered by
https://www.desmos.com/calculator
Page 1 of 1
CHAPTER 4. PITCHED SOUNDS
39
Exercise 4.9. Sketch a sine wave with amplitude = 3 and period = 4 seconds. Be sure to label both
the horizontal and vertical axes of your graph. Find the formula for this wave and use a graphing
program to check your answer.
Solution 4.7. In this formula, A = 0.4 is the amplitude, the frequency is 100π/(2π) = 50 Hz, and the
period is the reciprocal of the frequency, 1/50 = 0.02 seconds.
Solution 4.8. Amplitude = 1 = A and period = 0.5 seconds, so F = 2 and the formula is
y = sin(4πt).
Solution 4.9. The formula is y = 3 sin(2πt/4) = 3 sin(πt/2).
Check out Worksheet A for more practice problems.
4.5
The Superposition Principle
The Superposition Principle explains how multiple sounds combine. It is the result of a basic
principle of physics stating that the impact of two forces at a point is the sum of those forces.
De inition 4.4 (The Superposition Principle). The waveform of two sounds that reach the
same place at the same time is the sum of the waveforms of each sound on its own.
The Superposition Principle has many implications for acoustics. Its most basic result is that
if two sounds with the same period, P , combine, the result is a sound that repeats every P
seconds. Normally this means that the sum of sounds with the same frequency, f , is a sound
with frequency f . (It is possible to cook up examples where the sum’s frequency is some
multiple of f , or where the sounds combine to produce silence.)
Constructive interference. Suppose we start with two sinusoids with the same frequency
and phase. In this situation, the amplitude of the sum of the waves equals the sum of the
amplitudes of the waves. This phenomenon is called constructive interference. For example,
if the 50 Hz waveforms 2 sin(100πt) and 3 sin(100πt) combine, the result is 5 sin(100πt),
which has amplitude 5 and frequency 50 Hz. Notice that frequencies do not add—the
frequency of the combination tone is the same as the original tones’ frequency.
Destructive interference. Can two sounds cancel each other out? Yes! Any vibration can
be cancelled by an equal and opposite vibration. This is called destructive interference.
Noise-cancelling headphones use destructive interference. They don’t muf le sound—
instead, they produce sound with an equal but opposite waveform to the surrounding sound.
Adding the two sounds together due to superposition results in sound cancellation.
Mathematically, if a sound is represented by a function f (t), then adding the function −f (t)
will cancel the original function (that is, f (t) + (−f (t)) = 0). Graphically, the relationship
CHAPTER 4. PITCHED SOUNDS
40
1.0
0.5
0
0.5
1.0
1.5
2.0
t
K
0.5
K
1.0
Figure 4.13: Two functions that cancel each other out. Their sum is zero.
between f (t) and −f (t) is a “ lip” around the horizontal axis. In Figure 4.13, f (t) is
represented by the solid line and −f (t) is the dotted line.
“Mixing” and addition of waveforms. When adding waveforms, you add their
y-coordinates. Recording engineers call this process mixing. In Audacity, if you select two
waveforms, then choose Tracks → Mix and Render To New Track, the sum of the waveforms
will appear in a new track. Figure 4.14 shows the result of mixing a 440 Hz sine wave with
amplitude 0.4 (top) and a 330 Hz square wave with amplitude 0.3 (middle) to produce a new
wave (bottom), which is their sum.
Figure 4.14: Demonstration of mixing using Audacity. The two top waves are “mixed” (added) to produce the
bottom one.
Demonstration: The Disappearing Tone. Let’s use Audacity to demonstrate destructive
interference. In Audacity, you can multiply a sound wave by −1 using Effect → Invert.
Produce a tone, then use Tracks → Add New → Audio Track to add a new track. Copy and
paste your original tone into the new track so the two tracks are identical. Select a portion of
the new track and invert it, as shown in Figure 4.15. If you solo the second track, the inverted
portion will sound the same (except you will hear a “pop” where the inversion happens). You
will hear pops where the inversion starts and ends.
CHAPTER 4. PITCHED SOUNDS
41
Figure 4.15: Two sinusoidal waves, with the middle portion of the bottom waveform inverted.
However, when you play the two tracks together, the sound will disappear at the inversion!
Select-All and Tracks → Mix and Render produces the waveform in Figure 4.16. Before the
inversion happens, the combined tone is twice the amplitude of the original; during the
inversion, the amplitude is zero (silence).
Figure 4.16: The sum of the waveforms in Figure 4.15
Demonstration: Phase shift and the secret message. If your original sound is a sinusoid,
shifting the wave by half a cycle in the horizontal direction has the same effect as inversion.
This corresponds to setting its phase equal to π. Create a stereo track and generate two
identical sine waves. Split the track. Zoom in so you can see the waveform clearly and
experiment with shifting one tone using the Move button (↔) so it cancels the other one out.
Now record a “secret message” and a sine tone in stereo and mix them together. You should
be able to split the stereo track and move one track slightly so that the tone disappears and
you can hear the secret message.
Exercise 4.10. Generate a 100 Hz sawtooth wave with amplitude 0.5 on Audacity, then choose Tracks
→ Add New → Audio Track. Select the new track and generate a 1000 Hz sine wave with amplitude
0.5 in that track. You can experiment with the Solo button to hear the tracks separately and together
and zoom in to see the waveforms. Then Select All and choose Tracks → Mix and Render. The new
sound created is the sum of the original waveforms. What does it sound like? What does it look like?
Exercise 4.11. Choose the correct answer and explain why the other answers are incorrect. The
Superposition Principle says that if two sound waves reach the same point at the same time, the
resulting waveform has (a) a frequency equal to the sum of their frequencies, (b) an amplitude equal
to the sum of their amplitudes, (c) a waveform equal to the sum of their waveforms, or (d) all of the
above.
CHAPTER 4. PITCHED SOUNDS
42
Exercise 4.12. In pop music with a single vocalist, the singer’s voice is usually recorded equally
in both stereo tracks, while the instruments are “panned” to the left or right, meaning that each
individual instrument is louder in one track that the other. Explain how the two stereo tracks can be
combined to cancel out the vocal without cancelling the instrumental sounds.
Solution 4.10. The original tracks and the mix track (the sum) are shown. The 100 Hz sawtooth
wave is a low buzzing sound. The 1000 Hz sine wave is a high, pure tone. The combination of the two
produces a buzzing sound that mostly sounds like the 100 Hz tone. It’s hard to hear the high tone in
the mix.
Solution 4.11. It follows from the de inition of the Superposition Principle that (c) is true. An
example that shows that (a) is false is constructive interference: when two sinusoids with the same
phase and frequency combine, the resulting wave has a greater amplitude but the same frequency.
The phenomenon of destructive interference shows that (b) is false. In that case, two waves can
combine to produce silence (amplitude = 0). Therefore, the correct answer is (c).
Solution 4.12. The trick is to separate the stereo tracks, invert one track, then mix the two tracks
together. Any waveform that was equally present in both left and right tracks will be cancelled out.
4.6
Beats
The phenomenon of “beats” is a direct result of the Superposition Principle. Historically,
beats were used to tune keyboard instruments before the advent of electronic tuners. Beats
are also important in the theory of dissonance and explain the timbre of some instruments.
CHAPTER 4. PITCHED SOUNDS
43
Let’s perform the following experiment with Audacity:
1. Generate a 440 Hz tone with amplitude 0.4 that lasts for ten seconds. Open a new
track and generate a 10 second, 442 Hz tone with amplitude 0.4 that starts about ive
seconds after the irst one.
2. Play each track individually, using the Solo button. Most people—even trained
musicians—can’t hear a difference between 440 Hz and 442 Hz tones.
3. Listen to both tracks together. You should hear a tone for about ive seconds, then a
pulsing effect as the two tracks sound simultaneously, and inally a tone that sounds
like the irst one. The frequency stays (approximately) the same.
4. Select-All and Tracks → Mix and Render to mix the two tracks into one. Figure 4.17
shows the combination track. There appear to be ten pulses, or “beats,” where the two
tones overlap. Therefore, the frequency of the beats is 2 Hz (that is, 10 cycles per
5 seconds). The maximum amplitude of the envelope that produces the beats is double
that of the individual tones.
Figure 4.17: The result of mixing 440 Hz and 442 Hz sinusoids that overlap from the 5 to the 10 second mark.
This pulsation is an example of beats and their frequency is predicted by the Beats Formula.
A mathematical derivation of the formula is on page 44.
De inition 4.5. Beats are pulses that are heard when two pitched sounds with close but not
equal frequency combine due to superposition.
De inition 4.6 (The Beats Formula). Suppose two pitched sounds with close but not equal
frequencies, f1 and f2 , combine and produce beats. Then
• The combination sound is pitched and its frequency is the average of the two individual
frequencies, (f1 + f2 )/2.
• The frequency of the beats is the difference of the two individual frequencies, |f1 − f2 |.
You might be wondering what “close” means in this context. You are welcome to experiment
with Audacity and try different combinations of frequencies. When the tones are less than
20 Hz apart, I hear the combination tone as beats, while I hear two different frequencies
when the tones are more than 40 Hz apart. There’s a grey area where the combination
tone sounds “gritty” without distinctly hearing beats. This is related to the phenomenon of
dissonance (literally, bad sound), which we will discuss later in the course. I’ll use “less than
20 Hz apart” as the threshold for “close” in the de inition.
CHAPTER 4. PITCHED SOUNDS
44
Figure 4.18: The waveforms of two approximately 440 Hz accordion reeds are shown (top and middle).
Beats of about 12 Hz are visible in the mixed waveform (bottom).
Example. When pitched sounds of frequency 200 Hz and 204 Hz combine, the result is a
202 Hz tone that pulsates at a rate of 4 Hz. The computations are (200 + 204)/2 = 202 (their
average) and |204 − 200| = 4 (their difference).
Example. Suppose two pitched sounds combine to produce a 1000 Hz tone that pulses
at a rate of 7 Hz. What are the frequencies of each sound individually? The answer is
1000 + (7/2) = 1003.5 Hz and 1000 − (7/2) = 996.5 Hz.
Application: Why does an accordion sound “bad”? The accordion is especially designed
to produce a strident tone that some people don’t like (other people, like me, love it!). In the
era before ampli ication, this was a way to make the sound of the instrument cut through
other noises and be heard. Each note on the instrument is actually produced by several
reeds, each of which sounds at a different frequency. To produce an A note on my accordion,
two reeds sound approximately 440 Hz, one sounds 220 Hz, and the fourth sounds 880 Hz.
The frequencies 220 Hz, 440 Hz, and 880 Hz form octaves, which sound so similar that a
high-voiced and a low-voiced person singing the same melody will naturally sing in octaves.
However, the two reeds that are close to 440 Hz are purposely tuned to different frequencies,
producing beats. Recordings of my instrument are shown in Figure 4.18. This warbling
sound is a characteristic of the accordion—love it or hate it!
Justi ication for the Beats Formula. The mathematical explanation for beats lies in a
formula you might have seen in a high school trigonometry class:
(
)
(
)
a+b
a−b
sin a + sin b = 2 sin
cos
.
2
2
CHAPTER 4. PITCHED SOUNDS
45
The sum of the sines of two angles equals twice the sine of the average angle times the cosine
of half the difference in the angles.
Now let’s suppose f1 and f2 are frequencies. For simplicity, let’s suppose they both have the
same amplitude, A, and zero phase, although a similar formula applies if they don’t.
( (
) )
( (
) )
f1 − f2
f1 + f2
t cos 2π
t .
A sin(2πf1 t) + A sin(2πf2 t) = 2A sin 2π
2
2
The combination of sinusoids with frequencies f1 and f2 and equal amplitudes is a sinusoid
whose frequency is their average frequency times a sinusoid whose frequency is half the
difference in their frequencies. The maximum amplitude of the resulting wave is twice
that of the individual sounds. Because its envelope is re lected in the horizontal axis and
cos x is symmetrical around the vertical axis, the frequency of the pulses is two times
|f1 − f2 |/2—that is, the absolute difference in frequencies, |f1 − f2 |.
Exercise 4.13. Describe what happens when a 780 Hz tone and a 774 Hz tone combine due to
superposition.
Exercise 4.14. Suppose you hear a 300 Hz tone that pulses at a rate of 4 Hz. Describe how this sound
could be produced using beats.
Exercise 4.15. An untuned piano often has a different timbre from a tuned piano, even when you
play only one note and the correct frequency is heard. What is one reason this might happen? How
might a piano tuner correct it? Hint: Look at the inside layout of a piano’s strings.
Exercise 4.16. Predict what happens when a 780 Hz tone and a 280 Hz tone combine.
Exercise 4.17. Describe what happens, according to the Superposition Principle, when you are
wearing headphones and listen to a 300 Hz tone in your right ear and a 308 Hz tone in your left ear.
Use Audacity to produce 300 Hz and 308 Hz tones and adjust the pan (the balance between left and
right ears or speakers) so each ear hears a different tone. Listen with headphones. What do you hear?
Research binaural beats and their relationship to this experiment.
Solution 4.13. The result is a 777 Hz tone that pulses at a rate of 6 Hz (equivalently, you hear 6 beats
per second). The maximum amplitude of the beats is 0.6.
Solution 4.14. This sound is produced when a 298 Hz tone and a 302 Hz tone combine due to
superposition.
Solution 4.15. Most notes on the piano are made when two or three strings of identical frequencies
are hit. When these strings are slightly out of tune with each other, beats create a pulsing sound
which is particularly unpleasant when the beats are fast. For example, suppose you strike the A
below middle C on a badly tuned piano and hear a 220 Hz tone that pulses at a rate of 10 Hz. It is
produced by two strings whose average frequency is 220 Hz. The difference in their frequencies is
10 Hz. Therefore, their frequencies are 220 + (10/2) = 225 Hz and 220 − (10/2) = 215 Hz. The piano
tuner must tune both strings to 220 Hz and the beats will disappear.
Solution 4.16. You would not hear beats or pulses because the frequencies are too far apart. Instead,
you perceive two different frequencies. As a rule of thumb, only tones less than 20 Hz apart produce
CHAPTER 4. PITCHED SOUNDS
46
beats.
Solution 4.17. You would not hear beats because superposition does not occur. You would hear
two separate tones. They are so close in frequency that you would probably perceive them as the
same pitch. The “theory” of binaural beats states that your brain is able to combine the sounds and
mentally perceive beats if you try hard enough—or at least, the beats have some effect that may or
may not get you high, cure cancer, etc. There is no scienti ic evidence for binaural beats.
4.7
The wave equation
How do musical instruments produce pitched sounds? Although the mechanism of each
type of instrument is different, there is one particular physical equation, the one-dimensional
wave equation, that applies to stringed and wind instruments. It relates the physical
properties of a string or a wind instrument’s air column to its patterns of vibration and
predicts which frequencies will be sounded.²
Vibration of a string
Let’s start with a vibrating string. Plucking a string sets it in a back-and-forth motion that
causes vibration both in the air and in the body of the instrument, which ampli ies its sound.
The wave equation says that a vibrating string produces a periodic waveform that is a sum
of sinusoids. Moreover, the frequencies of these sinusoids depend not only on the length,
tension, and heaviness of the string but also on the shape (called the “harmonic”) that the
string makes when it vibrates. Here is a summary of the predictions of the wave equation.
Frequency formula for a vibrating string. A vibrating string produces a periodic
waveform that is a sum of sinusoids. The frequency of each sinusoid is
f=
an
2L
where
• n is a positive integer that indicates the harmonic of the string
• L is the length of the string
• a is a number that depends on the tension and material of the string.
There are three ways to alter the frequency of a string: change n, change L, or change a by
changing its tension. An additional way to change a is to use a string with a heavier or lighter
material. Each of these is explained in more detail below.
²The wave equation is an example of a partial differential equation, meaning that it’s a relationship between
partial derivatives, which you might have studied in a Calculus class. A mathematical explanation of the wave
equation is found in Section ??.
CHAPTER 4. PITCHED SOUNDS
47
Harmonics of a string (n). The effect produced by changing n in the wave equation
can be observed using slow-motion photography. As it vibrates at certain frequencies, the
string forms sinusoidal shapes, called standing waves, which are the its modes of vibration.
Figure 4.19 shows the irst seven modes of vibration. The string moves back and forth
between the two sinusoids that intersect to make each picture. The points at which the
string appears stationary are called nodes. The two ends of the string are also nodes.
Between the nodes are half waves—individual “humps” of the sinusoid. There must be a
whole number of half-waves spanning the entire length of the string, and this number equals
n in the frequency formula.
You can force a guitar string to produce a particular standing wave by lightly touching your
inger to the string at a point that is a node of that wave, but not a node of any longer wave,
and then plucking the string. For example, the third mode of vibration results when a node
is created either 1/3 or 2/3 of the way along the string. Guitarists call this technique “playing
harmonics.”
Figure 4.19: Standing waves for the irst seven modes of vibration of a string.
What is the relationship between standing waves and frequency? Suppose the irst mode
of vibration has frequency f . Since the irst mode has one half-wave, n = 1, and therefore
f = a/(2L). For the second mode of vibration, n = 2 and a and L are the same. So the
frequency of the second mode is 2a/(2L), which equals 2f . In general, if the frequency of the
irst mode of vibration is f , then the second mode has frequency 2f , the third mode has
frequency 3f , and so on.
De inition 4.7 (Harmonics). The harmonics of a string are the frequencies f, 2f, 3f, 4f, . . .
which correspond to its modes of vibration. The frequency of the irst harmonic, f , is called the
fundamental frequency.
Harmonics are also called overtones, but the numbering is different: the second harmonic is
the irst overtone, and so on.
When we don’t force it to vibrate in a particular mode, the wave equation predicts that
a string actually vibrates in a sum of the waveforms of its harmonics. However, the
CHAPTER 4. PITCHED SOUNDS
48
fundamental frequency is normally the strongest—the component of the sound that has
the largest amplitude—while the higher harmonics are quieter. For this reason, we tend to
perceive the higher harmonics as part of the quality of the sound rather than as individual
tones. However, overtone singers are able to amplify certain harmonics to the extent that we
do hear them as separate tones.
Example. Suppose a string vibrates with a fundamental frequency of 300 Hz. The
frequencies of its irst four harmonics are 300 Hz, 600 Hz, 900 Hz, and 1200 Hz.
Length of a string (L). Harmonics only partly explain the frequencies that a stringed
instrument can produce. The most common way that a player makes a different frequency
is to press the string down with a inger of their left hand so it can’t vibrate at that point. A
guitar has frets—little metal bars along the neck of the instrument—to make this easier.
When the string is pressed onto a fret, the fret acts like a new end of the string. The effect
would be the same if the string were cut shorter and ixed at both ends with the same
tension.
Example. The fundamental frequency of a violin’s A string is 440 Hz and the length of
the part of the string that vibrates is 327 mm. Then 440 = a/(2 · 327), so a = 440 · 2 · 327.
Suppose a player then presses the string down near its end, so that only 291 mm of the
string can vibrate. The fundamental frequency produced is
f=
440 · 2 · 327
an
=
≈ 494 Hz.
2L
2 · 291
Note that the fundamental frequency of the shortened string is found by multiplying 440 Hz
≈ 440
.
by the reciprocal of the ratio of lengths, 291/327. Therefore, 291
327
494
In fact, since the value of a cancels out, this relationship is always true, and gives us a handy
formula:
Length-frequency ratio formula. If the length of a string is changed without changing its
tension, the ratio of lengths equals the reciprocal of the ratio of fundamental frequencies. That
is,
L1
f2
= ,
L2
f1
where the original string has length L1 and fundamental frequency f1 and the altered string
has length L2 and fundamental frequency f2 .
Example. The fundamental frequency of a cello’s A string is 220 Hz. Where would the
player need to press down to make the string vibrate at 330 Hz? In the formula, use f1 = 220
and f2 = 330. Since the ratio of frequencies is 330/220 = 3/2, the ratio of lengths is 2/3. The
player presses down at the point two-thirds of the distance from the top end of the string.
(Note that we don’t need to know the string’s length to make this computation.)
CHAPTER 4. PITCHED SOUNDS
49
Tension and mass of a string (a). In addition to altering a string’s length or mode of
vibration, you can change its tension or choose a string made of heavier or lighter material.
These last two choices affect the value of a, which represents the speed at which a wave
travels along the stretched
string. If T is the tension of the string and m is its mass per unit
√
length, then a = T /m. Therefore, assuming that L and n don’t change, increasing the
tension results in higher frequency, while increasing the mass of the string results in a lower
frequency.
Stringed instruments have tuning pegs that the player turns to change the string’s tension.
However, there is only so much that tuning pegs can do—strings can break if they’re
stretched too tight, while they lop if they’re not tight enough. To avoid this, a lighter string
will be used for a higher sound, so that you change the mass per unit length, m, in the
formula for a. This is why instrument strings are not interchangeable: strings intended for
lower sounds are weightier than those intended for higher sounds.
Wind instruments
In a wind or brass instrument, standing waves are bands of low and high pressure; the
pressure at an open end of the instrument is held constant, creating a node, just like the ends
of the string are ixed by the bridges. Though the physics of wind instruments is different
from those of stringed instruments, it turns out that the one-dimensional wave equation still
holds, and (most) wind instruments have standing waves that divide the instrument into a
whole number of half-waves. The frequencies produced are still an/(2L), but a depends on
the shape of the instrument and air pressure.
There is some more complicated behavior that explains the clarinet and a few other
instruments. In these instruments, there is a node at one end of the air column and an
antinode—the midpoint between two nodes—at the other. The frequencies produced are
an/(4L), where n is a positive odd integer. If you play both the clarinet and saxophone, for
example, you might have noticed that opening the register key takes you up an octave for the
saxophone and an octave plus a ifth for the clarinet. This is because the clarinet’s second
mode of vibration has a frequency 3 times the fundamental frequency.
The two- and three-dimensional wave equations
Modes of vibration and harmonics are the same thing only if an instrument obeys the
one-dimensional wave equation. Some instruments have non-harmonic modes of vibration
that re lect their two- or three-dimensional geometry, as with timpani, bells, and singing
bowls, which are members of the bell family. The wave equation that we used for
stringed and wind instruments is called the one-dimensional wave equation because those
instruments are essentially “one-dimensional”—meaning that we only have one position
variable, x, in the equation. However, in a drum, you have two position variables (x and y)
because the drum head is two-dimensional, while in a bell you need three position variables
(x, y, and z). The two- and three-dimensional wave equations are needed for drums and bells.
The solutions to these equations do not result in harmonics.
One way to understand this difference visually is to think of the standing waves that
are possible in a string. We know these result in frequencies that are multiples of the
CHAPTER 4. PITCHED SOUNDS
50
Figure 4.20: Chladni plate experiment from the UCLA Physics Lab demonstrations web site and Chladni
patterns published by John Tyndall in 1869.
fundamental. However, a drum vibrates in a different way, and its modes of vibration are
much more complex. There are several excellent demonstrations on YouTube. Try searching
for “drum modes vibration” and “Chladni plates.” Chladni plates are thin metal plates that
can be vibrated, either by bowing or electronically (see Figure 4.20). Different frequencies
and shapes produce different standing waves whose nodes are shown as white curves
and lines in Figure 4.20 (right). The Wikipedia page “Vibrations of a circular drum” has
computer-generated animations.
Exercise 4.18. Suppose a string vibrates with a fundamental frequency of 200 Hz. Find the
frequencies of the irst four harmonics, and sketch the standing waves corresponding to those
harmonics.
Exercise 4.19. Explain the difference in frequency between pressing the string down 1/3 of the way
along its length and lightly touching it at the same point.
Exercise 4.20. Suppose the fundamental frequency of a string is 50 Hz. List a few of its harmonics.
What is the minimum frequency difference between harmonics? Can harmonics combine with each
other due to superposition and produce beats? Is it possible for beats to be produced by harmonics,
in general?
Solution 4.18. The irst four harmonics are 200 Hz, 400 Hz, 600 Hz, and 800 Hz. Note that the irst
harmonic is always the same as the fundamental frequency. The standing waves look like the irst
four waves in the picture on page 47.
Solution 4.19. If you press the string down, you are creating a string that is effectively shorter. The
fundamental frequency of the longer 2/3 part of the string equals 3/2 times the original fundamental
frequency. If you lightly touch the string, you are creating a node and forcing the string to sound its
third harmonic, which has a frequency 3 times the original fundamental frequency and an octave
higher than the frequency made by pressing down.
Solution 4.20. The harmonics are 50 Hz, 100 Hz, 150 Hz, etc. Since the smallest distance between
harmonics is 50 Hz, beats cannot be produced. In general, the distance between successive harmonics
equals the fundamental frequency. Since the fundamental frequency of an audible sound is 20 Hz at a
minimum, the harmonics can never be close enough in frequency to produce beats.
CHAPTER 4. PITCHED SOUNDS
51
Figure 4.21: Waveforms of a violin (top) and lute (bottom).
4.8
Timbre and spectrum
Timbre (pronounced TAM-ber) is the quality of a musical sound that allows us to distinguish
between equal-frequency and equal-amplitude sounds played by different musical
instruments or sung by the voices of different people. This difference relates to the fact
that instruments actually vibrate in all their modes of vibration simultaneously and
the waveforms combine due to superposition. Two sounds with the same fundamental
frequency can have different higher frequency components to their sound.
Figure 4.21 shows two sound waves: the top represents a violin’s sound and the bottom is a
lute. Both instruments are playing a tone whose fundamental frequency is 440 Hz. Although
the shape of the wave is quite different, the two sound waves have the same period. The
shapes of the waveforms, and not their periods, re lect the instruments’ timbre.
A representation of frequencies present in a sound and their corresponding strengths is
called a spectrum (plural: spectra). The most common way to display a spectrum is to make
a plot with frequency in Hz on the horizontal axis and intensity in decibels, or dB, on the
vertical axis. Intensity is determined by amplitude, and higher amplitude means higher
intensity.
Metaphorically, the spectrum is a visual representation of the “recipe” of a sound. You can
make a number of different things from the same ingredients— lour, sugar, eggs, milk,
butter, and baking powder. The difference between pancakes, cake, and cookies is in the
proportions of the different ingredients. Likewise, the violin and lute playing 440 Hz have
the same harmonics, but these harmonics appear in different proportions. If you know the
“recipe” for a sound, you can use electronics to arti icially “cook up” the sound. This is how a
synthesizer works.
The irst recording I made is of a tuning fork. Its waveform looks like a sine wave
(Figure 4.10). Let’s highlight a portion of the sound in Audacity and use Analyze → Spectrum
to make a plot of the spectrum (Figure 4.22). There is a large spike at 440 Hz, a much smaller
spike at 880 Hz, and not much else interesting.
CHAPTER 4. PITCHED SOUNDS
52
Figure 4.22: Spectrum of the tuning fork’s waveform shown in Figure 4.10
Harmonic spectra. All the instruments that obey the one-dimensional wave equation
have harmonics—that is, their modes of vibration produce tones whose frequencies f , 2f ,
3f , etc. are multiples of the fundamental frequency, f . Any such instrument is said to have a
harmonic spectrum. Figure 4.23 displays graphs of the spectra corresponding to the violin
and lute. The comb-like spikes occur at multiples of 440 Hz—that is, the spikes represent
the instrument’s harmonics. The relative strength of each harmonic re lects the instrument’s
timbre.
Figure 4.23: Spectra for a violin (left) and lute (right)
Figure 4.24: Spectra for a sawtooth wave (left) and sine wave (right)
Figure 4.24 shows the spectra of two sounds generated by Audacity. The sawtooth wave’s
spectrum is harmonic and most like a violin. It sounds more like a violin than a lute. The
sine wave’s spectrum has only one spike, 440 Hz, because the wave is made of one sinusoid.
It looks more like the spectrum of a tuning fork.
Chapter 5
Intervals, Pitch, and Pitch Class
A musical interval is a relationship between two pitched sounds. Intervals form the
underpinnings of melody and harmony. In this chapter, we will investigate the role of
intervals in how we perceive frequency. This leads to the de inition of pitch—a measure
of frequency that corresponds to human perception—and pitch class, which models our
perception that pitched sounds an octave apart are fundamentally similar. Pitch class is
an example of the mathematical concept of equivalence, which applies in many musical
situations.
5.1
Intervals
De inition 5.1. An interval is a relationship between two frequencies that is determined by
the ratio of the higher frequency to the lower frequency.
If two pairs of frequencies form the same ratio, then they form the same interval. For
example, the interval between 120 Hz and 360 Hz is the same as the interval between 300 Hz
and 900 Hz, because both pairs of frequencies are in a 3:1 ratio¹
The most fundamental musical interval is the octave, which corresponds to a 2:1 frequency
ratio. The irst and second harmonics of a string form an octave. Their ratio is 2:1 because
2f
= 21 , where f is the fundamental frequency. Since the harmonics of a string (f , 2f , 3f , …)
f
are each multiples of the fundamental frequency, f cancels out when we reduce the ratios to
lowest terms. Therefore, the ratios formed by consecutive harmonics are 2:1, 3:2, 4:3, 5:4,
and 6:5. These intervals all have names that are familiar to musicians: 2:1 corresponds to an
octave, 3:2 to a perfect ifth, 4:3 to a perfect fourth, 5:4 to a major third, and 6:5 to a minor
third, as summarized in Figure 5.1.
2:1
3:2
4:3
5:4
6:5
f ←octave→ 2f ←perfect ifth→ 3f ←perfect fourth→ 4f ←major third→ 5f ←minor third→ 6f
Figure 5.1: Named intervals between the irst six harmonics.
¹Ratios work in the same way as fractions, so a 3:2 ratio is the same as a 6:4 ratio, because 64 = 32 . Ratios
should be reduced to lowest terms. When measuring intervals, I’ll write them with the larger number irst.
53
CHAPTER 5. INTERVALS, PITCH, AND PITCH CLASS
54
Intervals may also be combined. For example, the frequency 4f is an octave higher than 2f
and the frequency 8f is an octave higher than 4f . Therefore, 4f is two octaves higher than f
(a 4:1 ratio) and 8f is three octaves higher than f (an 8:1 ratio). Note that 4 = 22 and 8 = 23 .
In general, the frequency n octaves above f is 2n f and the interval of n octaves corresponds
to a frequency ratio of 2n : 1.
Example. The ratio between the second and ifth harmonics of a string is 5:2 because
the second harmonic has frequency 2f , the ifth harmonic has frequency 5f , and
(5f ) : (2f ) = 5 : 2. The third and sixth harmonics form an octave because (6f ) : (3f ) = 2 : 1.
Example. The frequencies that are one, two, three, and four octaves above 100 Hz are
200 Hz, 400 Hz, 800 Hz, and 1600 Hz. Each time, going up an octave corresponds to doubling
the frequency.
Exercise 5.1. Determine which pairs of frequencies form the same interval:
(a) 600 Hz and 400 Hz, (b) 600 Hz and 500 Hz, and (c) 600 Hz and 900 Hz. Generate a 15 second,
600 Hz tone in Audacity and a second tone that plays 5 seconds each of 400 Hz, 800 Hz, and 900 Hz.
Which pairs of tones sound most similar?
Exercise 5.2. Find a pair of frequencies that forms the same interval that 300 Hz and 400 Hz do. Find
the formulas for all the frequencies that form a 4:3 ratio with a frequency f .
Exercise 5.3. Name the interval between 45 Hz and 720 Hz.
Exercise 5.4. Find the frequencies that are one, two, and three octaves below 100 Hz. What is the
general formula for the frequency n octaves below a frequency f ?
Solutions to exercises
Solution 5.1. Pair (a) forms the same interval as pair (c), because
600/400 = 3/2
600/500 = 6/5
900/600 = 3/2
I hear (a) and (c) as more similar, because my musical training means that I recognize them as perfect
fourths and (b) as a major third. Someone might also hear (a) and (b) as similar because, in both of
them, 600 Hz is the higher frequency. Or maybe you have a different reaction…
Solution 5.2. Since 400/300=4/3, multiply any frequency by 4/3 or 3/4 to ind frequencies that
form a 4:3 ratio with your chosen frequency. Starting with 600 Hz, 600 · 43 = 800, so 600 Hz and
800 Hz form the same interval as 300 Hz and 400 Hz. We could also multiply 600 · 34 = 450, so 600 Hz
and 450 Hz form the same interval as 300 Hz and 400 Hz. The frequencies that form a 4:3 ratio with a
frequency f are (3/4)f and (4/3)f .
Solution 5.3. Since (720/45) = 16 = 24 , the interval is four octaves.
Solution 5.4. Divide by two each time: 50 Hz, 25 Hz, and 12.5 Hz. The general formula is f /2n
(equivalently, 2−n f ).
CHAPTER 5. INTERVALS, PITCH, AND PITCH CLASS
5.2
55
Measuring intervals using semitones
Our ears don’t judge musical “distance” between two notes (pitched sounds) in the same
way as we measure frequency. Suppose you start singing the song “Somewhere Over the
Rainbow” at 220 Hz. Since the relationship between the irst and second notes of the song
is an octave, the second note’s frequency is 440 Hz. If you start singing at 440 Hz, then the
second note of the song is 880 Hz. Both pairs of frequencies are exactly the same distance
apart on the piano—twelve keys. The relevant computation is the 2:1 octave relationship,
not the differences in frequencies, which are 220 Hz and 440 Hz, respectively.
De inition 5.2. A semitone is a unit of measurement of intervals that corresponds to human
perception. There is one semitone between adjacent² keys on the piano, and one octave equals
12 semitones. There are 100 cents per semitone.
An interval may be speci ied either by a ratio of frequencies or by a distance, in semitones.
If two pairs of pitched sounds are separated by the same number of semitones, their
frequencies form the same ratio, and vice versa.
The following table gives some insight into how frequency ratios and semitones correspond
when measuring octaves. The variable r is the fractional version of a ratio, so that f : g
corresponds to r = f /g. Remember that the number of octaves between two frequencies
is equal to the power of 2 in their ratio. To ind the number of semitones between them,
multiply that power by 12.
octaves
1
2
3
ratio r
2 : 1 21
4 : 1 22
8 : 1 23
semitones
12 = 1 · 12
24 = 2 · 12
36 = 3 · 12
For intervals that are not octaves, we need the logarithm base 2 (log2 ), which is a
mathematical function that inputs a power of 2, 2n , and outputs its exponent n. For example,
log2 (8) = log2 (23 ) = 3 and log2 ( 21 ) = −1 because 2−1 = 21 . A number doesn’t have to be an
integer power of 2 for its logarithm to be computed: log2 (7) ≈ 2.807 because 22.807 ≈ 7. Most
calculators won’t compute log2 directly, but you can use the formula log2 (a) = log(a)/ log(2).
Measuring intervals in semitones.
frequency ratio f : g is
The measure, s, in semitones, of the interval with
s = 12 log2 (r) = 12 log(r)/ log(2), where r = f /g.
Example. An octave is a 2:1 ratio, so r = 2 and s = 12 log(2)/ log(2) = 12 semitones. This
con irms something we already know: there are 12 semitones per octave. The frequency
ratio between the second and third harmonics of a string is 3:2. It corresponds to a difference
in pitch of 12 log(3/2)/ log(2) ≈ 7.020 semitones.
²“Adjacent” is somewhat ambiguous, since the black keys are shorter than the white ones. I’ll say that keys
are adjacent if they are next to each other on the ends farthest from the pianist’s ingers, or as if the black keys
extended out to the edge of the keyboard, as in Figure 5.2.
CHAPTER 5. INTERVALS, PITCH, AND PITCH CLASS
56
Converting from semitones to ratios. Suppose you are given a distance, s, in semitones
and want to know what frequency ratio it represents. Solving the previous equation for r
shows that the ratio corresponding to s semitones is
ratio = r : 1, where r = 2s/12
Example. There are 12 semitones in an octave; substituting s = 12 in the formula gives us
r = 212/12 = 2, con irming that the frequency ratio for an octave is 2:1. We can also convert
7 semitones into a ratio: r = 27/12 ≈ 1.498, so the ratio is 1.498 : 1, or approximately 3 : 2.
Exercise 5.5. We showed that a 2:1 ratio of frequencies corresponds to exactly 12 semitones and a
3:2 ratio corresponds to approximately 7.020 semitones. Find the measure, in semitones, of the other
ratios formed by the irst six harmonics: 4:3, 5:4, and 6:5. Round your answers to two decimal places.
Exercise 5.6. Find ratios approximating the distances of (a) 5 semitones, (b) 4 semitones, and
(c) 3 semitones. Round your answers to two decimal places.
Solutions to exercises
Solution 5.5. The 4:3 interval spans 12 log(4/3)/ log(2) ≈ 4.98 semitones, the 5:4 interval spans
12 log(5/4)/ log(2) ≈ 3.86 semitones, and the 6:5 interval spans 12 log(6/5)/ log(2) ≈ 3.16 semitones.
Solution 5.6. (a) r = 25/12 ≈ 1.33; (b) r = 24/12 ≈ 1.26; (c) r = 23/12 ≈ 1.19. In each case, the ratio
is r : 1. Note that (a) is close to 4/3 = 1.3; (b) is close to 5/4 = 1.25; and (c) is close to 6/5 = 1.2.
5.3
Application: the major scale
A scale is a set of notes from which melodies are selected. The white keys on the piano form
a scale, as do the black keys. Many other scales are commonly used, some even containing
notes that aren’t on the piano. The notes of a scale, written in ascending order, typically form
a pattern of intervals that repeats every octave.
The notes in any major scale form a repeating pattern of seven intervals, whose sizes,
measured in semitones, are 2, 2, 1, 2, 2, 2, 1, after which the same list of intervals repeats.
Notice that 2 + 2 + 1 + 2 + 2 + 2 + 1 = 12, so the pattern starts over again an octave higher. If
you play the white notes on a piano, starting at middle C and going up to your right, you play
a major scale; however, a major scale can start at any frequency, even one not on the piano,
as long as it has the 2, 2, 1, 2, 2, 2, 1 pattern of intervals.
Because the scale is determined by its intervals, not its starting note, there are many major
scales on the piano. Therefore, we don’t give letter names (like “A” or “C”) to the notes in a
generic major scale. Instead, musicians use the solfege system
do
re
mi
fa
sol
la
ti
pronounced
doe
ray
me fah
sew
la
tea
which you might recognize from the song “Doe, a deer” (Rodgers & Hammerstein, 1959):
CHAPTER 5. INTERVALS, PITCH, AND PITCH CLASS
Doe, a deer, a female deer
Ray, a drop of golden sun
Me, a name I call myself
Far, a long long way to run
57
Sew, a needle pulling thread
La, a note to follow sew
Tea, I drink with jam and bread
That will bring us back to doe.
The starting note of the major scale is always called do, then re is two semitones above do, mi
is two semitones above re, etc. Because the pattern of intervals repeats after an octave, the
names are recycled—the last line of the song, “that will bring us back to doe” actually ends
on the note that is one octave higher than the original starting note. The fact that solfege
names are recycled means that there is no single meaning for “the interval between do and
sol.” Rather, we must specify a direction. The interval from do up to sol is 7 semitones, while
the interval from do down to sol is 5 semitones, which is the shortest interval between any
do and any sol.
fa
sol
la
ti do
5 semitones
re
2
mi fa
2
1
7 semitones
sol
2
la
2
ti do
2
re
mi fa
1
Exercise 5.7. Find the number of semitones from do up to mi and from do down to mi.
Exercise 5.8. Find all the pairs of solfege syllables that are separated by seven semitones. Hint: there
are six pairs, and (do, sol) is one of them.
Solutions to exercises
Solution 5.7. 4 and 8
Solution 5.8. (do, sol); (re, la); (mi, ti); (fa, do); (sol, re); (la, mi)
5.4
Absolute pitch
So far, I have de ined the semitone as a unit of measurement, like an inch or a liter. In
circumstances such as tuning a piano, you would want to have an “absolute” way of locating
frequencies so that the distances between them correspond to semitones, in accord with
human perception. Any logarithmic measure of frequency is called pitch. There are several
different systems in use. We’ll be using MIDI note numbers, in which each key on the piano
corresponds to a whole number from 21 to 108. MIDI stands for Musical Instrument Digital
Interface and is an industry standard for electronic instruments.
De inition 5.3. MIDI note numbers are a system of measuring pitch. They are assigned so that
note 69 corresponds to 440 Hz and an interval of one semitone corresponds to a unit difference
in MIDI note number.
Figure 5.2 shows a piano keyboard with the note names, MIDI note numbers, and frequencies
labeled. Two important keys are highlighted: key 69 (A440) and key 60 (middle C). I’ll use
the words “pitch” and “MIDI note number” to mean the same thing.
CHAPTER 5. INTERVALS, PITCH, AND PITCH CLASS
58
note MIDI frequency
C 108 4186.009
B 107 3951.066
A#/Bb 106 3729.310
A 105 3520.000
G#/Ab 104 3322.438
G 103 3135.963
F#/Gb 102 2959.955
F 101 2793.826
E 100 2637.020
D#/Eb
99 2489.016
D
98 2349.318
C#/Db
97 2217.461
C
96 2093.005
B
95 1975.533
A#/Bb
94 1864.655
A
93 1760.000
G#/Ab
92 1661.219
G
91 1567.982
F#/Gb
90 1479.978
F
89 1396.913
E
88 1318.510
D#/Eb
87 1244.508
D
86 1174.659
C#/Db
85 1108.731
C
84 1046.502
B
83 987.767
A#/Bb
82 932.328
A
81 880.000
G#/Ab
80 830.609
G
79 783.991
F#/Gb
78 739.989
F
77 698.456
E
76 659.255
D#/Eb
75 622.254
D
74 587.330
C#/Db
73 554.365
C
72 523.251
B
71 493.883
A#/Bb
70 466.164
A
69 440.000
G#/Ab
68 415.305
G
67 391.995
F#/Gb
66 369.994
F
65 349.228
E
64 329.628
D#/Eb
63 311.127
D
62 293.665
C#/Db
61 277.183
middleC
60 261.626
B
59 246.942
A#/Bb
58 233.082
A
57 220.000
G#/Ab
56 207.652
G
55 195.998
F#/Gb
54 184.997
F
53 174.614
E
52 164.814
D#/Eb
51 155.563
D
50 146.832
C#/Db
49 138.591
C
48 130.813
B
47 123.471
A#/Bb
46 116.541
A
45 110.000
G#/Ab
44 103.826
G
43
97.999
F#/Gb
42
92.499
F
41
87.307
E
40
82.407
D#/Eb
39
77.782
D
38
73.416
C#/Db
37
69.296
C
36
65.406
B
35
61.735
A#/Bb
34
58.270
A
33
55.000
G#/Ab
32
51.913
G
31
48.999
F#/Gb
30
46.249
F
29
43.654
E
28
41.203
D#/Eb
27
38.891
D
26
36.708
C#/Db
25
34.648
C
24
32.703
B
23
30.868
A#/Bb
22
29.135
A
21
27.500
Figure 5.2: Note names, MIDI note numbers, and frequencies for the piano keyboard.
To ind the frequency of MIDI note number p, measure its distance from note 69 in semitones,
convert that measurement to a frequency ratio, and ind the frequency that forms the
desired ratio with 440 Hz. For example, let’s calculate the frequency for note 60 (middle C).
The distance between 60 and 69 is 9 semitones. Using the semitones-to-ratio conversion,
r = 29/12 , which tells us that the ratio between 440 Hz and the frequency of middle C is
23/4 : 1. To ind the answer, divide 440 by 23/4 to obtain approximately 261.626 Hz. Figure 5.2
shows that this is correct.
This computation is summarized by the pitch-to-frequency conversion formula. The
frequency corresponding to pitch p is
f = 440 · 2(p−69)/12 Hz.
The frequency-to-pitch conversion formula is found by solving the above formula for p. The
pitch corresponding to frequency f is
(
p = 69 + 12
)
log(f /440)
.
log(2)
MIDI note numbers don’t have to be whole numbers. Non-integer pitches are called
microtones—they fall “in the cracks” between piano keys.
Exercise 5.9. Choose any key on the keyboard in Figure 5.2 and use the pitch-to-frequency formula
to check that the frequency listed is correct.
Exercise 5.10. Find two pairs of pitches that are six semitones apart. Look up their frequencies in
Figure 5.2 and compute the frequency ratio for each pair.
Exercise 5.11. List the six frequencies that are one, two, or three octaves above or below 440 Hz. In
Figure 5.2, mark all the piano keys corresponding to these frequencies. Find the MIDI note numbers
of the piano keys you marked.
Exercise 5.12. Approximate to two decimal places (a) the pitch of 660 Hz; (b) the frequency of pitch
1; (c) the frequency of pitch 13; (d) the pitch whose frequency is in a 9:4 ratio above 440 Hz
Exercise 5.13. Suppose a major scale starts at pitch 50. Use the 2212221 interval pattern to list the
irst eight pitches in the scale by their MIDI note numbers. Mark the keys on the keyboard.
Exercise 5.14. List the MIDI note numbers of all the keys named “C” on the keyboard, from smallest
to largest. How are these numbers related mathematically? List the numbers of all the “G”s. How are
these numbers related?
CHAPTER 5. INTERVALS, PITCH, AND PITCH CLASS
59
Solutions to exercises
Solution 5.9. Here’s an example: suppose I choose key 38. Then f = 440 · 2(38−69)/12 ≈ 73.416 Hz.
Solution 5.10. Suppose I choose 60 and 66 as my irst pair and 44 and 50 as my second pair. For the
irst pair, the frequency ratio is approximately 369.994/261.626 ≈ 1.414 and for the second, the ratio
is 146.832/103.826 ≈√1.414. No matter which pair you pick, the ratio should be approximately 1.414
(the exact answer is 2).
Solution 5.11. Multiply 440 by 2−3 , 2−2 , 2−1 , 21 , 22 , and 23 to get 55 Hz, 110 Hz, 220 Hz, 880 Hz,
1760 Hz, and 3520 Hz. The MIDI note numbers are 33, 45, 57, 81, 93, and 105. They are all congruent
to 9 modulo 12.
Solution 5.12. (a) 76.02; (b) 8.66 Hz (not audible); (c) 17.32 Hz; (d) 83.04
Solution 5.13. 50, 52, 54, 55, 57, 59, 61, 62. If you’re a musician, you might recognize this as a D
major scale.
Solution 5.14. The C’s are 24, 36, 48, 60, 72, 84, 96, and 108. They are all multiples of 12. The G’s are
31, 43, 55, 67, 79, 91, 103, 115. They are all congruent to 7 modulo 12.
5.5
Pitch classes
As we have seen both with the letter names of keys on the piano and with the solfege
syllables, pitch names may be reused after an octave. This is because our brains are
hard-wired to hear pitches that are an octave apart as similar.
De inition 5.4. The pitch class number of a pitch p equals p mod 12. Pitch classes also have
letter names, as in Figure 5.3. A pitch class (pc) is a set of all pitches with the same pitch class
number.
As/Bf
A
B
Cs/Df
0
Cs Ds
Df Ef
10
9
Gs/Af
11
C
1
Fs Gs As
Gf Af Bf
3
C D E F G A B
8
G
7
6
Fs/Gf
5
D
2
4
F
E
Figure 5.3: The pitch class circle
Ds/Ef
CHAPTER 5. INTERVALS, PITCH, AND PITCH CLASS
60
Every pitch class has in initely many members, because you could theoretically keep going
up or down by octaves forever. Because the system of pitch classes repeats after an octave,
they are often depicted on a circle, called the pitch class circle, shown in Figure 5.3. Pitch
class C is at the top and numbered 0. Pitch classes are the analog of beat classes in a 12-beat
measure: two pitches are in the same pitch class if they are in the same position in an octave,
just as two beats are in the same beat class if they are in the same position in a measure.
Although there are twelve pitch classes represented on a piano, there are actually an in inite
number of pitch classes, because there are microtonal pitches “between the cracks” whose
MIDI numbers are not whole numbers.
Pitch classes corresponding to black keys on the piano are labelled with accidental signs:
the lat symbol ♭ means “minus one semitone” and the sharp symbol ♯ means “plus one
semitone.” In normal piano tuning, the black keys on the piano may be labeled in two
different ways, one using a sharp sign and the other using a lat sign. For example, C♯ and D♭
are the same pitch class. There are musical reasons that one label is preferred over another
in a particular piece, but they both refer to the same key.
For example, the pitch class of pitch 51 contains the pitches
. . . 27, 39, 51, 63, 75, . . .
To ind the pitch class number, compute 51 mod 12 = 3. Pitch 51 belongs to pc 3, or D♯/E♭.
Any pitch p is in the same pitch class as pitch 51 if p mod 12 = 3. Pitch 100 is not in the same
pitch class because 100 mod 12 ̸= 3.
Intervals between pitch classes (pc intervals). Measuring intervals between pitch
classes presents a problem, because there are two ways to measure distance on a
circle—clockwise (ascending in pitch) and counterclockwise (descending in pitch).
De inition 5.5. A pc interval is the shortest distance between any members of two pitch
classes on the pitch class circle, in semitones.
Pc intervals are numbers between 0 and 6. For example, although the interval between
pitches 60 and 78 is 9 semitones, the shortest distance between any C (pitch in pc 0) and any
A (pitch in pc 9) is 3 semitones. Therefore, the pc interval between pc 0 and pc 9 is 3.
Application: Pitch classes in a major scale. We can use the 2212221 pattern to ind
the major scale starting on any particular pitch. For example, the C major scale starting on
middle C is 60, 62, 64, 65, 67, 69, 71, 72. The pitch classes in a C major scale (use do = C) are
numbered {0, 2, 4, 5, 7, 9, 11} = {C, D, E, F, G, A, B}. Note that the pitch class interval between
pc 0 and 11 is 1, not 11, completing the 2212221 pattern. Pitch classes for the G major scale
are numbered {7, 9, 11, 0, 2, 4, 6} = {G, A, B, C, D, E, F♯}. When naming the pitch classes in a
scale using letters, every letter must be used once. So the seventh pitch class is called F♯, not
G♭.
Exercise 5.15. List MIDI numbers of ive C’s and ive C♯’s.
Exercise 5.16. Compute the pc number and letter of pitches (a) 80, (b) 84, (c) 100.
CHAPTER 5. INTERVALS, PITCH, AND PITCH CLASS
61
Exercise 5.17. Find the pc intervals for the following pairs of pitch classes: (a) 3 and 0, (b) 1 and 11,
(c) F and B
Exercise 5.18. If the interval between two pitches is 19 semitones, what is the pc interval between
their pitch classes?
Exercise 5.19. Find the pitch class numbers in a D major scale (use do = D). Use the “every letter
must be used once” rule to ind their letter names. Find the pitch class numbers and letters in a C♯
major scale (you have to be creative here).
Solutions to exercises
Solution 5.15. C = {. . . , 48, 60, 72, 84, 96, . . .} and C♯ = {. . . , 49, 61, 73, 85, 97, . . .}
Solution 5.16. (a) 80 mod 12 = 8 = G♯/A♭; (b) 84 mod 12 = 0 = C; (c) 100 mod 12 = 4 = E
Solution 5.17. (a) 3, (b) 2, (c) 6
Solution 5.18. Answer: 5. They are separated by 7 ascending (clockwise) semitones on the pitch
class circle, because 19 mod 12 = 7. However, the shortest distance between their pitch classes is 5
semitones.
Solution 5.19. D major = {2, 4, 6, 7, 9, 11, 1} = {D, E, F♯, G, A, B, C♯}; C♯ major = {1, 3, 5, 6, 8, 10, 0} =
{C♯, D♯, E♯, F♯, G♯, A♯, B♯} (Yes, E♯ is a pitch class. It equals F. Likewise, B♯ = C.)
Chapter 6
Transformation
6.1
Clapping math
Watch the video of Steve Reich’s “Clapping Music.” “Clapping Music” uses a single repeated
rhythm pattern. The two players start by clapping the 12-beat pattern together in unison.
After eight measures, one performer shifts the pattern forward by one beat and they clap the
new pattern for eight measures. This process is continued until the two patterns align again.
Reich says that the piece was inspired by the clapping patterns in traditional lamenco music
of the Roma people in Andalusia, Spain.
Steve Reich’s “Clapping Music” is all about the rhythm that results from playing multiple
simultaneous patterns. Let’s develop some mathematics to describe this concept, learn
three different mathe-musical notations, then analyze “Clapping Music.”
We can use drum tab to visualize “Clapping Music.” Player A claps the same rhythm over and
over until the piece ends.
Player A:
x x x . x x . x . x x .
This pattern can also be written as a set¹ of beat classes: { 0, 1, 2, 4, 5, 7, 9, 10 }. Player B
progresses through a number of patterns, playing each one eight times. Here are the irst
four, written in drum tab and as sets of beat classes.
1.
2.
3.
4.
x
x
x
.
x
x
.
x
x
.
x
x
.
x
x
.
x
x
.
x
x
.
x
.
.
x
.
x
x
.
x
x
.
x
x
.
x
x
.
x
x
.
x
x
.
x
x
x
{ 0, 1, 2, 4, 5, 7, 9, 10 }
{ 0, 1, 3, 4, 6, 8, 9, 11 }
{ 0, 2, 3, 5, 7, 8, 10, 11 }
{ 1, 2, 4, 6, 7, 9, 10, 11 }
Each drum tab pattern is formed by shifting the previous one forward and moving the irst
hit or rest to the end. This shift is called cyclic permutation by mathematicians. It will take
12 changes of pattern for the second player to sync up with the irst player again in unison,
meaning that they play the same pattern at the same time.
As sets of beat classes, the patterns are related by subtraction on the beat class circle (so
that 0 − 1 = 11). The subtraction is more obvious when you write the members of the sets in
different orders. Remember that order doesn’t matter in a set.
¹A set is de ined in Section ?? to be a non-repeating collection of objects, called members. Curly brackets “{”
and “}” enclose a list to indicate that it is a set.
62
CHAPTER 6. TRANSFORMATION
{ 0,
{ 11,
{ 10,
{ 9,
63
1,
2, 4, 5, 7, 9, 10
0,
1, 3, 4, 6, 8, 9
11 , 0, 2, 3, 5, 7, 8
10, 11 , 1, 2, 4, 6, 7
}
}
}
}
Here are the irst four patterns played by player B shown with circular notation. The
progression from one pattern to the next corresponds to a 30 degree rotation of the pattern
counterclockwise.
1
2
3
4
Analysis of “Clapping Music.” As we have seen, the progression from one pattern to the
next involves repeatedly shifting the pattern clapped by the second player, producing a
change in the resultant rhythm. The tension created by the two musicians clapping different
patterns is resolved when the second player’s rhythm completes a full cycle, bringing the
two players back into unison and giving closure to the piece. However, the fact that the piece
ends exactly as it began suggests that “Clapping Music” might continue inde initely, following
the same pattern as before.
Though I don’t know “why” Reich chose this particular rhythm pattern and not another,
there are a few features that give extra interest to the music. Trying the same process with a
different rhythm can also give us insight into why this pattern works so well. Here are a few
observations I have:
1. The fact that the rhythm has no adjacent rests means that in the penultimate pattern
(the one before the inal unison), the resulting rhythm is equal to the set of all 12 beat
classes in the measure. This gives the maximum feeling of tension before the end.
2. Moreover, the resultant rhythm is usually not equal to the set of 12 beat classes, but
something more interesting. In fact, that situation only happens three times. Can you
ind them?
3. The Clapping Music rhythm cannot be divided into smaller repeating patterns of, say, 6
beats. If it could, the players would come into unison after only 6 different patterns.
Exercise 6.1. Write all shifts of the rhythm xx.x. in drum tab, as sets of beat classes, and in circular
notation.
Exercise 6.2. How many times do you have to shift ....xx to make ..xx..? Write these rhythms
as sets of beat classes. What is the mathematical relationship between these sets?
.
6.2
Rhythm class
Flamenco is a dance form from Andalusia, Spain that is associated with the Roma peoples.
Watch the videos of Flamenco dancers, an explanation of the rhythm patterns, and a
demonstration of clapping.
We saw earlier that beats that occupy the same position in a measure belong to the same
beat class. The notion of two rhythms being similar if one is a shift of the other gives us a
concept of similarity for rhythms.
CHAPTER 6. TRANSFORMATION
64
De inition 6.1 (Rhythm class). Suppose there are n beats per measure. The rhythm class of a
given rhythm pattern is the set of all the shifts of that pattern. Equivalently, two n-beat
rhythms belong to the same rhythm class if and only if one can be repeatedly shifted to align
with the other. A rhythm class has at most n elements.
Example. The rhythm class of the ive-beat pattern x..x. is a set of ive rhythms. I’ve
written them without the normal commas between members, because commas are easy to
confuse with periods in drum tab.
{ x..x.
..x.x
.x.x.
x.x..
.x..x}
Example. There are eight three-beat rhythms:
{ xxx
xx.
x.x
x..
.xx
.x.
..x
...}
These patterns may be divided into four rhythm classes:
{ xxx },
{ xx.
x.x
.xx },
{ x..
..x
.x. }, and { ... }
Example. In “Clapping Music,” the second player claps all the members of the rhythm class
of the basic pattern. There are 12:
{ xxx.xx.x.xx.
xx.xx.x.xx.x
… .xxx.xx.x.xx }
Example. Clapping is an essential part of lamenco music. Within the genre of lamenco,
there are several different styles and dances that are distinguished by their characteristic
rhythm patterns. The bulería pattern is x..x..x.x.x. while x.x.x..x..x. is the
siguiriyas pattern. Shifting the bulería eight times produces the siguiriyas. Therefore, the
two patterns belong to the same rhythm class.
The relationship between bulería and siguiriyas is perhaps easier to recognize using circular
notation. Here are the two patterns. Flamenco musicians often represent rhythms on a
clock: bulería starts at 12:00 and siguiriyas at 8:00.
bulería
siguiriyas
Exercise 6.3. What are the patterns in the rhythm class of xxx.x..? Do the patterns xxx.x.. and
..x.xxx belong to the same rhythm class?
Exercise 6.4. What are the patterns in the rhythm class of x..x..? Under what circumstances will
a rhythm class of 6-beat rhythms have fewer than six members?
CHAPTER 6. TRANSFORMATION
65
Solution 6.1. Shifts: xx.x. x.x.x .x.xx x.xx. .xx.x
Sets of beat classes: { 0, 1, 3 }; { 4, 0, 2 }; { 3, 4, 1 }; { 2, 3, 0 }; { 1, 2, 4 }. I solved this problem by
subtracting 1 each time. Note that sets may be written in any order. For example, { 4, 0, 2 } = { 0, 2, 4 }.
Solution 6.2. You must shift 2 times. The sets are { 4, 5 } and { 2, 3 }. Mathematically, you subtract 2
from the irst set to get the second.
Solution 6.3. The rhythm class of xxx.x.. is
{ xxx.x.. xx.x..x x.x..xx .x..xxx
pattern ..x.xxx does not belong to this class.
x..xxx.
..xxx.x
.xxx.x. } The
Solution 6.4. The rhythm class is { x..x.. .x..x. ..x..x }. If a rhythm contains repeats, its
rhythm class will have fewer than 6 members. Therefore, it is possible for a rhythm class in a 6-beat
measure to have 1, 2, 3, or 6 members. In general, if n is the number of beats in a measure, then the
sizes of rhythm classes are divisors of n, and every divisor of n is represented.
6.3
Transposition
When you play the piano, you can sound more than one key at a time, producing a set of
pitches (remember that the order of the members of a set does not matter). Two sets of
pitches sound similar if their members form the same set of pitch classes. For example,
the pitch sets { 60, 64, 67 } and { 55, 72, 76, 84 } are similar because they both contain the
same pitch classes: { 0, 4, 7 } = C, E, and G. A chord is a set of pitch classes. We can label this
chord using pitch classes: { C, E, G }. Notice that, because elements of a set can’t repeat, the
number of pitches played in a pitch set might not equal the number of pitch classes in the
corresponding chord.
Some chords sound more or less the same, just higher or lower in pitch. For example, the
chords { C, E, G } and { D, F♯, A } sound alike. This is because, although they contain different
pitch classes, the pc intervals between pitch classes are the same. We can de ine a type of
similarity on chords: transposition.
De inition 6.2. To transpose a pitch, pitch class, set of pitches, chord, scale, or melody means
to move all its pitches up or down by the same number of semitones, n, producing the
transposition of that set by n semitones. The sign of n (+ or −) indicates whether the pitches
are moved up (positive) or down (negative).
Example. The transposition of pitch 60 by 5 semitones is pitch 65. The transposition of
pitch 60 by −5 semitones (equivalently, its transposition down by 5 semitones) is pitch
55. The transposition of pc 0 by 5 semitones is pc 5, and the transposition of pc 0 by
−5 semitones (or 7 semitones) is pc 7. Since pitch classes “live” on a circle, one normally
uses only positive numbers for pc transpositions, since transposition by −n, 12 − n, or any
number congruent to −n modulo 12 produce the same result.
Example. The transposition of the pitch set { 60, 64, 67 } by 5 semitones is { 65, 69, 72 }.
The transposition of { 60, 64, 67 } by −5 semitones is { 55, 59, 62 }.
CHAPTER 6. TRANSFORMATION
66
Example. The transposition of the chord { 0, 4, 7 } by 5 semitones (or −7 semitones)
is { 5, 9, 0 } (equivalently, { 0, 5, 9 }). The transposition of { 0, 4, 7 } by 7 semitones (or
−5 semitones) is { 7, 11, 2 } (equivalently, { 2, 7, 11 }).
Example. The transposition of the C major scale { 0, 2, 4, 5, 7, 9, 11 } by 5 semitones is
{ 5, 7, 9, 10, 0, 2, 4}, which is the F major scale. The transposition of the C major scale by
7 semitones is { 7, 9, 11, 0, 2, 4, 6 }, which is the G major scale.
De inition 6.3. Two sets of pitches, chords, scales, or melodies are transpositionally
equivalent if one of them can be transposed by some number of semitones to produce the other.
The easiest way to determine transpositional equivalence between chords or scales is to
draw two pitch class circles and mark the pc’s, one chord or scale on each circle. If one set of
pitch classes is a rotation of the other, the two chords or scales are equivalent.
Example. The pitch sets { 46, 57, 60 } and { 40, 51, 54 } are transpositionally equivalent
by a transposition of −6 semitones. The chords { 0, 3, 5 }, { 6, 9, 11 }, and { 2, 4, 11 } are
transpositionally equivalent. Spotting the relationship between { 0, 3, 5 } and { 2, 4, 11 } is
sometimes dif icult if you just look at the pc’s. The circular representation helps.
Example. The chords { C, E, G } and { D, F♯, A } are transpositionally equivalent because one
is the rotation of the other by two semitones. Both chords are transpositionally equivalent
to { D♯, F♯, B }.
De inition 6.4. The chord type of a given chord is the set of all the chords that are
transpositionally equivalent to it.
Chord types often have names. Here are a few, each listed alongside one representative of
their chord type.
• { C, E, G }
major triad
• { A, C, E }
minor triad
• { C, E, G, B♭}
• { C, E, G♯}
dominant seventh
augmented triad
• { C, D♯/E♭, F♯/G♭, A }
diminished seventh
Exercise 6.5. Find (a) the transposition of the pitch set { 34, 52, 64 } by −5 semitones (b) the
transposition of the pitch class set { 3, 7, 9 } by 5 semitones (c) the transposition of a D major scale by
2 semitones. Which major scale is this?
Exercise 6.6. List a representative of each chord type that is not shown above.
When a chord is transposed, what is ixed? The answer is the intervals between pitches or
pitch classes.
CHAPTER 6. TRANSFORMATION
6.4
67
Solutions to exercises
Solution 6.5. (a) { 29, 47, 59 }; (b) { 8, 0, 2 } (or { 0, 2, 8 }); (c) { 4, 6, 8, 9, 11, 1, 3 }, which is the
E major scale.
Solution 6.6. Answers vary. Here are some examples: Major triad: { G, B, D }. Minor triad: { C, D♯/E♭,
G }. Dominant seventh: { F, A, C, D♯/E♭ }. Augmented triad: { D, F♯, A♯}
Appendix A
Worksheets
68
APPENDIX A. WORKSHEETS
69
Poetic meters of ixed duration
Syllables in Sanskrit poetry are either short or long. The duration of a long syllable is exactly twice that of a
short syllable. This means that we can measure the total duration of a line of poetry. For example, the duration
of SSLS is 5:
duration(SSLS) = 1 + 1 + 2 + 1 = 5
The Indian scholar Hemachandra (c. 1100 A.D.) investigated the following question: how many poetic meters
are there of any given duration? We can igure out what he did by creating a list:
Duration
1
2
3
4
5
Meters
S
SS, L
Number
1
2
1. Complete the table.
2. How many meters have duration 6?
Duration 7?
3. Explain how you ind the next number in the sequence.
4. What is the Western name for this sequence of numbers?
5. Hemachandra actually found an algorithm for listing all the meters that have some given number of
beats. It is somewhat similar to Pingala’s algorithm for listing all the meters that have a given number of
syllables. Summarize the algorithm:
6. Use the algorithm to list all the meters with 6 beats.
APPENDIX A. WORKSHEETS
70
Musical rhythms with 2- and 3-beat notes
70 Musical patterns are typically classi ied by their number of beats. In this worksheet, you will derive the
sequence counting meters of a given duration formed from notes of length 2 and 3.
Duration
1
2
3
4
5
6
7
8
9
Rhythms
(none)
2
3
2+2
2+3, 3+2
3+3, 2+2+2
2+2+3, 2+3+2, 3+2+2
Number
0
1
1
1
2
2
3
1. Complete the table.
2. Look for a pattern in the right-hand column. How many rhythms have duration 10?
11?
Duration
3. Explain how you ind the next number in the sequence.
4. Write the recursive rule for the sequence using the notation Pn for the nth number in the sequence.
5. Explain how to write all the rhythms with 10 beats using the patterns you have written in the table.
6. Explain why your recursive rule is correct.
APPENDIX A. WORKSHEETS
71
The expanding mountain of jewels
In each box, enter the number of meters that it the description. The number in the right-hand column should
equal the sum of the numbers in that row. This pattern was irst noticed by Pingala when solving a slightly
different problem. He called it the meruprastāra, or “Expanding Mountain of Jewels.”
Number of short syllables
Total number of syllables
0
1
2
3
4
5
6
7
1
2
2
4
3
8
4
16
5
32
6
64
7
128
1. Describe an algorithm that tells you how the ind the nth row in the table, if you know the previous row.
2. What is the name for this array in Western mathematics?
3. What are other things that the numbers in this array count?
4. How is this array related to the Hemachandra numbers?
APPENDIX A. WORKSHEETS
72
Patterns in the meruprastāra
There are some other cool recursive patterns in the meruprastāra. Shade the odd numbers and leave the even
numbers white.
1
1
1
1
1
1
1
1
1
6
7
8
2
3
4
5
1
1
3
6
1
4
10 10
1
5
15 20 15
1
6
21 35 35 21
28 56 70 56 28
1
7
1
8
1
In fact, you can predict which numbers will be odd and which numbers will be even, just by knowing how odd
and even numbers add:
• odd + odd =
• odd + even =
• even + even =
So you don’t need to calculate the numbers to shade in the hexagons. See the template on the next page →
APPENDIX A. WORKSHEETS
73
Make a patterned meruprastāra using the hexagonal graph paper. Turn the paper sideways. Start by coloring
the hexagons that would contain 1’s, then color the others using the addition rules for odd and even numbers.
Research Sierpinski’s Triangle. How is this picture related? What is a fractal?
Free Hexagonal Graph Paper from http://incompetech.com/graphpaper/hexagonal/
APPENDIX A. WORKSHEETS
74
Sinusoids
1. Find the amplitude, frequency, period, and sketch the waveform y = 2 sin(4πt). Use correct units in
your answer.
Desmos Graphing Calculator
https://www.desmos.com/calculator
3/9/17, 12:46 PM
2. Find the formula for this wave. The numbers on the horizontal axis are seconds.
3. Sketch a sine wave with amplitude = 0.5 and period = 0.05 seconds. Be sure to label both the horizontal
and vertical axes of your graph. Find the formula for this wave and use a graphing program to check
your answer. Is this the waveform of an audible sound?
by
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4. Sketch on the same axis and describe the relationship between the graphs of y = sin(t), y = 2 sin(t),
and y = 3 sin(t).
APPENDIX A. WORKSHEETS
75
5. Sketch on the same axis and describe the relationship between the graphs of
y = sin(t) and y = − sin(t).
6. In general, the sum of two waveforms (or any two mathematical functions) is found by adding their y
values. That is, the graph of the sum of y = f (t) and y = g(t) is the graph y = f (t) + g(t). If the
waveform y = sin(t) is added to y = 2 sin(t), what is the result? This is an example of constructive
interference.
7. If the waveforms y = sin(t) and y = − sin(t) are added, what is the result? This is an example of
destructive interference.
8. Suppose the graphs of y = sin(2t) and y = sin(3t) are added. Sketch both waves and their sum on the
same axes. This example is neither constructive nor destructive, because the waves have different
frequencies.
9. Is the sum of two periodic waves with the same frequency also periodic? Does it have the same
frequency as well?
APPENDIX A. WORKSHEETS
76
Intervals and pitch
1. Here are the intervals between the successive harmonics of a string.
2:1
3:2
4:3
5:4
6:5
f ←octave→ 2f ←perfect ifth→ 3f ←perfect fourth→ 4f ←major third→ 5f ←minor third→ 6f
(a) Identify the interval between 70 Hz and 105 Hz.
(b) List two frequencies that are a perfect fourth apart.
(c) Identify the interval between the second and fourth harmonics of a string.
2. Find the frequency that is ive octaves above 200 Hz. Find the frequency that is two octaves below
200 Hz.
3. Measure the interval between 500 Hz and 200 Hz in semitones. Round your answer to three decimal
places.
4. Find the number of semitones between the third and ifth harmonics of a string. Round your answer to
three decimal places.
5. If two notes are 2 semitones apart, what is the relationship between their frequencies? Round your
answer to three decimal places.
APPENDIX A. WORKSHEETS
6. List the six frequencies that are one, two, or three octaves above or below
440 Hz. Mark all the piano keys corresponding to these frequencies. Use the
chart to ind the MIDI note numbers and letter names of the piano keys you
marked.
7. Approximate to two decimal places
(a) the frequency of pitch 127
(b) the frequency of pitch 0
(c) the pitch of the lowest audible frequency, 20 Hz
(d) the pitch of the highest audible frequency, 20,000 Hz
8. List the MIDI note numbers of all the keys named “C” on the keyboard, from
smallest to largest. How are these numbers related mathematically? How are
the numbers of the keys named “D” related?
9. Suppose a major scale starts at pitch 55. Use the 2212221 interval pattern to
list the irst eight pitches in the scale by their MIDI note numbers. Mark the
keys on the keyboard.
77
note MIDI frequency
C 108 4186.009
B 107 3951.066
A#/Bb 106 3729.310
A 105 3520.000
G#/Ab 104 3322.438
G 103 3135.963
F#/Gb 102 2959.955
F 101 2793.826
E 100 2637.020
D#/Eb
99 2489.016
D
98 2349.318
C#/Db
97 2217.461
C
96 2093.005
B
95 1975.533
A#/Bb
94 1864.655
A
93 1760.000
G#/Ab
92 1661.219
G
91 1567.982
F#/Gb
90 1479.978
F
89 1396.913
E
88 1318.510
D#/Eb
87 1244.508
D
86 1174.659
C#/Db
85 1108.731
C
84 1046.502
B
83 987.767
A#/Bb
82 932.328
A
81 880.000
G#/Ab
80 830.609
G
79 783.991
F#/Gb
78 739.989
F
77 698.456
E
76 659.255
D#/Eb
75 622.254
D
74 587.330
C#/Db
73 554.365
C
72 523.251
B
71 493.883
A#/Bb
70 466.164
A
69 440.000
G#/Ab
68 415.305
G
67 391.995
F#/Gb
66 369.994
F
65 349.228
E
64 329.628
D#/Eb
63 311.127
D
62 293.665
C#/Db
61 277.183
middleC
60 261.626
B
59 246.942
A#/Bb
58 233.082
A
57 220.000
G#/Ab
56 207.652
G
55 195.998
F#/Gb
54 184.997
F
53 174.614
E
52 164.814
D#/Eb
51 155.563
D
50 146.832
C#/Db
49 138.591
C
48 130.813
B
47 123.471
A#/Bb
46 116.541
A
45 110.000
G#/Ab
44 103.826
G
43
97.999
F#/Gb
42
92.499
F
41
87.307
E
40
82.407
D#/Eb
39
77.782
D
38
73.416
C#/Db
37
69.296
C
36
65.406
B
35
61.735
A#/Bb
34
58.270
A
33
55.000
G#/Ab
32
51.913
G
31
48.999
F#/Gb
30
46.249
F
29
43.654
E
28
41.203
D#/Eb
27
38.891
D
26
36.708
C#/Db
25
34.648
C
24
32.703
B
23
30.868
A#/Bb
22
29.135
A
21
27.500
APPENDIX A. WORKSHEETS
78
Answers to worksheet: Intervals and pitch
1. Here are the intervals between the successive harmonics of a string.
2:1
3:2
4:3
5:4
6:5
f ←octave→ 2f ←perfect ifth→ 3f ←perfect fourth→ 4f ←major third→ 5f ←minor third→ 6f
(a) Identify the interval between 70 Hz and 105 Hz. ANSWER: 105:70 = 3:2 = a perfect ifth
(b) List two frequencies that are a perfect fourth apart. ANSWER: any frequency f and either (4/3)f
or (3/4)f
(c) Identify the interval between the second and fourth harmonics of a string. ANSWER: 4f : 2f = 2:1
= an octave
2. Find the frequency that is ive octaves above 200 Hz. Find the frequency that is two octaves below
200 Hz. ANSWER: 25 · 200 = 6400 Hz and 200/22 = 50 Hz
3. Measure the interval between 500 Hz and 200 Hz in semitones. Round your answer to three decimal
places. ANSWER: 12 log(500/200)/ log(2) ≈ 15.863 semitones
4. Find the number of semitones between the third and ifth harmonics of a string. Round your answer to
three decimal places. ANSWER: 12 log(5f /3f )/ log(2) ≈ 8.844 semitones
5. If two notes are 2 semitones apart, what is the relationship between their frequencies? Round your
answer to three decimal places. ANSWER: r = 22/12 ≈ 1.122 so their ratio is ≈ 1.122 : 1.
6. List the six frequencies that are one, two, or three octaves above or below 440 Hz. Mark all the piano
keys corresponding to these frequencies. Use the chart to ind the MIDI note numbers and letter names
of the piano keys you marked. ANSWER: multiply 440 by 2−3 , 2−2 , 2−1 , 21 , 22 , and 23 to get 55 Hz,
110 Hz, 220 Hz, 880 Hz, 1760 Hz, and 3520 Hz. The MIDI note numbers are 33, 45, 57, 69, 81, 93, and
105.
7. Approximate to two decimal places
(a) the frequency of pitch 127 ANSWER: 12543.85 Hz
(b) the frequency of pitch 0 ANSWER: 8.18 Hz
(c) the pitch of the lowest audible frequency, 20 Hz ANSWER: MIDI note number 15.49
(d) the pitch of the highest audible frequency, 20,000 Hz ANSWER: MIDI note number 135.08
8. List the MIDI note numbers of all the keys named “C” on the keyboard, from smallest to largest. How are
these numbers related mathematically? How are the numbers of the keys named “D” related? ANSWER:
The C’s are 24, 36, 48, 60, 72, 84, 96, and 108. They are all multiples of 12. The D’s are 26, 38, 50, 62, 74,
86, and 98. They are all congruent to 2 modulo 12.
9. Suppose a major scale starts at pitch 55. Use the 2212221 interval pattern to list the irst eight pitches
in the scale by their MIDI note numbers. Mark the keys on the keyboard. ANSWER: 55, 57, 59, 60, 62,
64, 66, 67. If you’re a musician, you might recognize this as a G major scale.
Appendix B
Assignments
79
APPENDIX B. ASSIGNMENTS
80
Homework for Chapter 1
1.1. Install software that allows you to record and analyze sound. I recommend Audacity for a
computer, TwistedWave for iPhone, and MixPad for Android, all of which are free.¹ Make two
ive-second recordings of “found sounds” that are not intended to be music. One recording should be
a sound that you perceive as highly musical and the other should be something that doesn’t sound
like music to you.
(a) Identify and describe the sounds that you recorded. Why did you choose them?
(b) Comment on the differences between the pattern (or lack of pattern) in their waveforms, both
zoomed out and zoomed in.
(c) Extra Credit: If you don’t have one already, create an account at SoundCloud.com. Upload your
sounds and add the hashtag #thesoundofnumbers. Send me email at [email protected] with
links to your sounds. Please identify yourself in the email so I know who you are.
1.2. Prove that each of these statements is false.
(a) If n is an integer, then n × 1/n = 1.
(b) All musical compositions involve musicians making sound.
(c) Every musical composition has a speci ic beginning and end.
1.3. De inition: An integer a is divisible by an integer b if a = bk, where k is an integer.
Explain why the following is NOT a proof of the statement “The square of any even number is divisible
by 4.”
Proof. A number is even if it equals 2n, where n is an integer. Suppose the number is 6, which is even
because 6 = 2 · 3. Then 62 = 36, which is divisible by 4 because 36 = 4 · 9. The square of any even
number is divisible by 4 for the same reason.
2
Homework for Chapter 2
2.1. Consider the collection of poetic meters with seven syllables which may be long or short.
How many total meters are there?
How many of these meters have
(a) no short syllables?
(e) four short syllables?
(b) one short syllable?
(f) ive short syllables?
(c) two short syllables?
(g) six short syllables?
(d) three short syllables?
(h) seven short syllables?
¹Audacity is installed on all the computers in the Digital Media Zone, which is located on the second loor of
the Post Learning Commons side of the library. If you would like help installing Audacity on your own computer,
go to Science Center 129.
APPENDIX B. ASSIGNMENTS
81
Hint: this question is extremely tedious if you try to write down all the meters and count them. The
best way to answer it is to use the relationship between poetic meters and the meruprastāra (Pascal’s
Triangle).
2.2. Suppose you lip a coin seven times and write down the sequence of heads (H) and tails (T).
There is a exact correspondence between sequences of heads and tails and poetic meters: just replace
H with L and T with S. How many possible patterns are there?
Use your answers to the previous question to determine the percentage of the patterns, rounding to
two decimal places, that have
(a) no tails
(e) four tails
(b) one tail
(f) ive tails
(c) two tails
(g) six tails
(d) three tails
(h) seven tails
Assuming your coin is fair, each of the patterns of heads and tails is equally likely. The percentages
you have computed equal the probabilities that you will get each number of tails when you lip the
coin seven times.
2.3. How many meters have the same duration as SSSSSS (including SSSSSS)? How have the same
duration as LLLLLL? Your answers should be Hemachandra numbers.
2.4. Suppose a drummer wants to take a solo that is eight beats long and made up of 1-beat notes,
2-beat notes, and 4-beat notes. How many patterns are possible? Hint: I don’t recommend trying
to list the patterns, because there are a lot. Rather, ind base cases and a recursive rule, write the
sequence of number of patterns, and ind the eighth number in that sequence. If you’re a musician,
the question is, “How many ways can you ill two measures in 4/4 time with quarter notes, half notes,
and whole notes?”
2.5. Add the numbers between the diagonal lines in Pascal’s Triangle. For example, the irst few
sums are 1, 1, 1+1=2, 1+2=3, and 1+3+1=5. What’s the pattern?
1
1
1
1
1
1
1
1
1
6
7
8
2
3
4
5
1
1
3
6
1
4
10 10
1
5
15 20 15
1
6
21 35 35 21
28 56 70 56 28
1
7
1
8
1
APPENDIX B. ASSIGNMENTS
82
2.6. Demonstrate that the seventh row of the meruprastāra may be found using Bhaskara’s method.
Show your work.
2.7. A prime number is an integer greater than 1 whose only positive divisors are 1 and itself. For
example, 2, 3, and 5 are prime but 6 is not. Study the irst four prime number rows in Pascal’s
Triangle—that is, the rows whose second number is prime. What do all the numbers greater than 1
in those rows have in common? Make a conjecture about the 11th row and all prime rows.
2.8. Extra Credit: Complete the worksheet “Patterns in the meruprastāra” in the text posted online.
Note that the worksheet has two pages and you must print both of them.
APPENDIX B. ASSIGNMENTS
83
Solutions (Chapters 1 & 2)
1.1. Answers vary. Your choices of sounds should show that you know the de inition of a found
sound (that is, a sound that is not intended to be music by whoever or whatever created it). Sounds
that seem more musical tend to be more structured: either the overall shape of the waveform has a
pattern that indicates the sound is rhythmic, or the waveform has a repetitive pattern when you zoom
in; however, this observation isn’t a hard and fast rule.
2 points based on completeness
1.2. You have to show a counterexample to each of the statements: (a) 0 is an integer, but 0 × 10 is
unde ined because 10 is not a number. (b) Answers vary. Any composition where no musician is
making a sound is a counterexample. Computer music, found sounds, and Cage’s 4’4” are some of the
possible answers. (c) Answers vary. Some types of music don’t have beginnings and endings. African
drumming, algorithmic music, and jam band music are some of the possible answers.
3 points (1 point each)
1.3. This is not a proof because it is only an example showing that the square of 6 is divisible by 4.
There are an in inite number of examples, so you can’t prove the statement by working out examples.
Instead, you have to make a logical argument that applies to all even numbers.
2 points. A good answer says something like “you can’t prove by examples” or “there are an in inite
number of even numbers and you can’t test them all.”
2.1. There are 27 = 128 meters. The numbers of meters of each type are found in row 7 of the
meruprastāra: (a) 1, (b) 7, (c) 21, (d) 35, (e) 35, (f) 21, (g) 7, (h) 1.
2 points
2.2. There are 27 = 128 patterns. The percentages of patterns of each type are found by dividing
row 7 of the meruprastāra by 128: (a) 0.78%, (b) 5.47%, (c) 16.41%, (d) 27.34%, (e) 27.34%,
(f) 16.41%, (g) 5.47%, (h) 0.78%.
2 points
2.3. SSSSSS has duration 1+1+1+1+1+1=6, so the answer is the 6th Hemachandra number: 13.
LLLLLL has duration 12 so the answer is the 12th Hemachandra number: 233.
2 points
2.4. The technique is to follow the procedure used in the worksheet on page 70. There are
1 pattern with duration 1 beat: 1
2 patterns with duration 2 beats: 1+1, 2
3 patterns with duration 3 beats: 1+2, 1+1+1, 2+1
6 patterns with duration 4 beats: 4, 1+1+2, 2+2, 1+2+1, 1+1+1+1, 2+1+1
This is the base case. The recursive rule is that each number in the sequence is the sum of the
numbers that are one, two, and four places before it. In mathematical notation, if Sn is the number
of patterns with n beats, then Sn = Sn−1 + Sn−2 + Sn−4 . The proof of this is similar to the proof of
Theorem 2.2. Using the recursive rule, the sequence is 1, 2, 3, 6, 10, 18, 31, 55, . . . so the answer is 55,
the eighth number.
3 points
2.5. The answers are Hemachandra/Fibonacci numbers: 1, 1, 2, 3, 5, 8, …
APPENDIX B. ASSIGNMENTS
84
1 point
2.6.
7
1
6
2
5
3
4
4
3
5
2
6
1
7
The irst number in the row is 1 (this is true for every n). Obtain the other numbers in the row by
succesively multiplying and dividing by the numbers you have written:
7
6
5
4
3
2
1
1 = 7; 7 · 2 = 21; 21 · 3 = 35; 35 · 4 = 35; 35 · 5 = 21; 21 · 6 = 7; 7 · 7 = 1. This tells us that the
seventh row is 1 7 21 35 35 21 7 1.
2 points
2.7. All the entries in these rows that are greater than 1 are divisible by the prime. All the entries in
the row 11 that are greater than 1 are divisible by 11.
1 point
2.8. The pattern you get when coloring the hexagons on the second page should be something like a
portion of Sierpinski’s Triangle:
Maximum of 2 bonus points
APPENDIX B. ASSIGNMENTS
85
Homework for Chapter 3
Directions: Write your homework neatly on the front side of the paper. Please do not write
on the back. Staple your papers together and cut off any messy edges.
3.1. Draw four beat class circles and notate the patterns
x...
.x..
..x.
...x
on the circles. How are the pictures related?
3.2. Suppose a piece of music has 12 beats per measure. Assume this is true for problems (a)-(h).
(a) Beat 0 is the irst downbeat. List four other beats that are downbeats.
(b) List four beats that belong to beat class 5.
(c) List four beats that have the same beat class as beat 270.
(d) To which beat class does beat 100 belong?
(e) To which beat class does beat 120 belong?
(f) Do beats 45 and 165 belong to the same beat class? Show work.
(g) Do beats 45 and 265 belong to the same beat class? Show work.
(h) Suppose A and B are beats and A − B = 36. Do beats A and B belong to the same beat class?
How do you know?
3.3. Determine whether the following are true or false. Show work.
(a) T/F: 357 ≡ 379 (mod 11)
(b) T/F: 30 ≡ 44 (mod 4)
3.4. Suppose N is a positive whole number and 0 ≡ 8 (mod N ). What are the possible values of N ?
(Hint: there are four possible values.)
3.5. List four numbers that are congruent to 0 (mod 2) and four numbers that are congruent to 1
(mod 2). What is the common name for the numbers that are congruent to 0 (mod 2)? The numbers
that are congruent to 1 (mod 2)?
3.6. Suppose A is a positive integer. Explain why A mod 10 equals the last digit of A.
3.7. Suppose there are eight beats per measure and you clap every sixth beat, starting with beat 0.
List the beat classes that you clap on, in the order you clap them. Do you clap on all the possible beat
classes?
3.8. Extra Credit. Suppose there are seven beats per measure. Use the tihai equation to show that
a 3-beat pattern and a 3-beat gap work for a tihai ending. Find a gap length that works for a 4-beat
pattern in a 7-beat measure. Check your work by writing the patterns in drum tab. Find another gap
length that works for a 4-beat pattern in a 7-beat measure.
APPENDIX B. ASSIGNMENTS
86
Solutions (Chapter 3)
3.1. You should draw four “four-hour clocks” with number 0 on top. Each circle has one blob. The
circles are related to each other by 90◦ rotations.
2 points
3.2.
(a) 12, 24, 36, 48 (or any numbers that are multiples of 12).
(b) 5, 17, 29, 41 (or any numbers that are 5 more than a multiple of 12).
(c) 270, 282, 294, 306, and any number equalling 270 plus some integer times 12.
(d) Since 100 ÷ 12 = 8r.4, the beat class is bc 4.
(e) Since 120 ÷ 12 = 10r.0, the beat class is bc 0.
(f) Yes, since (165 − 45)/12 is an integer.
(g) No, since (265 − 45)/12 is not an integer.
(h) Yes, because they are separated by three measures (alternately, because
(A − B)/n = 36/12 = 3 is an integer).
2 points for each part (16 points total)
3.3.
(a) True, because (379 − 357)/11 is an integer.
(b) False, because (44 − 30)/4 is not an integer.
2 points each
3.4. Any divisors of 8 work, because you need (8 − 0)/N to be an integer. The possible values are 1, 2,
4, and 8.
2 points
3.5. If a ≡ 0 (mod 2), then (a − 0)/2 is an integer, so a is divisible by 2. Examples include 2, 4, 6, 8, ….
If a ≡ 1 (mod 2), then (a − 1)/2 is an integer, so a − 1 is divisible by 2. Examples include 1, 3, 5, 7, ….
They are the even numbers and the odd numbers.
4 points
3.6. Since every positive integer A can be written as the sum of a multiple of 10 and the number’s last
digit, which is less than 10, the last digit is the remainder when A is divided by 10, and, by de inition,
the remainder equals A mod 10. For example 278 mod 10 = 8 because 278 = 270 + 8 = 27 · 10 + 8,
so 278 ÷ 10 = 27r.8. 2 points. The students don’t need to show an example.
3.7. The beats you clap on are 0, 6, 12, 18, 24, 30, …. Reduce them mod 8 to get the beat classes: 0, 6,
4, 2, 0, 6, …. You do not clap on all possible beat classes—just the even ones. 2 points
APPENDIX B. ASSIGNMENTS
87
3.8. (3 · 3 + 2 · 3) mod 7 = 15 mod 7 = 1. For a 4-beat pattern, (3 · 4 + 2 · 5) mod 7 = 22 mod 7 = 1.
In drum tab:
x x x x . . .|. . x x x x .|. . . . x x x|x
Another gap length that works is 12 beats, because (3 · 4 + 2 · 12) mod 7 = 36 mod 7 = 1. maximum
4 points
APPENDIX B. ASSIGNMENTS
88
Homework for Chapter 4
Directions: Write your homework neatly. Show your work clearly. Staple your papers
together and cut off any messy edges.
4.1. Sketch the periodic function whose period is
the graph shown.
4.2. Compute (use correct units in your answers):
(a) the frequency of a vibration whose period is 0.025 seconds, to the nearest whole number
(b) the period of a 1758 Hz vibration, to four decimal places
(c) the period of a vibration of 4 cycles per second
4.3. For each of the graphs shown,
(a) Sketch a cycle
(b) Estimate the period (in seconds, to four decimal places)
(c) Estimate the frequency (in Hz, to two decimal places).
(d) For the graph that is a sinusoid (there is one), write the formula in the form A sin(2πf t).
The numbers on top of the graph are seconds.
4.4. Find the amplitude, period, and frequency of the following waveforms and sketch the waveform.
Assume t is measured in seconds. Round to three decimal places, and be sure to use correct units in
your answer.
(a) 4 sin(65πt)
(b) 8 sin(250t)
APPENDIX B. ASSIGNMENTS
89
4.5. Open or record some music in Audacity, select a portion of the waveform that is a few seconds
long, and experiment with using the Effects menu to change it in various ways.
(a) Describe what each of the following options on the Effects menu do: Change Pitch, Change
Speed, and Change Tempo.
(b) What is the difference between Change Speed and Change Tempo?
(c) What does Reverse do?
(d) What does Invert do? (Hint: select a very small portion of the ile and zoom in so you can see
the waveform, then select Effects → Invert.)
4.6. Suppose that two sounds represented by the two waveforms shown are combined using the
superposition principle.
What is the name of the phenomenon
that happens at the beginning of the
combination wave? At the end? What will
you hear at the beginning and at the end?
4.7.
(a) Suppose a 800 Hz tone and a 804 Hz tone are played simultaneously. Describe what you would
hear, and make a sketch of the irst second of the resulting sound ile.
(b) Suppose a 800 Hz tone and a 1600 Hz tone are played simultaneously. Describe what you
would hear.
(c) Suppose you are wearing headphones and hear a 800 Hz tone in your right ear and a 804 Hz
tone in your left ear. Describe what you would hear.
4.8. Describe how the phenomenon of beats can be used to produce an 1000 Hz tone that pulses at a
rate of 3 times a second.
4.9. Suppose a string has a fundamental frequency of 200 Hz.
(a) Find the frequencies of the irst, second, and third harmonics and sketch the standing waves
corresponding to these harmonics.
(b) Suppose a 300 Hz fundamental frequency is produced by shortening the string. Find the ratio
of the length of the shortened string divided by length of the original string.
APPENDIX B. ASSIGNMENTS
90
Solutions (Chapter 4)
4.1.
4.2. (a) frequency = 1/period = 1/0.025 = 40 Hz. (b) period = 1/frequency = 1/1758 ≈ 0.0006
seconds. (c) frequency = 4, so period = 0.25 second.
4.3. (a) Here are the cycles:
(b) The periods appear to be about 0.0022 seconds and 0.0200 seconds. (c) The frequencies are
about 1/0.0022 ≈ 454.55 Hz and 1/0.0200 = 50.00 Hz. (d) The irst graph is a sinusoid. Its amplitude
is about 0.8, and its formula is y = 0.8 sin(909.09πt) ≈ 0.8 sin(2855.99t)
4.4. (a) Amplitude = 4; period = 2/65 = 0.031 seconds; frequency = 65/2 = 32.500 Hz. (b) Amplitude
= 8; period = 2π/250 = 0.008π ≈ 0.025 seconds; frequency = 250/(2π) ≈ 39.789 Hz. Graphs:
(a)
(b)
4.5. (a) Change pitch makes the sound higher or lower, change speed makes it higher and faster
or lower and slower, and change tempo makes it faster or slower. (b) Change speed changes both
the speed and pitch, while change tempo doesn’t change the pitch. (c) Reverse lips the selected
waveform over a vertical axis, so it plays backward. (d) Invert lips the waveform over the horizontal
axis. It doesn’t seem to change the sound.
4.6. Constructive interference happens at the beginning and destructive interference happens at the
end. You would hear a loud sound at the beginning and a very quiet sound or silence at the end.
4.7. (a) You hear a 802 Hz tone that has beats; it pulses at a rate of four times a second (4 Hz). 802 Hz
is the average of the two frequencies and 4 Hz is their difference. A picture is below (any picture that
has four beats per second is ine). (b) You don’t hear beats because the frequencies are not close.
Instead, you hear a 800 Hz tone and 1600 Hz tone separately. They form the interval of an octave.
(c) Even though 800 Hz and 804 Hz are close, you don’t hear beats because superposition does not
happen.
APPENDIX B. ASSIGNMENTS
'
0.10
1.0
'
0.20
'
0.30
'
0.40
91
'
0.50
'
0.60
'
0.70
'
0.80
'
0.90
'
1.(
0.5 -
-0.5 -1.0
4.8. If two sounds have frequencies 1001.5 Hz and 998.5 Hz, they will combine to produce a tone that
is 1000 Hz (their average) and has 3 Hz beats (their difference).
4.9. (a) Multiply the fundamental frequency (also known as the irst harmonic) by the number
of the harmonic. The irst harmonic is 200 Hz, the second harmonic is 400 Hz, and the third
harmonic is 600 Hz. The standing waves are represented by
,
. (b) The ratio is 2/3; it’s the reciprocal of the ratio of frequencies, 300/200.
, and
APPENDIX B. ASSIGNMENTS
92
Homework for Chapter 5
Directions: This assignment is due April 13. Write your homework neatly. Show your work
clearly. Staple your papers together and cut off any messy edges.
5.1. Determine which pairs of frequencies form the same interval: (a) 200 Hz and 300 Hz; (b) 600 Hz
and 400 Hz; (c) 300 Hz and 400 Hz
5.2. Find a pair of frequencies that forms the same interval as 200 Hz and 500 Hz.
5.3. Use the information in Section 5.1 to ind the name of the interval between 81 Hz and 108 Hz.
5.4. Find the frequency that is four octaves above 55 Hz.
5.5. Find the pitch that is four octaves above MIDI note number 30.
5.6. Find the measure, in semitones, of the interval between the eighth and ninth harmonics of a
vibrating string. Approximate your answer to two decimal places.
5.7. Using solfege syllables, list the pairs of notes in the major scale that are separated by four
semitones (for example, the pair (do, mi) is one).
5.8. Convert the following frequencies to pitches, rounding your answers to two decimal places:
(a) 50 Hz
(b) 100 Hz
5.9. Convert the following MIDI note numbers to frequencies, rounding your answers to two decimal
places:
(a) 110
(b) 61.98
5.10. List the irst eight pitches in the major scale that starts on pitch 50.
5.11. Find the pitch classes of the following MIDI note numbers
(a) Pitch 40
(b) The note that is 2 octaves above pitch 40
(c) Pitch 120
5.12. Name the pitch classes in the A♭ major scale.
5.13. Find the interval class of the interval from do to la.
APPENDIX B. ASSIGNMENTS
93
Solutions (Chapter 5)
5.1. (a) and (b) because they form a 3:2 ratio.
5.2. Any pair whose ratio is 5:2. For example, 1000 Hz and 400 Hz.
5.3. Since 108/81 = 4/3, the interval is a perfect fourth.
5.4. 55 · 24 = 880 Hz
5.5. 30 + (4 · 12) = 78
5.6. r = (9f )/(8f ) = 9/8, so s = 12 · log(9/8)/ log(2) ≈ 2.04 semitones
5.7. (do, mi), (fa, la), (sol, ti)
5.8. (a) 69 + 12 log(50/440)/ log(2) ≈ 31.35; (b) 69 + 12 log(100/440)/ log(2) ≈ 43.35. Alternately,
since 100 Hz is one octave higher than 50 Hz, add 12 to 31.35 to get 43.35.
5.9. (a) 440 · 2(110−69)/12 ≈ 4698.64 Hz; (b) 440 · 2(61.98−69)/12 ≈ 293.33 Hz
5.10. 50, 52, 54, 55, 57, 59, 61, 62
5.11. (a) 40 mod 12 = 4 (E); (b) 4 (E), because pitch classes repeat after an octave;
(c) 120 mod 12 = 0 (C)
5.12. 8, 10, 0, 1, 3, 5, 7 = A♭, B♭, C, D♭, E♭, F, G
5.13. 3, which is the shortest distance between any do and any la.
Appendix C
Review Problems
94
APPENDIX C. REVIEW PROBLEMS
95
Review Problems for Test 1
Words to the wise. Bring your calculator, pencils, and an eraser to class for the test.
Phones will not be allowed. You should review your homework, these problems, the text, and
your quizzes. You will be asked to show work on many of the test problems. Test problems
will be more dif icult than quizzes.
1. The YouTube star “Nora the Piano Cat” is a cat who likes to make noise on the piano. Are
these noises music? There is no single right answer, but you need to justify your answer using the
de initions of music we discussed in class.
2. Suppose there are seven beats per measure.
(a) List four downbeats.
(b) List four beats that belong to the same beat class as beat 100.
(c) Determine whether beats 60 and 81 are in the same beat class.
(d) Explain why, for every number A, beat A is in the same beat class as beat A + 21.
(e) To which beat class does beat 200 belong?
3. Suppose a sequence of numbers S1 , S2 , S3 , . . . is de ined recursively by
Sn = Sn−1 + Sn−3
If the irst four numbers in the sequence are 1, 1, 2, 3, what is the ifth number?
4. Evaluate (a) 60 mod 5 and (b) 60 mod 8.
5. Prove that the statement “if beat b is a downbeat, then b mod 4 = 0” is false.
6. Determine whether the following modular congruences are true. Give evidence for your answers.
(a) −16 ≡ 56 (mod 10)
7. The sixth row of the meruprastāra is:
(b) 101 ≡ 11 (mod 9)
1
6
15
20
15
6
1
(a) Show how it may be found using Bhaskara’s method.
(b) How many six-beat rhythms have three hits and three rests? How do you know?
(c) How many six-beat rhythms are there in total?
(d) What percentage of six-beat rhythms have three hits and three rests? Your answer should be
correct to two decimal places.
(e) Find the seventh row using any method.
APPENDIX C. REVIEW PROBLEMS
96
8. Explain the difference between the tactus, the tatum, a beat, and a beat class.
9. How many ive-beat rhythm patterns are there? How many of them have four hits and one rest?
10. Write all the three-beat rhythm patterns in drum tab.
11. Suppose there are four beats per measure. Write the pattern of playing the of beats as drum tab
and in circular notation.
12. If there are n beats per measure, beats a and b belong to the same beat class if
(a) a ≡ b (mod n)
(b) (b − a)/n is an integer.
(c) a mod n = b mod n.
(d) All of the above.
13. Suppose two drummers play the patterns xx... and x...x together. Find the resultant
rhythm.
14. What information does a waveform’s overall shape when you zoom out tell you?
15. True or False. A theorem is a mathematical statement that may or may not be true.
16. True or False. An axiom is a mathematical statement that has been proven to be true.
17. George and Ira Gershwin’s 1924 song “Fascinating Rhythm” features the catchy rhythm shown
below. The song is performed in “swing eighths,” meaning that the “&’s” happen later than the
midpoint between the numbered counts.
1 & 2 & 3 & 4 &|1 & 2 & 3 & 4 &|1 & 2 & 3 & 4 &|1 &
x x x x x x . x|x x x x x . x x|x x x x . x x x|x x
fascinating rhythm,
you’ve got me on the go,
fascinating rhythm,
I’m all a-quiver
The irst three phrases (“Fascinating rhythm, you’ve got me on the go, fascinating rhythm”) are the
beginning of a polyrhythm, though the pattern is broken in the fourth phrase. Which polyrhythm?
How long would the pattern have to continue to complete a whole number of measures?
18. Suppose poetic meters are made of short syllables (S) worth 1 beat, medium syllables (M), worth
2 beats, and long syllables (L), worth 3 beats. Write all the meters that have total duration equal to 4
beats.
APPENDIX C. REVIEW PROBLEMS
97
Solutions to Review Problems for Test 1
1. A good answer is well written and refers to at least one of the three de initions of music found in
Chapter 1:
(a) Music is the art of organizing sound in time.
(b) Music is the creation of a musician.
(c) Music is determined by the listener.
Questions you might consider are: Can a cat “organize” sound? Does Nora think she is a musician? Do
people hear Nora’s piano “playing” as music? There is no single right answer.
2.
(a) 0, 7, 14, 21, …
(b) 100, 107, 114, 121, …
(c) Since (60 − 81)/7 is an integer, they are in the same beat class.
(d) Beat A and beat A + 21 are separated by 21 beats, which equals three measures. Therefore,
they are in the same beat class.
(e) Calculate 200 mod 7: since 200 ÷ 7 = 28r.4, the beat class is bc 4.
3. The rule says that S5 = S5−1 + S5−3 = S4 + S2 = 4.
4. (a) 60 mod 5 = 0 because 60 ÷ 5 = 12 (with 0 remainder); (b) 60 mod 8 = 4 because 60 ÷ 8 = 7r.4.
The modulus is the remainder.
5. A counterexample is: Suppose there are 7 beats per measure. Then beat 7 is a downbeat, but
7 mod 4 = 3.
6.
(a) False, because (−16 − 56)/10 = −72/10, which is not an integer
(b) True, because (101 − 11)/9 = 10, which is an integer.
7.
6 5 4 3 2 1
The irst number in the row is 1 (this is true for every n). Obtain
1 2 3 4 5 6
the other numbers in the row by succesively multiplying and dividing by the numbers you have
written:
6
5
4
3
2
1
1 = 6; 6 · 2 = 15; 15 · 3 = 20; 20 · 4 = 15; 15 · 5 = 6; 6 · 6 = 1. This tells us that the sixth row is
1 6 15 20 15 6 1.
(a) Write:
(b) There are 20 because that is the middle entry in the sixth row.
APPENDIX C. REVIEW PROBLEMS
98
(c) There are 26 = 64 in total.
(d) The percentage is 20/64 = 31.25%.
(e) Either use Bhaskara’s method to ind the seventh row, or add the adjacent numbers in the sixth
row to get: 1 7 21 35 35 21 7 1
8. The tactus of a piece of music is the pulse where you normally tap. The tatum is the smallest level
of subdivision of the tactus. A beat is the length of time between pulses in the tatum. A beat class
indicates how long a beat falls after the beginning of measure, with beat class 0 corresponding to the
downbeat.
9. There are 25 = 32 ive-beat rhythm patterns. Five of them have four hits and one rest (this answer
comes from the ifth row of Pascal’s Triangle; you can also just list the patterns).
10. There are 23 = 8:
xxx
xx.
x.x
x..
.xx
.x.
..x
...
11. The of beat pattern is .x.x and its circular notation looks like a four-hour “clock” with blobs on
1 and 3 (0 is at the top).
12. (d) All of the above
13. xx..x
14. It tells you how the volume of the sound changes.
15. False. A theorem is a statement that has been proven to be true.
16. False. Axioms cannot be proven. They are accepted to be true without proof.
17. The repeating pattern xxxxxx. has seven beats, while the measure has eight beats. A 7-against-8
polyrhythm repeats after lcm(7, 8) = 56 beats, which equals seven measures.
18. SL, SSM, MM, SSSS, SMS, MSS, LS
APPENDIX C. REVIEW PROBLEMS
99
Review Problems for Test 2
Words to the wise. Bring your calculator, pencils, and an eraser to class for the test.
Phones will not be allowed. You should review your homework, these problems, the text,
and your quizzes. You will be asked to show work on some of the test problems or explain
how concepts are related to each other. The formula sheet on page 105 will be provided.
Test problems will be more dif icult than quizzes.
Section 4.1
Given the graph of a sound wave, you should be able to
1. Explain the relationship between periodic vibrations or waveforms and pitched sound.
2. Determine whether or not a graph is periodic.
3. If a graph is periodic, sketch a cycle and estimate its period in seconds.
Section 4.2
1. Convert between period and frequency, using appropriate units of measurement.
2. Distinguish between high and low sounds and know the relationship to frequency.
3. Determine whether a frequency is within the range of human hearing.
Section 4.3
1. Know the de initions of amplitude and envelope.
2. Be able to predict what its envelope tells us about a sound.
Section 4.4
1. Given an equation y = A sin(2πf t), be able to ind its amplitude, period, and frequency
and sketch a few cycles of its graph.
2. Write the formula for a sinusoid with a given amplitude and frequency, a given
amplitude and period, or a given graph.
3. Know what the variables in y = A sin(2πf t) measure.
Section 4.5
1. Understand the superposition principle and when it applies.
2. Know the de inition of mixing and its relationship to superposition.
3. Identify constructive and destructive interference visually or from a description.
APPENDIX C. REVIEW PROBLEMS
100
Section 4.6
1. Use the Beats Formula to predict how sounds of two different—but
close—frequencies combine to produce beats.
2. Given a tone with a given frequency that has beats of another frequency, describe
which two sounds combine to produce that tone.
3. Know when the Beats Formula applies and does not apply, its relationship to
superposition, and how it relates to the sounds of particular instruments.
Section 4.7
1. Know the formula f = an/(2L), what each variable stands for, and for which types of
instruments it applies.
2. Be able to list the frequencies of the harmonics of a wind or stringed instrument and
sketch the standing waves corresponding to those harmonics.
3. Know the relationship between length and frequency and be able to solve problems
relating the two.
4. Understand that tension and mass of a string affect frequency (you don’t have to know
the formula).
Section 5.1
1. Understand the relationship between ratios, frequencies, and intervals; be able to
name intervals when looking at a chart
2. Know the de inition of an octave and be able to ind frequencies that are octaves above
or below a given frequency.
Section 5.2
1. Be able to convert between ratios and semitones.
2. Understand what semitones measure, and why musicians prefer to measure intervals
in semitones rather than use ratios
Section 5.3
1. Know the 2212221 interval pattern for a major scale and be able to compute intervals
between any two notes of the scale. You do not have to memorize the names of the
notes.
Section 5.4
1. Know what a MIDI note number is and understand the relationship between
frequency and pitch.
2. Be able to convert between pitch and frequency.
3. Be able to ind the pitches in the major scale that starts on any given MIDI note number.
APPENDIX C. REVIEW PROBLEMS
101
Section 5.5
1. Be able to interpret the pitch class circle diagram, which will be on the formula sheet.
2. Be able to compute the pitch class number and letter of a pitch.
3. Be able to measure intervals between pitch classes.
4. Find the pitch classes in any major scale. Write them as numbers and letters.
Review Problems 1. Here are two sound waves. Identify which is periodic. For the periodic
wave, sketch a cycle, measure the period, and calculate the frequency. Use correct units in your
answer. Is the frequency within the range of human hearing?
2. Find the amplitude, period, and frequency of each sinusoid, and sketch a cycle:
(a) y = 3 sin(1000πt) (b) y = 2 sin(πt).
3. Write the formulas for the following sinusoids in the form y = A sin(2πf t), where t is measured in
seconds, with (a) amplitude = 5 and frequency =3 Hz (b) amplitude = 10 and period = 0.005 seconds
(c) the following waveform.
4. Suppose a 1000 Hz tone has the following envelope. Describe in words what you would hear.
4f
1.0
2.0
3.0
4.0
I
s.o
I
1.0
0.5 -
•
•
•
•
0.0-0.5 -1.0
5. What happens when two tones from different sources reach your right ear at the same time? What
is the relationship between the waveform(s) you hear and the waveform of each tone? Is the result
different if you are wearing headphones and hear one tone in the right ear and the other in the left?
6. Prove that the statement “if two sounds reach your ear at the same time, the result is a sound that
is louder than each of the two individual sounds” is false. Clearly explain your answer.
7. In each example, sketch the result of mixing the two waveforms in the third track. Does
constructive or destructive interference occur in either example?
APPENDIX C. REVIEW PROBLEMS
102
8. (a) Suppose a 75 Hz tone and a 78 Hz tone are played simultaneously. Describe what you would
hear, and make a sketch of the irst second of the resulting sound ile. (b) Now suppose a 75 Hz tone
and a 150 Hz tone are played simultaneously. Describe what you would hear.
9. Explain, in as much detail as possible, under what circumstances two sounds will combine to
produce a 75 Hz tone that pulses 8 times a second.
10. Suppose a string has a fundamental frequency of 400 Hz. Calculate the frequencies of the second
and third harmonics and sketch the standing waves corresponding to these harmonics.
11. The following statement is true: “the frequencies of the standing waves of a clarinet are equal
to (an)/(4L), where a is a constant relating to air pressure, n is an odd integer, and L is length.”
Suppose a clarinet has fundamental frequency of 100 Hz. If the length and a are not changed, ind the
frequencies of its second, third, and fourth standing waves. Clearly explain your work.
12. Suppose that when a string is 6 inches long, its fundamental frequency is 800 Hz. Assuming the
tension and material of the string are not changed, (a) If its length is decreased to 5 inches, what is the
fundamental frequency that results? (b) What length would produce a 1600 Hz sound? Alternately, if
the length is kept at 6 inches, what is another way to produce a 1600 Hz sound?
13. Find the following frequencies: (a) four octaves above 600 Hz (b) four octaves below 600 Hz.
14. The timbre of the sound of an instrument is determined by its (a) volume, (b) frequency,
(c) period, or (d) none of the above. Explain.
15. Explain why, when you choose any two pitches that are 6 semitones apart, calculate their
frequencies, and divide the larger frequency by the smaller, you always get the same answer.
16. Find, rounding to three decimal places: (a) the pitch of 500 Hz (b) the frequency of MIDI
note number 79 (c) the interval between 60 Hz and 75 Hz, in semitones (d) the ratio between the
frequencies of two sounds that are 7 semitones apart.
17. Determine the pitch class number and letter of pitch 38. Show your work. List the pitches of two
other members of this pitch class.
18. Find the pc intervals between (a) pc 2 and pc 9 (b) the third note (mi) and the ifth note (sol) of
any major scale (c) any pitch and the pitch that is 14 semitones above it
19. List the pitch classes in an E♭ major scale as both numbers and letters.
APPENDIX C. REVIEW PROBLEMS
103
Solutions to Review Problems for Test 2
1. The periodic wave is on the left. Its cycle looks like this:
and its period is 0.01 seconds.
Therefore, its frequency is 1/0.01 = 100 Hz. Since this is between 20 Hz and 20,000 Hz, it is within
the range of human hearing.
2. (a) amplitude = 3, freq = (1000π)/(2π) = 500 Hz, period = 1/500 = 0.002 seconds. (b) amplitude
= 2, freq = (π)/(2π) = 0.5 Hz, period = 1/0.5 = 2 seconds.
Sketches:
(a)
(b)
3. (a) y = A sin(2πf t) = 5 sin(2π · 3 · t) = 5 sin(6πt) (b) y = 10 sin(400πt) (c) amplitude = 0.5 and
frequency = 2 Hz, so y = A sin(2πf t) = 0.5 sin(2π2t) = 0.5 sin(4πt).
4. You would hear a 1000 Hz tone that fades in (that is, goes from silent to loud) from 0 to 1 seconds,
then holds a steady volume from 1 to 2 seconds, then fades out quickly. There is a short burst around
4 seconds, then the tone fades back in at 5 seconds.
5. Because of the superposition principle, the two waveforms combine; the resulting waveform is
the sum of the two original waveforms. Yes; if you are wearing headphones, superposition does not
occur, so you hear different tones in each ear.
6. The statement is false because it is possible for two sounds to combine, due to superposition, to
produce a quieter sound or even silence. For example, if a waveform is added to its inverse, the result
is silence. This is an example of destructive interference.
7. In the irst example, there is destructive interference: the two waves cancel each other out and the
sum is silence, represented by a horizontal line at y = 0. In the second example, there is constructive
interference; the sum of two identical waves is a wave with twice the amplitude of the original, which
looks like this:
8. (a) Assuming superposition happens, the two tones are close enough in frequency to
produce beats. You would hear a 76.5 Hz tone that pulses 3 times a second. The graph is
.
(b) The frequencies are too far apart to form beats. Instead, you hear two separate tones. Their ratio,
150:75, equals 2:1, so they form an octave.
APPENDIX C. REVIEW PROBLEMS
104
9. The Superposition Principle says that two sounds combine when they are at the same place at the
same time. The Beats Formula says that when two frequencies that are approximately equal combine,
they produce a tone whose frequency is their average; the tone has beats (pulses) whose frequency is
their difference. In this case, sounds with frequencies 71 Hz and 79 Hz produce the a 75 Hz tone with
8 Hz beats.
10. Multiply the fundamental frequency (also known as the irst harmonic) by the number of the
harmonic. The second harmonic is 800 Hz and the third harmonic is 1200 Hz. The standing waves
are represented by
and
.
11. For the fundamental frequency, n = 1, so (a · 1)/(4L) = 100. The next standing wave has n = 3,
because n must be an odd integer. Its frequency is (a · 3)/(4L), which equals 300 Hz because a and
L have not changed. For the same reason, the third and fourth standing waves produce frequencies
500 Hz and 700 Hz.
12. (a) Using the length-frequency ratio formula, 6/5 = f2 /800. Then f2 , the frequency that results, is
800 · 6/5 = 960 Hz; (b) Use the same formula with 6/L2 = 1600/800, so L2 = 3 inches; (c) 1600 Hz
is the frequency of the second harmonic, which is produced by lightly touching the string at its
midpoint.
13. (a) 600 · 24 = 9600 Hz; (b) 600/(24 ) = 37.5 Hz
14. The timbre of the sound of an instrument is determined by (d) none of the above. Timbre is the
quality of sound that differentiates the sounds of instruments, voices, etc. when their frequencies and
amplitudes are equal.
15. Saying that “two pitches are 6 semitones apart” is equivalent to saying that the interval between
them measures 6 semitones. Every interval corresponds to both a number of semitones (in pitch)
and a ratio of frequencies.√The semitones-to-ratio conversion formula says that if s = 6 semitones,
then r = 26/12 = 21/2 = 2 ≈ 1.414. Alternately, we could convert the pitches to frequencies irst,
then ind their ratio, which must also equal 21/2 .
16. (a) p = 69 + 12 log(500/440)/ log(2) ≈ 71.213; (b) f = 440 · 2(79−69)/12 ≈ 783.991 Hz;
(c) s = 12 log(75/60)/ log(2) ≈ 3.863 semitones; (d) r = 27/12 ≈ 1.498 (the ratio is 1.498:1, close to
3:2)
17. 38 mod 12 = 2 = D because 38 ÷ 12 = 3r.2. Two other D’s are 50 and 62.
18. (a) 5 semitones; (b) 3 semitones; (c) 2 semitones.
19. 3, 5, 7, 8, 10, 0, 2 = E♭, F, G, A♭, B♭, C, D
APPENDIX C. REVIEW PROBLEMS
105
Formula Sheet for Test 2
B
y = A sin(2πf t)
f = an/(2L)
As/Bf
L1 /L2 = f2 /f1
s = 12 log(r)/ log(2)
r = 2s/12
f = 440 · 2(p−69)/12
p = 69 + 12 log(f /440)/ log(2)
A
Cs/Df
0
Cs Ds
Df Ef
10
9
Gs/Af
11
C
1
Fs Gs As
Gf Af Bf
3
C D E F G A B
8
G
7
6
Fs/Gf
5
D
2
4
F
E
Ds/Ef
APPENDIX C. REVIEW PROBLEMS
106
Review for Final Exam (Part 1)
The test covers ALL the material we covered in class (Chapters 1-6). You will receive the
same formula sheet as for Test 2. These sample problems cover Chapters 4, 5, and 6. I’ll
write some more for Chapters 1, 2, and 3.
1. List the pitch classes in an E♭ major scale as both numbers and letters.
2. When an accordion player plays an A above middle C, four reeds sound simultaneously. Their
frequencies are
220 Hz
437 Hz
443 Hz
880 Hz
Discuss the relationship between these frequencies, and predict what sound will be produced.
3. Estimate the period and the frequency of the following wave, where the numbers on the horizontal
axis are seconds. Is the frequency within the range of human hearing?
4. Find the amplitude, frequency, and period of the waveform described by y = 6 sin(70πt). Round
your answers to three decimal places.
5. Explain how to ind the formula for a sinusoid if its graph is shown.
6. Suppose guitar string is 25 in long and has a irst harmonic of 200 Hz. (a) Calculate the frequency
of the third harmonic and sketch the standing wave corresponding to this harmonic. (b) If the string
is shortened to 20 in and the tension is not changed, calculate the fundamental frequency. (c) If the
tension is increased, the resulting sound is (i) louder in volume (ii) higher in frequency (iii) softer in
volume (iv) lower in frequency.
7. Suppose two sound waves combine due to the Superposition Principle. Which of the following are
added: (a) their amplitudes, (b) their frequencies, (c) their waveforms, (d) all of the above.
8. Find the pitch number corresponding to 1000 Hz. Round your answer to two decimal places.
9. Find the frequency of pitch 80. Round your answer to two decimal places.
10. Determine the pitch class number and letter of pitch 55. List the pitch numbers of two other
members of this pitch class.
11. Determine the number of semitones between 300 Hz and 200 Hz. Round your answer to two
decimal places.
APPENDIX C. REVIEW PROBLEMS
107
12. Determine the pc interval formed by pitch 50 and pitch 100.
13. Give an example of two different chords that are transpositionally equivalent.
14. How many different chords are transpositionally equivalent to { 0, 2, 4, 6, 8, 10 }? List them.
15. How many different rhythm patterns belong to the rhythm class of x..x..? List them.
16. The chord { 0, 3, 6, 9 }, or { C, E♭, G♭, A }, is called a diminished seventh chord. Give an example
of a different chord that belongs to the same chord type, written in numbers. How many different
diminished seventh chords are there?
17. Here’s a relationship between four-beat rhythm patterns: “Pattern A is similar to pattern B
if pattern A can be shifted some number of times to form pattern B.” Using this relationship, xx.x
is similar to .xxx because shifting the irst pattern twice gives the second pattern. Give another
example of two different four-beat patterns that are similar to each other.
18. Give a counterexample to the FALSE statement “If there are 6 beats per measure, every rhythm
class has 6 members.”
19. Explain why listing examples is not suf icient to prove a universal statement.
20. How many examples are needed to disprove a universal statement?
21. Find the following frequencies: (a) two octaves above 660 Hz (b) an octave below 660 Hz (c) a
frequency that forms a 3:2 ratio with 660 Hz
22. Determine whether the modular equivalences are true:
(a) 38 ≡ 138 (mod 7)
(b) −20 ≡ 29 (mod 7)
23. What is the relationship between the harmonics of a plucked string and the octave? the ifth (a
3:2 ratio)?
24. What is the difference between constructive and destructive interference?
25. Suppose a plucked string is touched lightly at its midpoint to create a node. Is the sound higher
or lower than the original string? By how much?
26. Under what circumstances will two audible, sinusoidal waveforms combine and produce a
waveform that sounds like (a) silence (b) a pulsation (c) a sound that you perceive as two different
tones (d) a louder tone that has a sinusoidal waveform and does not pulse.
APPENDIX C. REVIEW PROBLEMS
108
Review for Final Exam (Part 2)
The test covers ALL the material we covered in class (Chapters 1-6). You will receive the
same formula sheet as for Test 2. These sample problems cover Chapters 1, 2, and 3. In
addition to these problems, study the review sheet for Test 1.
27. Suppose there are eight beats per measure.
(a) List four downbeats.
(b) List four beats that belong to the same beat class as beat 50.
(c) Determine whether beats 50 and 100 are in the same beat class.
(d) Find the beat class of beat 32.
(e) If beat A is in beat class 2, to which beat class does beat A + 14 belong?
28. Suppose a sequence of numbers S1 , S2 , S3 , . . . is de ined recursively by
Sn = Sn−1 + Sn−3
T or F: The 100th number in the sequence is the sum of the 99th number and the 98th number.
29. Evaluate (a) 100 mod 7 and (b) 100 mod 5.
30. Find the sixth row of the meruprastāra using Bhaskara’s method. Check your work using the
addition algorithm, starting with the ifth row, which is 1 5 10 10 5 1.
31. Write the pattern xxx.. in circular notation and as a set of beat classes.
32. Suppose two drummers play the patterns x..x. and x...x together. Find the resultant
rhythm.
33. Suppose poetic meters are made of medium syllables (M), worth 2 beats, and long syllables (L),
worth 3 beats. Write all the meters that have total duration equal to 6 beats.
Appendix D
Quizzes
109
APPENDIX D. QUIZZES
110
Quiz 1. (1/19) Listen to the bouncing balls video, which has been edited to create interesting
rhythmic effects. Is this music? Explain.
Quiz 2. (1/26)
1. Find a counterexample to the statement: The sum of two odd numbers is an odd number.
2. Suppose a student was asked to prove the statement “The sum of three odd numbers is an odd
number.” The student wrote,
1+3+5=9
− 3 + 1 + (−3) = −5
7 + 7 + 7 = 21
“All odd numbers follow this pattern. You can add any three odd numbers and you will always
get an odd number.”
Explain why the student’s answer does not deserve full credit.
3. Explain, in your own words, the difference between an axiom and a theorem.
Quiz 3. (2/2) Suppose poetic meters can be made of syllables of length S (1 beat), L (2 beats), and E
(3 beats). Suppose Tn is the number of meters of total duration n beats.
1. Find T1 , T2 , and T3 .
2. If Tn = Tn−1 + Tn−2 + Tn−3 , ind T4 and T5 .
Quiz 4. (2/16; class was cancelled on 2/9)
1. De ine tactus and measure.
2. How many beats are required for a 6-against-10 pattern? OR How many beats are required for
an 8-against-10 pattern?
Quiz 5. (2/23)
1. Find 2017 mod 21
2. If there are 12 beats per measure, ind the beat class of beat 100.
OR
1. Find 2017 mod 22
2. If there are 8 beats per measure, ind the beat class of beat 55.
Quiz 6. (3/9)
1. Find the period and frequency of this waveform.
APPENDIX D. QUIZZES
111
2. Sketch a periodic waveform with a frequency of 2 Hz.
Quiz 7. (3/23)
1. The formula for a sinusoid is y = A sin(2πf t). Explain what y, A, π, f , and t stand for.
2. Write the formula for the following sinusoid.
by
powered
powered by
https://www.desmos.com/calculator
Quiz 8. (3/30) Suppose a 30 cm string has a fundamental frequency equal to 400 Hz. If the string is
shortened to 25 cm without changing its tension, what is its new fundamental frequency?
Quiz 9. (4/11) If two pitched sounds’ frequencies are in a 7:5 ratio, how far apart are they in
semitones?
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APPENDIX D. QUIZZES
112
Solutions to Quizzes
1. E♭ major scale: 3, 5, 7, 8, 10, 0, 2, or E♭, F, G, A♭, B♭, C, D
2. Since 437 Hz and 443 Hz are close (less than 20 Hz apart), they combine to produce a tone of
440 Hz (the average of 437 and 443) that pulses with beats at a rate of 6 Hz (the difference of 440
and 437). The other two tones are an octave below 440 and an octave above 880. So you hear three
A’s, and the middle one is pulsing six times a second.
3. The period appears to be 15.030 − 15.020 = 0.01 seconds. Since frequency = 1/period, the
frequency is 100 Hz, which is within the range of human hearing, which is 20 Hz to 20,000 Hz.
4. Amplitude = 6, frequency = 70π/(2π) = 35 Hz, period = 1/35 ≈ 0.029 seconds.
5. The formula is y = A sin(2πf t). The amplitude, A, is the height of the wave above the horizontal
axis. The frequency, f , is found by measuring the period of the wave (the time it takes to complete
one full cycle) and inding the reciprocal of the period, which equals the frequency.
6. (a) The frequency of the third harmonic is 3 · 200 = 600 Hz. The standing wave has two nodes
that divide the length of the string into three parts.
(b) Since L1 /L2 = f2 /f1 , we have
25/20 = f2 /200, so f2 = 250 Hz. (c) (ii)
7. (c) Only their waveforms are added.
8. 69 + 12 · log(1000/440)/ log(2) ≈ 83.21
9. 440 · 2(80−69)/12 ≈ 830.61 Hz
10. Since 80 ÷ 12 = 6 r.8, the pitch class number is 8 and the letter is G♯/A♭. Two other numbers of
this pitch class are 80 + 12 = 92 and 80 + 24 = 104.
11. s = 12 log(300/200)/ log(2) ≈ 7.02 semitones.
12. The pitches are separated by 50 semitones, which is equal to 2 semitones on the pitch class circle
because 50 mod 12 = 2. Alternately, you can solve the problem by inding the pitch classes of 50 and
100, which are 2 and 4, and inding the distance between them on the circle, which is 2 semitones.
13. Answers vary. Choose a random set of pitch classes to make a chord, choose some number of
semitones, then transpose the chord by adding that number to each of its pitch classes, remembering
to add on the pitch class circle. For example, suppose I chose { 3, 7, 8, 9 } as my chord and I
transposed it by 3 semitones to get { 6, 10, 11, 0 }. This shows that { 3, 7, 8, 9 } and { 6, 10, 11, 0
} are transpositionally equivalent. I recommend drawing the chords on the pc circle to see their
relationship.
14. There are two: { 0, 2, 4, 6, 8, 10 } and { 1, 3, 5, 7, 9, 11 }
15. There are only three different shifts: x..x.., ..x..x, and .x..x..
APPENDIX D. QUIZZES
113
16. A different chord is { 1, 4, 7, 10 } or { 2, 5, 8, 11 }. There are three different diminished seventh
chords.
17. Answers vary. Choose a four beat pattern and shift it to ind a pattern that is similar. For example,
if I chose x... I could say x... and ..x. are similar. However, since the patterns are supposed to
be different, I would not choose xxxx or .... because these patterns are the same as all their shifts.
18. The rhythm class of xxxxxx has only one member.
19. A universal statement is one that is true for in initely many examples. It is impossible to con irm
that the statement is true for every example because it would take forever. Therefore, logic must be
used.
20. Only one (called a counterexample)
21. (a) 660 × 4 = 2640 Hz = two octaves above 660 Hz; (b) 660/2 = 330 Hz = an octave below 660 Hz;
(c) Both 660 × 3/2 = 990 and 660 × 2/3 = 440 are correct.
22. (a) (138 − 38)/7 is not a whole number, so FALSE. (b) (29 − (−20))/7 is a whole number, so TRUE.
23. If f is the irst harmonic of a string, the irst, second, third, etc. harmonics are f , 2f , 3f , etc. The
ratio of the second harmonic to the irst harmonic is 2f : f , which equals 2 : 1, which is an octave.
The ratio of the third harmonic to the second harmonic is 3f : 2f , which equals 3 : 2, which is a ifth.
24. Constructive interference occurs when two waveforms that have the same frequency and phase
add due to the superposition principle. The result is a louder sound. Destructive interference
occurs when the waveforms cancel each other out because their peaks and valleys point in opposite
directions. The result is a quieter sound or silence.
25. The new sound is exactly an octave higher than the original sound, because it is the second
harmonic of the fundamental frequency and therefore twice the fundamental frequency.
26. The Superposition Principle says that, in order for waveforms to combine (add), they need to
be in the same place at the same time. (a) Silence is produced when the waveforms have the same
frequency but opposite phase, so that they are vibrating in opposite directions and cancel each other
out (this is destructive interference). (b) When the sinusoids’ frequencies are close—less than 20 Hz
apart—but not equal, they combine to produce beats, which sound like a single, pulsating tone.
(c) When the sinusoids’ frequencies are more than 20 Hz apart, you hear two different sounds at
the same time. (d) A louder sound is produced when the waveforms have the same frequency and
phase, so that they are vibrating in the same direction and reinforce each other (this is constructive
interference).
27.
(a) 0, 8, 16, 24, …
(b) 50, 58, 66, 74, …
(c) Since (100 − 50)/8 is not an integer, they are not in the same beat class. Alternately, 50 is in
beat class 2 and 100 is in bc 4.
APPENDIX D. QUIZZES
114
(d) Calculate 32 mod 8: since 32 ÷ 8 = 4 r.0, the beat class is bc 0.
(e) Answer: beat class 0, because 14 beats equals one full measure plus 6 beats. Alternately,
suppose A = 2. Then A + 14 = 16, which is in bc 0.
28. False. The 100th number is the sum of the 99th number and the 97th number, because if n = 100,
n − 1 = 99 and n − 3 = 97.
29. (a) 100 mod 7 = 2 because 100 ÷ 7 = 14 r.2; (b) 100 mod 5 = 0 because 100 ÷ 5 = 20 r.0. The
modulus is the remainder.
30. Write
6
1
5
2
4
3
3
4
2
5
1
6
The irst number in the row is 1 (this is true for every n). Obtain the other numbers in the row by
succesively multiplying and dividing by the numbers you have written:
6
5
4
3
2
1
1 = 6; 6 · 2 = 15; 15 · 3 = 20; 20 · 4 = 15; 15 · 5 = 6; 6 · 6 = 1. This tells us that the sixth row is
1 6 15 20 15 6 1. Adding the neighboring numbers in the ifth row gives the same answer.
31. The beat classes are { 0, 1, 2 }. In circular notation, you divide the circle into 5 parts and put blobs
on the top and the two to the right of it.
32. x..xx
33. MMM, LL
Solution to Quiz 1. A good answer is well written and refers to the de inition “Music is the art of
organizing sound in time.”
Solution to Quiz 2.
1. A counterexample is 1 + 1 = 2.
2. You can’t prove a universal statement by examples. There are in initely many odd numbers and
you can’t test them all, so you must use logic to prove the theorem.
3. An axiom is a mathematical statement that is accepted to be true but cannot be proven. A
theorem is a mathematical statement that has been proven to be true.
Solution to Quiz 3.
1. T1 = 1 because the only meter is S. T2 = 2 because the meters are SS and L. T3 = 4 because the
meters are SSS, SL, LS, and E.
2. T4 = T4−1 + T4−2 + T4−3 = T3 + T2 + T1 = 1 + 2 + 4 = 7. T5 = T4 + T3 + T2 = 7 + 4 + 2 = 13.
Solution to Quiz 4.
APPENDIX D. QUIZZES
115
1. The tactus is the basic pulse of a piece of music. It’s where you naturally tap your foot. A
measure is the primary grouping of beats in the tactus.
2. lcm(6, 10) = 30 OR lcm(8, 10) = 40.
Solution to Quiz 5.
1. 2017 mod 21 = 1. The procedure is to divide 2017 by 21 and ind the remainder. If you didn’t
get the right answer but did show that you knew to divide 2017 by 21, you get 1 point partial
credit.
2. If there are 12 beats per measure, ind the beat class of beat 100. The answer is 4 because 4 is
the remainder when 100 is divided by 12. Again, you get partial credit if you show some
progress on the problem but didn’t get the right answer.
OR
1. 2017 mod 22 = 15. The procedure is to divide 2017 by 22 and ind the remainder. If you didn’t
get the right answer but did show that you knew to divide 2017 by 22, you get 1 point partial
credit.
2. Find beat class of beat 55 if there are 8 beats per measure. The answer is 7 because that’s the
remainder when you divide 55 by 8. Again, you get partial credit if you show some progress on
the problem but didn’t get the right answer.
Solution to Quiz 6.
1. The period is 8 seconds and the frequency is 1/8 = 0.125 Hz. Note that frequency is not always
a whole number. “The frequency is 0.125 Hz” means that there are 0.125 cycles per
second—that is, one second equals one eighth of a cycle, as shown in the graph.
2. Any periodic waveform with a period of 0.5 seconds is correct.
Solution to Quiz 7.
1. y is displacement, A is amplitude, π is the number 3.14159 . . ., f is frequency in Hz, and t is
time in seconds.
2. The amplitude is 0.5 and the period is 0.2, so the frequency is 1/0.2 = 5 Hz. The formula is
y = 0.5 sin(2π · 5 · t) = 0.5 sin(10πt).
Solution to Quiz 8. Use the length-frequency ratio formula L1 /L2 = f2 /f1 . So
30
f2
=
25
400
which has solution f2 = 480 Hz.
Solution to Quiz 9. Use the ratio to semitones conversion formula, with r = 7/5:
s = 12 log(r)/ log(2) = 12 log(7/5)/ log(2) ≈ 5.825 semitones
Index
“22” (song), 28
“Doe, a deer”, 56
“Papa’s Got a Brand New Bag|hyperpage, 23
“Somewhere Over the Rainbow|hyperpage, 55
4’33|hyperpage, 1
chord, 65
chord type, 66
augmented triad, 66
diminished seventh, 66
dominant seventh, 66
major triad, 66
minor triad, 66
chords
transpositionally equivalent, 66
circular notation, 25
Clapping Music, 62
clarinet, 49
conjecture
de inition, 4
constructive interference, 39
counterexample
de inition, 4
cycle, 33
cyclic permutation, 62
accidental signs, 60
accordion, 44
algorithm
de inition, 4
amplitude, 36
Audacity, 2, 80
axiom
de inition, 3
Parallel postulate, 3
Peano axioms, 3
backbeat, 23
bar, 21
base case, 10
beat, 21, 25
beat class
de inition of, 26
downbeat, 26
formula, 26
theorem, 27
beat class circle, 26
beats, 43
formula, 43
bell, 49
binary number system, 8
binary pattern, 6
counting, 8
listing, 7
binaural beats, 45
Brown, James, 23
de Bruijn sequence, 14
de inition, 3
destructive interference, 39
Diophantine equation, 29
dissonance, 43
divisible, 80
downbeat, 23
drum tablature, 24
envelope, 36
even number, 3
example
de inition, 4
expanding mountain of jewels, 71
Fibonacci numbers, 9
irst harmonic, 47
lamenco, 62, 63
lat, 60
found sound, 2
Cage, John, 1
cent, 55
Chladni plates, 50
116
INDEX
fractal, 73
frequency, 34
converting to pitch, 58
fundamental frequency, 47
geometry
Euclidean, 3
spherical, 3
graph
directed, 19
groove, 23
harmonic spectrum, 52
Hemachandra numbers, 9
Hertz, 34
integer, 3
interference
constructive, 39
destructive, 39
interval, 53
between pitch classes, 60
converting ratio to semitones, 55
converting semitones to ratio, 56
de inition of, 53
octave, 53
major scale, 56
mathematics
as a science, 2
as a way of thinking, 2
de inition of, 2
maximal evenness, 29
measure, 21
meruprastāra, 12, 71
meter, 6
additive, 11
counting, 8
iambic pentameter, 6
listing, 7
naming, 13
microtiming, 23
microtones, 60
middle C, 57
MIDI note number, 57
mixing, 40
modes of vibration, 47
music
de inition of, 1
dif iculty of de ining, 1
musical instruments
117
bell, 49
Chladni plates, 50
singing bowl, 49
timpani, 49
node, 47
noise-cancelling headphones, 39
nonretrogradable, 11
note, 55
number
even, 3
integer, 3
odd, 4
prime, 82
number sequences
Padovan sequence, 11, 18
octave, 53
odd number, 4
off-beats, 23
on-beats, 23
overtones, 47
Padovan sequence, 11, 18
panning, 42
paradox, 3
parallel lines, 3
Parallel postulate, 3
Pascal’s triangle, 12, 71
pc (pitch class), 59
pc interval, 60
Peano axioms, 3
period, 32
periodic, 32
phase, 37
Pingala, 6
pitch, 57
converting to frequency, 58
higher or lower, 35
pitch class, 59
pitch class circle, 60
pitch class interval, 60
pitched sound, 32, 35
range of audible pitches, 34
polyrhythm, 28
Presley, Elvis, 29
prime number, 82
proof
de inition, 3
range of audible pitches, 34
INDEX
recursion, 10
recursive rule, 10
Reich, Steve, 62
relatively prime, 28
Resnik, Michael, 2
rest, 21
resultant rhythm, 24
rhythm class, 64
sampling, 2
sawtooth wave, 33
saxophone, 49
scale, 56
major, 56
semitone, 55
set
member, 62
sharp, 60
Sharp-Dressed Man, 36
Sierpinski’s Triangle, 73
singing bowl, 49
sinusoid, 37
solfege, 56
sound
difference from music, 1
found, 2
organized, 1
physics of, 32
pitched, 32, 35
spectrum, 51
harmonic, 52
standing wave, 47
statement, 3
de inition, 3
paradox, 3
universal, 4
Superposition Principle, 39
Swift, Taylor, 28
118
swing, 23
tactus, 21
tatum, 25
tempo, 21
tension of a string, 49
theorem
beat class, 27
de inition, 3
tihai, 29
timbre, 2, 51
time signature, 22
timeline, 26
timpani, 49
transposition, 65
tresillo, 29
tuning fork, 37
tuning pegs, 49
two- and three-dimensional wave equations,
49
two-against-three, 22
unison, 62
universal statement, 4
upbeat, 23
Varese, Edgar, 1
Volume, 36
wave equation, 46
higher dimensions, 49
in two or three dimensions, 49
waveform, 2
sawtooth, 33
white noise, 33
yamātārājabhānasalagā, 14
ZZ Top, 36