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MAT 117 Fall 2010
10/27/10 or 10/28/10
Worksheet 16
Section 8.1 & 8.2
Setting the Tone
This worksheet is loosely connected with sections 8.1 and 8.2,
but covers a variety of mathematical topics.
Preparation problems: Section 8.1, p. 359, #5, 7, 31; Section 8.2, p. 370, #1, 7, 15
Names:
In music, the frequency of a given sound determines the pitch. For example, a tone
with a frequency of 220 Hz (a Hertz is the number of cycles per second made by the
sound wave) makes the sound of the note called “A below middle C”. The next A, one
octave higher, has frequency 440 Hz. Doubling the frequency always results in a note
that is one octave higher. So doubling a pitch of 440 Hz to 880 Hz gives us yet another
A, this time two octaves higher than the original. Similarly, halving 220 Hz gives us
another A, this time with a frequency of 110 Hz, which is one octave below the original
note.
The tricky part is finding the frequencies of the notes in between. We have been
conditioned, by listening to music all our lives, to hear certain notes as ‘right’ or ‘in
tune’. It was Pythagoras (of the Pythagorean theorem) who discovered that when two
notes are played together and they create a pleasant sound, the ratio of their frequencies
could be written as the ratio of two small whole numbers. For example, a C with a
frequency of 264 Hz sounds very nice with a G of frequency 396 Hz. The fraction
396
264
3
simplifies to , so the ratio of their frequencies is 3:2. Generally, the smaller the number
2
in the simplified ratio, the more pleasant the sound.
The most common type of scale, a major scale, consists of the eight notes we
sometimes think of as do re mi fa so la ti do. In the key of C, these notes are C, D, E, F,
G, A, B, C’, or just one octave of the white keys on the piano. When counting from C to
D, we call that move one whole-step or two half-steps, since C# is between the two.
Since some of the notes in the major C scale do not have notes in between them, it is
easier to count in half-steps rather than in whole-steps. The move from C to G, for
example, is 7 half-steps.
1. If we use the ‘nice ratios’ that Pythagoras discovered, comparing the
frequencies of the different notes works fine as long as you stay in a particular
key. However, some of the frequencies do not match when comparing the same
note in different keys. Show this by completing Tables 1 and 2. The ratios
given in Table 1 are that of the frequencies between a given note and C. For
example, the ratio of the frequencies between D and C is
297
264
9
= . Table 2
8
gives the expected ratio between a given note and D. Round the frequencies to
the nearest 0.1 Hz.
Table 1
Note
C
D
E
F
G
A
B
C’
Number of half-steps
Ratio of frequencies Frequency (Hz)
above starting note
0
1:1
264
2
9:8
297
4
5:4
5
4:3
7
3:2
9
5:3
11
15:8
12
2:1
528
Table 2
Note
D
E
F#
G
A
B
C#
D
Number of half-steps
Ratio of frequencies Frequency (Hz)
above starting note
0
1:1
297
2
9:8
4
5:4
5
4:3
7
3:2
9
5:3
11
15:8
12
2:1
594
You can see that a problem immediately presents itself. What is called an E has a
frequency of 330 Hz in the key of C and a frequency of 334.1 Hz in the key of D. Being
in tune in one key would then sound out of tune in another. The problem continues for
every possible key—and there are twelve major scales alone! The impact of this on the
music world was significant. Musicians were not always able to play with each other. If
they started tuning with different notes they would sound out of tune. Because of this,
they would need to play every piece in the same key to always have the instrument in
tune. That turns out to be quite boring after awhile. It would also be torture for singers
who would sometimes be forced to sing above or below their range.
One solution to the tuning dilemma was to redefine what was considered ‘in tune’.
Instead of making all of the intervals nice ratios, it was decided around the year 1600 to
take the twelve notes of the scale and make the ratio of one note to the next the same. In
other words, take the frequency for A (110 Hz) and multiply it by some value, 𝑟, to get
the frequency for A#. Then take the frequency for A# and multiply it by the same 𝑟 to
get the same frequency for B. Continue this process until you get to the next A which is
220 Hz. This will give you a geometric sequence with 𝑟 as its constant ratio. The idea
didn’t hit with immediate acceptance. In fact, it took more than 200 years before it was
universally accepted. Bach helped matters along by writing a series of pieces in the early
1700’s for the newly tuned keyboard called the Well-Tempered Clavier. It wasn’t until
the 19th century that equal temperament was adopted by all countries and for all
instruments.
2. Let’s determine exactly how all twelve notes in a scale can be equally
tempered.
a. If we multiply by a constant ratio 𝑟 to get from a note to the note one halfstep higher, what should we multiply by to get from a note to the note one
octave, which is 12 half-steps, higher?
b. Doubling the frequency of any note (i.e., multiplying the frequency by 2)
will give the same note one octave higher. Use this information to
determine the exact value of 𝑟.
c. Using the constant ratio, 𝑟, determined in part b., complete Table 3 by
finding the frequencies for the missing notes from A (110 Hz) to A’ (220
Hz). Round to the nearest 0.1 Hz.
Table 3
Number of half-steps
Note Frequency (Hz)
above starting note
0
A
110
1
A#
2
B
3
C
4
C#
5
D
6
D#
7
E
8
F
9
F#
10
G
11
G#
12
A’
220
d. Find a function in which the number of half-steps, 𝑛, above the starting note,
A, is the input and the frequency of the note, 𝐹, is the output. This should be
an exponential function.
3. On different musical instruments various things are done to change the
frequency of the notes. One common method is to change the length of a tube
or pipe, as in organs or trombones, or to change the length of a string, as in
violins or guitars. If you take a string on a guitar and decrease its length by
half, the frequency of the note played is doubled. In this situation the
relationship between a note and its frequency is similar to that between a note
and the length of the string on which it is played. This relationship, however, is
backwards. As a note gets higher its frequency is given by a larger number, but
its string length is given by a smaller number. We measured the length of the A
string on a guitar and found it to be 61.2 centimeters. When a string is played
without reducing its length, the note is an A. When the length is reduced by
moving your finger on the fingerboard, the frequency of the tone increases.
When the string is half as long as the original, the frequency is doubled and the
first octave, or A’, is heard.
a. Halving the length of any guitar string will give the same note one octave
higher. Because of this, what would be the ratio of string lengths of a note
and the note one octave lower? Assume the ratio of string lengths, 𝑠,
between each of the 12 half-steps in each octave is the same. Determine the
exact value of 𝑠.
b. Using the constant ratio, 𝑠, determined in part a., complete Table 4 by
finding the missing lengths of the A string. They should form a geometric
sequence. If you were building a guitar these numbers would also determine
the positions of the frets. Frets are the metal ridges along the neck of a
guitar. When you push down on a string on the neck of a guitar, a fret will
‘cut’ the string off at just the right point to produce the note you want.
Table 4
Number of half-steps
Note String length (cm)
above starting note
0
A
61.2
1
A#
2
B
3
C
4
C#
5
D
6
D#
7
E
8
F
9
F#
10
G
11
G#
12
A’
30.6
c. Determine the function in which the number of half-steps above the starting
note, 𝑛, from Table 4, is the input and the length of the string, 𝐿, is the
output. This should be an exponential function.
4. The frequency of a specific guitar string is inversely proportional to its length.
Let’s show that this is true, at least for the A string on our guitar.
a. Combine your data from Tables 3 and 4 in Table 5.
Table 5
Number of half-steps
Note String length (cm) Frequency (Hz)
above starting note
0
A
61.2
110
1
A#
2
B
3
C
4
C#
5
D
6
D#
7
E
8
F
9
F#
10
G
11
G#
12
A’
30.6
220
b. If the frequency of the guitar string is inversely proportional to its length,
then the product of a string length and its corresponding frequency is a
constant. Using a few data values from Table 5, approximate that constant.
(Since rounding occurred in the determining the values in Table 5, you
won’t get exactly the same value for each product, but they should be close.)
c. Let 𝑓 represent your frequency function from question 2.d. and let 𝑙
represent your string length function from question 3.c. Which composition
of functions below would represent the function that would have the data for
the string length column of Table 5 as the input and the data in the frequency
column as the output? Explain your answer.
𝑓∘𝑙
𝑓 −1 ∘ 𝑙
𝑓 ∘ 𝑙 −1
𝑓 −1 ∘ 𝑙 −1
𝑙∘𝑓
𝑙−1 ∘ 𝑓
𝑙 ∘ 𝑓 −1
𝑙 −1 ∘ 𝑓 −1
d. Find the formula for the function composition that you determined was
correct in part c. Does your resulting function show that the frequency
of the guitar string is inversely proportional to the length? Explain.