Download 1 Simple harmonic motion related to circular motion

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PHYS 202
Notes on oscillations and waves
Crowell is rather terse in his treatment of simple harmonic motion, waves, and sound. I
would like to fill in some of the gaps.
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1.1
Simple harmonic motion related to circular motion
Mass and spring
The archetype of oscillating systems is a mass on a spring. For a system to oscillate, two
things are necessary: a mass or other source of inertia, and a restoring force. In the diagram
below, the mass slides on a frictionless surface, so no energy is lost.
Force
m
Equilibrium
position
xo
x
We know from previous studies that an ideal spring obeys the relation
F = −k(x − xo )
(1)
where F is the force exerted BY the spring in response to a displacement x − xo . (xo is
the relaxed position of the spring.) That is, the spring’s force is in a direction opposite to
the displacement direction. We say that it provides a restoring force, trying to “restore” the
mass (or whatever) to its equilibrium position. The equation above is called Hooke’s Law,
after Robert Hooke, a contemporary of Isaac Newton.
If you pull the mass to one side and let go, it oscillates. The spring pulls the mass to the
equilibrium position, but because of the mass’s inertia, it “overshoots” and ends up displaced
in the other direction. The spring reverses its force, stops, the mass, and accelerates it back to
the equilibrium position. The mass overshoots, and the cycle repeats. An object oscillating
in this way, when the spring obeys Hooke’s Law, is said to exhibit simple harmonic motion,
often abbreviated SHO.
1.2
Relation to circular motion
Let’s consider a system involving a mass sliding on a frictionless surface as before,
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but now suppose that a turntable of some kind is
placed over the mass, with its center above the
equilibrium position. The sun is directly overhead, and an object on the turntable casts a
shadow on the mass below. If we adjust the radius of the circle and the angular velocity of the
object, we can get the shadow to exactly follow
the motion of the oscillating mass. Therefore, we
can discribe the oscillation in terms of the radius
of the turntable and its angular velocity ω. The
linear motion of something undergoing simple harmonic motion is the projection, along one axis,
of an object moving in a circle.
LIGHT
shadow
ω
m
xo
Now let’s make this more quantitative.
1.3
Equations for simple harmonic motion
Take a look at the diagram on the next page. The circle has a radius A, so the mass below
can oscillate between positions −A and +A: we call A the amplitude.
For a particular angle θ (measured with reference
to the x-axis in our diagram), we can find the position x of the oscillating mass using the right triA
y
angle. The hypotenuse is A and the displacement
θ
is x, so x/A is the cosine of θ. Thus,
x
ω
x = A cos θ
Now, suppose that θ = 0 when time t is also zero.
Then by the definition of angular velocity, θ = ωt.
Therefore we can write
x = A cos (ωt)
m
(2)
−A
x=0
A
At this point we should note that the vertical projection of the dot on the circle would
also exhibit simple harmonic motion. The vertical displacement y in the diagram is equal to
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A sin θ, so we may also describe simple harmonic motion as
x = A sin (ωt)
(3)
Both expressions are used. Some textbooks prefer one, and some the other. I personally
prefer to use the sine function since that way, x = 0 at time t = 0. This seems simpler and
more elegant to me than using the cosine. But an engineer friend of mine says that no, if
you pull the mass to one side and let go at t = 0, x must have some nonzero value at t = 0.
So he prefers the cosine expression. It’s a religious argument.
1.4
Maximum velocity for SHO
The oscillating mass accelerates toward the equilibrium point, except that a = 0 right at
that point. Therefore when the mass is moving toward the point, it speeds up until it reaches
the equilibrium point. Then the maximum speed of the mass will occur at the equilibrium
point. To find this speed, we look at our diagram: the speed of the mass at the center of
travel is equal to the speed of the dot on the circle, either at the top or bottom of the circle.
That is because the dot is moving horizontally at that point: it has no vertical component
of velocity. This velocity is equal to the tangential velocity of the dot on the circle (i.e. its
speed) v = rω. Thus the maximum speed is
vmax = Aω
1.5
(4)
Maximum acceleration for SHO
The maximum acceleration occurs at the ends of travel, because the force of the spring is
greatest there. Looking at the circle, we see that this is when the dot crosses the x-axis. From
what we know about centripetal acceleration, we see that the dot is accelerating horizontally,
toward the center of the circle. This is the same direction as the acceleration of the oscillating
mass. This acceleration is simple ac = rω 2 , so the maximum accleration of the mass is
amax = Aω 2
(5)
Here, amax is a magnitude; it is a positive quantity. In a system, the sign of the actual
acceleration is opposite sign of the displacement. (See Equations 12 and 14, and the graphs
which follow them.)
1.6
Energy in a SHO system
A spring’s potential energy is
1
P E = kx2
2
as we learned last term. (One can readily show this by finding the area under the curve of
the spring’s force vs. displacement graph.) A mass oscillating on a spring has two kinds of
energy, KE and PE, and its total energy is the sum of these:
1
1
E = KE + P E = mv 2 + kx2
2
2
4
At the center, PE = 0, so all energy is kinetic. At the extremes of motion (x = ±A), KE
= 0, so all energy is potential. The total energy does not change if there is no friction. It is
clear, then, that
1 2
kA at the extremes, and
2
1 2
E =
mv
in the middle (equilibrium position)
2 max
E =
1.7
(6)
(7)
Finding the angular velocity of a mass/spring system
Let us exploit this knowledge to find yet another important relation. Look at Hooke’s law
(Equation 1): F = −kx. If the force on the oscillating mass is only due to the spring, we
may use Newton’s second law to write
Fnet = −kx = ma
Thus,
k
x
m
At the ends of travel, say when x = +A, the acceleration is maximum:
a=−
amax = −
(8)
k
A
m
But now let us go back and look at Equation 5: amax = −Aω 2 . Comparing these two
equations, we see that
k
= ω2
m
Therefore, the angular velocity is
r
k
(9)
ω=
m
Equations 8 and 9 are worth remembering. We will use them a lot this term. The
former, which relates acceleration and displacement, is often called the “equation of motion”
for a system.
1.8
Java applets which illustrates these concepts
This might be a good time to use a computer and view some animated applets which are
good illustrations of this discussion. I suggest the following URLs:
http://www.phy.ntnu.edu.tw/ntnujava/index.php?topic=148 This one also shows a graph
of position versus time, generating a sine-wave graph.
http://surendranath.tripod.com/Applets/Oscillations/SHM/SHMApplet.html This one
has better color and is simpler, but does not make a graph.
http://qbx6.ltu.edu/s schneider/physlets/main/mass spr vfric.shtml It does not
have a circle, but this is a fun applet since you can use the mouse to position the mass
on the spring, and you can add two types of friction to see what happens.
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1.9
The simple pendulum
A simple pendulum is a mass, considered as a point mass, hung ong the end of a string. The
mass is often called a “bob.” In the figure, the pendulum string of length ℓ is at an angle θ
from the vertical 1 . The pendulum bob is accelerating at a right angle to the string’s line.
Looking at the free-body diagram, we see that the component of the weight, mg, in this
direction is just mg sin θ. Therefore the torque on the pendulum is
τ = LF⊥ = Lmg sin θ
Putting this into the angular form of Newton’s second law gives us
τ = Iα
rF⊥ = Iα
ℓmg sin θ = (mℓ2 )α
g sin θ = Lα
T
θ
θ
mg
Now, let us restrict the motion to small angles, less than 10◦ . This is important, because
it allows us to use the small-angle approximation for the sine function. For small angles,
sin θ ≈ θ. (Of course, θ must be in radians.) Try this using your calculator. For angles less
than 0.2 radians, the approximation is not bad. At θ = 0.2 rad, sin θ is less than one percent
different from θ. For smaller angles, the approximation is better yet.
Using this approximation in our equation above, we can obtain
gθ = ℓα
g
α= θ
ℓ
Now, compare Equations 8 and 10. They have the same form. The only difference is
that the linear terms x and a are replaced with their angular counterparts, θ and α. But both
equations state that acceleration is proportional to displacement. In the case of Equation 8,
we found that the angular velocity
could be found from the constant of proportionality, k/m.
p
This is Equation 9: ω = k/m. Looking at our “angular” equation, we reason that the
angular velocity (or angular frequency) ω could be found the same way, using the constant
of proportionality:
r
g
Angular frequency of a simple pendulum
(10)
ω=
ℓ
1
We don’t use L, since that symbol is reserved for angular momentum in this material.
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1.9.1
Digression– another derivation
We could have done this another way. For small angles, the pendulum bob’s path does not
vary much from a straight horizontal line. The horizontal displacement is just x = ℓ sin θ.
Now, what is the restoring force? For small angles, the tension in the string does not change
much from the weight: T ≈ mg. Drawing a diagram, you should be able to convince yourself
that the horizontal component of the tension is
Fx = T sin θ
Since sin θ = x/ℓ and T ≈ mg, we can write
Fx = mg
x
ℓ
Now we apply Newton’s second law:
F = ma
x
mg = ma
ℓ
g
a= x
ℓ
Again, we have an equation in which acceleration is proportional to displacement. This looks
just like Equation 8 exceptpthat k/m is replaced by g/ℓ. We end up with the same expression
for angular velocity: ω = g/ℓ.
1.9.2
Keep this straight!
With a pendulum, there are two kinds of angular velocity.
1. The one we have just been dealing with. This is the constant angular velocity of a dot
on an imaginarypcircle, the kind you viewed in the Java applets. This is the ω which
we calculate as k/m. We use this to find the frequency, f = ω/(2π).
2. The angular velocity of the string holding up the pendulum bob. This angular velocity
varies, oscillating back and forth between two values. We sometimes need to know it,
such as cases where we calculate the kinetic energy of the pendulum.
1.10
The Physical Pendulum
The term “physical pendulum” can describe any pendulum in which an extended body is
swinging back and force on a pivot. Some textbooks restrict it to a rod pivoting about its
end, but the term is more general than that. For illustration, we will examine the case of a
rod.
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This rod pivots about its end. There are forces at the
pivot, Fx and Fy , but they cannot exert a torque about
the pivot, so they do not enter in to the analysis. The
torque is just
ℓ
τ = mg sin θ
2
We again use the small-angle approximation for the sine
function, so that
ℓ
τ ≈ mgθ
2
Fy
Fx
2
θ
θ
mg
We now need an “equation of motion,” which we obtain by writing Newton’s second law,
angular form:
τ = Iα
1
ℓ
mgθ = mℓ2 α
2
3
3g
α= θ
2ℓ
Using the analogy with Equation 8, as we did before, we find the angular frequency to be
r
3g
(11)
ω=
2ℓ
The frequency, as usual, would be ω/2π, or
1
f=
2π
r
3g
2ℓ
This, again, is not the only physical pendulum possible. You might imagine lots of other
configurations. For example, a hoop hanging on a peg and swinging back and force is a
physical pendulum. A square plate pivoting about one corner is another. A grandfather
clock’s physical pendulum has a large disk fastened near the bottom of a long rod. In each
of these cases, as long as you can figure out the rotational inertia I, you can find the angular
velocity ω and the frequency f .
1.11
Energy in a pendulum
For all oscillating mechanical systems, the total energy is the sum of kinetic and potential
energies. In a spring/mass system, the potential energy is that of the spring: (1/2)kx2 . With
a pendulum, the potential energy is gravitational in origin.
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Let us use the bottom of the pendulum’s swing
as our reference point for gravitational potential
energy: y = 0 when θ = 0. Then PE = mgh where
h is the height of the bob above our reference.
Referring to the diagram, we see that h = ℓ(1 −
cos θ). For a pendulum of total energy E, then,
we may write
cos( θ)
θ
1
h
E = mv 2 + mgℓ(1 − cos θ)
2
Example. Suppose we have a simple pendulum with string length ℓ = 1.0 m, bob mass
0.20 kg. When the bob is at the bottom of the swing, its velocity is 1.50 m/s. What is the
maximum angle to which the pendulum swings? (That is, what is the amplitude of θ?)
To solve this, we first find the total energy. This will allow us to find the maximum value
of h, from which we can find the maximum value of θ. The total energy is equal to the
kinetic energy at the bottom of the swing:
1
1
E = mv 2 = (0.20 kg)(1.50 m/s)2 = 0.225 J
2
2
At the maximum value of θ, all energy is potential, so
E = mgℓ(1 − cos θ)
E
1 − cos θ =
mgℓ
0.225 J
1 − cos θ =
(0.20 kg)(9.8 m/s2 )(1.0 m)
= 0.1148
cos θ = 1 − 0.1148 = 0.8852
θ = 28◦
We might note that this angle is quite a bit more than 10◦ , so the small-angle approximation
does not apply. That means that the restoring force (or torque) is not linearly proportional
to the angle, and this pendulum would not oscillate with purely simple harmonic motion.
It would be pretty close, but it’s graph of position vs. time would not be a “good” sine
wave. Even so, our calculation is still correct: our energy calculations do not depend on the
small-angle approximation.
1.12
Graphing simple harmonic motion
We already know that the general equation for simple harmonic motion is
x = A sin (ωt)
(12)
where A is the amplitude. Further, we know, from a previous argument, that the maximum
velocity of the oscillating mass is Aω. We might then expect that the velocity also obeys
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a “sine wave” shape, with an amplitude of Aω. However, it cannot be the same as the
displacement, since the displacement’s maximum (x = A) occurs at a time the velocity is at
a minimum (zero). So let’s go back to our knowledge from early in first term physics: the
velocity is the slope of the position graph. Plotting the sine wave and taking the slope at a
few points, we find indeed that we obtain a sinusoidal graph, with maxima occuring where
the sine wave is zero. This is a cosine wave, and we illustrate it in the next figure. The
velocity may be written
v = Aω cos (ωt)
(13)
Now, what about the acceleration? We have already shown that the maximum of the
acceleration is Aω 2 . Again, we expect some kind of sinusoidal shape to the graph. Taking the
slope of the velocity graph, we do obtain a sinusoidal shape. However, when the displacement
is at +A, the acceleration is negative, and when x = −A, the acceleration is positive. This
makes physical sense: the force on the mass is always opposite the sign of the displacement
from equilibrium, since the force is always trying to pull the mass back to equilibrium. The
acceleration is in the same direction as the net force, so the acceleration’s sign will always
be opposite that of x. Thus we find that
a = −Aω 2 sin (ωt)
These results are plotted on the next page.
(14)
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Plots of position, velocity, and acceleration for simple harmonic motion.
Here we use A = 1.0 m and ω = 1.0 rad/s.
1
A sin(ω t) where
ω = 1.0 rad/s
A = 1.0 m
Diplacement, m
0.5
0
−0.5
−1
0
2
4
6
8
time, seconds
10
12
14
10
12
14
10
12
14
Aω
Velocity, m/s
Aω cos(ωt)
0
−Aω
0
2
4
Acceleration, m/s/s
A ω2
6
8
time, seconds
A ω2sin(ωt)
0
A ω2
0
2
4
6
8
time, seconds
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1.13
Why is simple harmonic motion important?
Simple harmonic motion occurs whenever the restoring force is a linear function of displacement: that is, when F ∝ x. This is true for a spring. We found that it is very close to true
for a pendulum if the amplitude is small enough. This turns out to be true for almost any
system that has a stable equilibrium point.
We know that the potential energy function for a spring is (1/2)kx2 , which we found by
taking the area under the curve of the force function F = kx (recall, this is force on the
spring, so there is no negative sign.) Another way to describe conditions for SHO is that the
potential energy function is quadratic: it depends on x2 . For example, we have shown that
the potential energy function for a simple pendulum of length ℓ is mgℓ(1 − cos θ). The cosine
function can be described as an infinite series: cos θ = 1 − θ2 /2! + θ4 /4! − . . .. For small
angles, the terms in θ4 and above are so small that we can ignore them. Plugging (1 − θ2 /2)
into the potential energy function mgℓ(1 − cos θ) gives us
1
P E = mgℓθ2
2
Potential energy
This is a quadratic function and looks a lot like P E = (1/2)kx2 . So a simple pendulum has
the required quadratic PE for small displacements.
For an example less prosaic than a pendulum,
let’s look at the covalent bond between hydrogen and chloride in HCl. It can be described by
Distance r
ro
0
what is called a Lenard-Jones potential. This
potential has terms in 1/r6 and 1/r12 , where r is
Cl
H
the distance between nuclei. It is a strange funcr
tion indeed. We sketch the function at right. At
the equilibrium point ro , the potential energy is
a minimum and the force is zero. The force increases for displacements (r − ro ), away from the
equilibrium point.
The function itself is not quadratic in r − ro , but something interesting occurs if we
consider only small distances from the equilibrium point: the function is approxiximately
proportional to (r − ro )2 . That is, if you “blew up” the graph and looked at the region near
the bottom of the potential energy function, it is nearly a parabola: it is almost quadratic.
That means that for small oscillations, the system should exhibit simple harmonic motion.
Thus, we could use a “masses-on-a-spring” model for the molecule.
Now, we can’t really treat a molecule like this as little hard balls hooked by a spring.
Objects that small obey the rules of quantum mechanics, not Newtonian mechanics. (We
will get to quantum theory Spring term.) But there are some aspects of the behavior of
such molecules that do remind us of simple harmonic motion. The point of this discussion
is that any function that has a stable equilibrium point will be quadratic for small enough
displacements from equilibrium. Thus, we ought to see something akin to simple harmonic
motion showing up in a great many physical systems.
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2
Notes on Waves
2.1
Waves versus particles
So far in our course, we have dealt with particles or solid objects made of many particles.
Moving particles have kinetic energy and momentum, and particles which interact with each
other can be analyzed using the concept of potential energy. Waves also have energy and
momentum, even though waves are not, in themselves, particles.
A wave may be described as a disturbance propagating through a medium. For example,
water is the medium for surface water waves, and air is the usual medium for sound waves.
(Sound waves also may travel through liquids and solids, however.) A wave travels through
a medium, but the medium itself does not travel, except to vibrate back and forth a bit.
We should note here that the big exception to waves needing a medium is light. It is an
electromagnetic wave, and no medium is needed: it can travel through a vacuum. Radio
waves, microwaves, x-rays, and related waves are all the same kind of wave. The only
difference is wavelength and frequency.
Wave phenomena are everywhere in physics. A person who only studied particles would
miss out on roughly 2/3 of all phenomena in physics. Moreover, at the microscopic level (the
level of atoms, say), all objects have both wave and particle natures. You cannot understand
quantum theory without understanding waves.
Let us summarize some similarities and differences between waves and particles.
• A particle has a definite position in space, but a wave does not. We say that a wave is
not “localized.”
• A moving particle has energy and momentum. So does a wave.
• Two particles which collide with each other will interact strongly, bouncing off or
perhaps sticking together. Two waves colliding will simply pass through each other
and keep going on their respective ways.
• Particles can “bounce off” barriers put in their way. So can waves.
• Waves obey the superposition principle: if two waves (of the same kind, of course)
overlap, they form a resulting wave in which the displacements add together. For the
example of water waves, peaks adding to peaks will form a larger wave, whereas a
peak and a trough add to form a smaller, or perhaps zero, displacement. There is no
analogous principle for particles.
Often, we are interested in regularly oscillating waves, which we call periodic waves.
These waves are closely related to simple harmonic motion: a piece of the medium vibrates
back and forth with SHM. Music, optics, and other fields deal largely with period waves.
2.2
Describing waves.
Consider the profile of surface water waves traveling along with velocity v. When we say
“velocity v”, we mean that a point at a certain height of the wave travels at velocity v. For
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example, the peak of the wave does this. But you could pick any other point, and if you
follow it along, keeping the height of your point constant, it does the same thing. You might
say that the whole profile of the wave travels along at that velocity.
λ
v
We need to define another quantity to describe the wave, and that quantity is the wavelength, the peak-to-peak distance. This is usually denoted by the Greek letter λ, as in the
diagram. One could call it the distance for a complete cycle in space, much as the period is
a complete cycle in time.
In the figure, we put a “cork” or something else floating on the water. As the wave
passes, the cork goes up and down, with simple harmonic motion2 In one period T , the cork
will make a complete cycle of its motion. Further, on one period, the peak of the wave will
moving exactly one wavelength. Thus, the speed of the peak is just
v=
λ
T
(15)
Now, recall the relation between frequency and period. They are reciprocals: f = 1/T .
Substituting in Equation 15, we find
v = λf
(16)
We will use this relation often.
Example. Radio waves and light waves travel at the same velocity, which is the velocity
of light, the symbol of which is c. What is the frequency of a radio wave of wavelength 2.0
meters, and what is that for a light wave of wavelength 0.50 µm?
For the radio wave, we write
3.0×108 m/s
c
=
λ
2.0 m
= 1.5×108 Hz or 150 MHz
f=
Contrast this with the light wave:
3.0×108 m/s
c
=
λ
5.0×10−7 m
= 6.0×1014 Hz
f=
The frequency of the radio wave is less than the clock frequency of any modern desktop
computer. The frequency of the light wave, however, is unimaginably high.
2
Actually this cork would move in an ellipse, but we ignore that for simplicity. It is not central to this
argument, but is an interesting topic for further study on your own!
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2.3
Transverse and longitudinal waves
Waves are often classified as follows:
• Waves in which the medium’s displacement is at a right angle to the velocity are
called transverse waves. Surface water waves are transverse since the wave travels
horizontally, but the displacement is up and down. So are waves on a string, s-waves
of earthquakes, and light waves.
• Waves in which the displacement is parallel to the velocity are are called longitudinal
waves. The most familiar example is sound. Note that sound can travel in many media
besides air; for example water is a good carrier of sound. p-waves of earthquakes are
longitudinal, and we can produce such waves along a spring such as a “slinkly.”
Good computer demonstrations of longitudinal and transverse waves are found at these
sites:
http://surendranath.tripod.com/Applets/Waves/Lwave01/Lwave01Applet.html
http://www.phy.ntnu.edu.tw/ntnujava/index.php?topic=14.0
http://paws.kettering.edu/ drussell/Demos/waves/wavemotion.html
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2.4
Superposition and standing waves
When two waves (in the same medium) pass
through each other, the resulting wave is the sum
of the two waves where they overlap. For example,
look at the illustration below, showing a sequence
in which two waves pass through each other. These
are not periodic waves, but rather “pulses”. The
thin lines show the individual waves, while the thick
line shows the sum of the two waves. If the two
had the same amplitude but opposite signs, then
the fourth panel in the series, where the pulses are
centered over the same point, would show a flat line
for the sum.
For periodic waves, this superposition principle has
huge implications. If two waves of the same wavelength, but opposite directions, are superposed, the
sum of the two waves will produce a standing
wave. This is a wave which seems to stay in place.
Parts of the wave oscillate back and forth, while
other points do not move. The most familiar example of this is a plucked string, like a guitar string.
The ends do not move, while the middle oscillates
back and forth.
This would be a good point to look at some online
resources which demonstrate standing waves. First,
look at
http://physics.info/waves-standing/ .
This page illustrates standing waves for strings, but
also points out that standing waves can be twoand three-dimensional. Drum heads can exhibit
two-dimensional standing waves, while sound waves
in a cavity can produce three-dimensional standing
waves. The behavior of electrons in atoms can be
understood by treating the electrons as waves which
are standing waves in “allowed” states.
Now look at
http://www.physicsclassroom.com/Class/
waves/U10L4a.cfm
After looking at this page, follow the menu at
the top, taking the link labeled “Traveling waves
vs. standing waves.” This one has good animations of how standing waves are formed. Go to
all the other links as well: “Formation of standing
waves,” “Nodes and anti-nodes,” “Harmonics and
patterns,” and “Mathematics of standing waves.”
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2
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2.5
Standing waves in strings
A simple system for demonstrating standing waves is the bowed violin string. Such a string
is fixed on each end, so it must have a node on each end. When the string is bowed, it
ordinarily has an antinode in the middle, as shown in the diagram below.
A
N
N
L
In the diagram, “N” indicates a node and “A” an antinode. The solid and dotted lines
represent the positions of the string at two different times. It is clear that the distance L
is half a wavelength, not a complete wavelength. For a complete wavelength, one must go
through a complete cycle, as shown in this diagram:
A
N
N
A
N
λ
Here we show only one “snapshot” of the string in time, with the dotted line representing a
relaxed (non-moving) string. A complete cycle, then, goes N-A-N-A-N or A-N-A-N-A.
The lowest possible frequency, or the fundamental frequency, of a string fixed at both
ends therefore corresponds to a wavelength of 2L, twice the length of the string.
Suppose that the string’s length is 32 cm and the frequency is 660 Hz (the “E” of a violin
string.) We may find the speed of the wave along the string very easily:
v = λf = 2Lf = 2(0.32 m)(660 /s) = 422 m/s
This is faster than the speed of sound in air, but nontheless is realistic.
Now, suppose we bow the string while lightly touching it in the middle, producing the
first overtone or first harmonic (these are synonyms.) The node-antinode pattern looks like:
N
A
N
A
N
L
It is important to note that the speed of the wave along the string has not changed. We see
that the wavelength is now 1/2 of what it was before, so since
v = λf = constant
the frequency must double. Therefore the new frequency is twice the fundamental frequency:
2(660 Hz) = 1320 Hz. This is another “E”, but it is a musical octave above the first “E”.
Suppose now we touch the string 1/3 of the distance from the end, and bow it to produce
a pattern like this:
17
N
A
A
N
N
A
N
L
We see that the wavelength is now 1/3 of the original value, so the frequency must be 3 times
the fundamental frequency: 3(660 Hz) = 1980 Hz. This note is a “B”, a musical interval of
a fifth higher than the first harmonic.
We see that the pattern of frequencies is simply a series of integral multiples of the
fundamental frequency fo :
fo , 2fo , 3fo , 4fo , 5fo , . . .
The “naming” of these overtones is sometimes confusing. The first overtone has a frequency
2fo , and the second overtone has a frequency 3fo , etc.
Since strings must have nodes at the ends, where they are fixed, the overtone series is
simple. However, we now turn to a system where there may be antinodes at the ends.
2.6
Standing waves of sound in pipes
Suppose we have a pipe closed on both ends, and we are somehow able to inject sound of
some frequency into the pipe. If a standing wave is to be set up inside the pipe, it must
have a node at each end, because the air molecules are against the wall ends and cannot
move. There must be at least one antinode between two nodes. An antinode consists of air
molecules moving back-and-forth with some large amplitude. This is different than waves
on a string, since a string wave is transverse, and the motion is side-to-side. As in a string,
the lowest possible frequency will occur when there is only one antinode between the two
ends, as shown in this diagram.
Maximum motion back−and forth at antinode
N
A
N
L
String equivalent
A standing wave will only occur at resonance, a situation where the the reflected waves
“ build up” one another. In other words, the waves reflecting back and forth are intefering
constructively. If we vary the injected sound frequency, only specific frequencies will create
a standing wave. At hese resonant frequencies, energy is stored in the standing wave.
18
As an example, what would be the resonant frequency of a 1.0-meter-long tube: Looking
at the diagram above, we see that the wavelength is 2.0 m. The speed of sound at room
temperature is 343 m/s, so
v = λf
343 m/s = (2.0 m)f
f = 172 Hz
For the next highest frequency, we shorten the wavelength by adding another node and
antinode:
N
A
N
A
N
L
String equivalent
We see that in this case, λ = L, so
343 m/s
v
=
= 343Hz
λ
1.0 m
or twice the previous frequency. As in strings, the frequency of the first overtone is twice
that of the fundamental.
Therefore, in analogy to the case with strings, the overtone series for a tube closed on
both ends is just
f=
fo , 2fo , 3fo , 4fo , 5fo , . . .
Next we example a tube open on both ends. It is in some ways surprising that such a
tube can set up a standing wave. How can a wave be reflected off an open end? It’s easy
to visualize a wave reflecting off a wall, because high pressure will build up when the wave
hits, and “push back” on the air, causing reflection. Briefly, a similar thing happens at the
open end of a tube, except that it is the low pressure part of the wave which acts as a partial
vacuum to “suck” some air back into the tube, causing a wave to go in the other direction
along the tube.
A tube open on both ends would correspond to a string somehow held under tension, but
free to move vertically without friction. In practice this is hard to do, but we can imagine it
and model it. A Java applet here:
http://surendranath.tripod.com/Applets/Waves/TwaveRefTran/TwaveRefTranApplet.html
will allow you to observe reflection of waves on a string, in both the fixed-end and free-end
cases.
Since air is free to move at the open ends, there must be antinodes at both ends. The
fundamental mode would look as follows:
19
A
N
A
L
String equivalent
This has a wavelength equal to twice the length of the tube. Therefore the frequency is the
same as the case for a tube closed on both ends. The first overtone (or harmonic) would
look like this:
A
N
A
N
A
L
String equivalent
As in the case for the tube closed on both ends, the wavelength is simply the length of the
tube. It will therefore have the same frequency as for a tube closed on both ends: f = 2fo .
Indeed, the whole series of harmonics is the same for a tube open on both ends, as for a tube
closed on both ends.
The only other case to examine is a tube open on one end and closed on the other. There
must be a node at the closed end, and an antinode at the open end. The lowest frequency
case is illustrated here:
N
A
L = λ /4
String equivalent
20
Now, from node to antinode is a distance of a quarter of a wavelength. Therefore, for this
fundamental frequency, L = λ/4. The frequency is therefore
fo =
v
v
=
λ
4L
This is a frequency half the fundamental of a tube closed on both ends (or open on both
ends.) For large instruments employing pipes, such as an organ, it is advantageous to use
this kind of arrangement rather than the other. It takes half the length of pipe to produce a
given frequency. This is especially important for the low notes, which may have wavelengths
of more than 10 meters!
Now, what about the next overtone? We have to add another node and antinode, as
shown:
N
A
N
A
L = 3 λ /4
String equivalent
You can see that the wavelength has been divided by a factor of three. The length of the
tube covers 3 quarter-wavelengths: L = (3/4)λ or λ = 4L/3. Therefore the new frequency
is
v
v
v
=
=3
λ
4L/3
4L
= 3fo
f=
The new frequency is not just twice the fundamental, it is three times fo .
Let’s do one more to see the pattern:
N
A
N
A
L = 5 λ /4
String equivalent
N
A
21
This time, there are 5 quarter wavelengths inside the tube, so one wavelength is 4/5 the
length of the tube: λ = 4L/5. Then
v
v
v
=
=5
λ
4L/5
4L
= 5fo
f=
We now see the pattern: for a tube closed on one end and open on the other, the overtone
frequencies are odd multiples of the fundamental:
fo , 3fo , 5fo , 7fo , . . .
You need not remember every formula presented in this last discussion. However, you
should understand and be able to reproduced each of the diagrams. From these you can
figure out the frequency of the fundamental and those of the overtones, for any length string
or tube.
2.7
The decibel scale
Definition of a logarithm. For base 10 logs, we may state:
If y = 10x
then x = log y
so for example, the log of 104 is 4. The log of 5×104 is 4.7; this is because 5×104 =
104.7 .
Intensity is the amount of power passing through a cross-sectional area. The units are
then watt/m2 . How much power actually gets to the eardrum? To find this, we take
the intensity (energy per unit area) and multiply by the area of the eardrum, which is
about 0.6 square centimeters. At the threshold of hearing, the energy getting to the
eardrum is (10−12 W/m2 )(0.6×10−4 m2 ) = 6×10−17 watts! This is a really tiny amount
of power.
The decibel scale is just a convenient way of expressing a large range of intensities. It is
defined
β = 10 log
I
Iref
where β represents sound power level in dB, and Iref is a reference intensity, usually
the threshold of hearing which is 10−12 W/m2 . Therefore if I = 10−5 W/m2 , then
β = 10 log
I
Iref
10−5
= 10 log −12
10
= 10 log 107
= 70 dB
22
What if you increase the intensity by multiplying it by 10? Looking at the equations
above and substituting 10−6 for 10−5 , you can show yourself that the result would be
80 dB. Thus, when I is multiplied by 10, the sound power level increases by adding 10
dB. If I is multiplied by 1000, the sound power level increases by 30 dB.
A handy thing to remember is: if we double the intensity I, the sound power level goes
up 3.0 dB. (This is because 10 log (2) = 3.) So if we increase the intensity by 4, the
power level goes up 6 dB. Conversely if we cut the intensity in half, we subtract 3 dB
from the sound power level.
23
We may make a table:
Intensity in W/m2
1
10−1
10−2
10−3
10−4
10−5
10−6
10−7
10−8
10−9
10−10
10−11
10−12
Sound power level
120 dB
110 dB
100 dB
90 dB
80 dB
70 dB
60 dB
50 dB
40 dB
30 dB
20 dB
10 dB
0 dB
Usually, 0 dB is considered the threshold of hearing, and 120 dB is the threshold of
pain. Some examples of sound levels are:
• Near total silence – 0 dB
• A whisper – 15 dB
• Normal conversation – 60 dB
• Busy city traffic – 80 dB
• A lawnmower (or hair dryer near the head) – 90 dB
• Chainsaw, snowmobile – 100 dB
• A rock concert or a jet engine – 120 dB
• A gunshot or firecracker – 140 dB
Frequency dependence of apparent loudness. The human ear is most sensitive to sounds
at a frequency of about 1000 Hz. At lower or higher frequencies, the ear is less sensitive. This is often shown in a graph of “constant apparent loudness.” Such a graph is
shown on the next page.
The lines are contours of constant apparent loudness, and for this a unit called the phon
was invented. One phon is equal to one dB at a frequency of 1000 Hz. 50 phons, say,
is the same loudness at any frequency. But because the ear’s sensitivity changes with
frequency, the power level in dB to produce a loudness of 50 phons changes depending
on frequency. 50 phons is 50 dB at 1000 Hz, but (looking at the chart) at 100 Hz it
takes almost 60 dB of sound power to produce a loudness of 50 phons. At 30 Hz it
takes 83 dB to produce a 50 phon loudness.
24
Another scale for loudness is the sone scale. One sone is defined as 40 phons at any
frequency. Sones are supposed to be representative of apparent loudness. Suppose you
are listening to a sound at some level. It turns out that to turn up the level so that
this sound doubles in apparent loudness, you have to increase the power level by 10
dB, or 10 phons. Therefore 2.0 sones has double the apparent loudness of 1.0 sone and
is an increase of 10 phons, giving 50 phons. 4.0 sones is double the apparent loudness
again, adding another 10 phons, for a total of 60 phons. 8.0 sones would double the
loudness again, and so on. The chart has both phons and sones.
Noice-induced hearing loss. Exposure of the ear to sounds at 85 dB or above will cause
damage to the hair cells in the cochlea, resulting in permanent loss of sensitivity to
sound. Usually we lose the ability to hear the very high frequencies first, but abuse of
the ear will result in hearing loss over all frequencies. (High frequencies are especially
effective at damaging the hair cells.) At 85 dB, exposure to hours of sound will cause
damage. At higher intensities, the amount of time to cause damage shortens. At 100
dB, less than a minute will cause some damage. At a rock concert (110 to 120 dB) the
damage starts in seconds.
No American over 15 years of age has escaped damage to his/her hearing. The addiction
of the young to very loud music and the constant use of MP3 players with cheap ear
buds has produced a large population with damaged hearing. The stereotype of old
rockers with bad hearing is sadly accurate.
25
2.8
The speed of a wave on a string
A stringed instrument, such as a guitar or violin, has strings of different thicknesses. The
lower strings (lower in pitch or frequency, that is) are thicker than the higher strings: they
have a larger mass for a given length. If you think about a string vibrating back and forth
as a standing wave, it seems clear that a more massive string would vibrate more slowly,
because more mass is harder to accelerate.
Also, it is well known to string players that tightening the string (increasing the tension)
raises the frequency. Increasing the tension increases the restoring force, which will increase
the rate at which the mass accelerates.
These observations can be quantified by a relationship between tension F , mass per unit
length µ, and the velocity of the wave on a string:
s
F
v=
(17)
µ
For example, consider a cello string 70 cm long. The high “A” vibrates at 220 Hz.
Therefore the speed of the wave on that string is
v = λf = 2(0.70 m)(220 Hz) = 308 m/s
This string has a mass of approximately 4 grams. We can then use the relationship of
Equation 17 to find the tension necessary for this string to be in tune. The mass per unit
length is
kg
4 grams
= 5.7×10−3 kg/m
µ=
0.70 m 1000 g
Rearranging Equation 17,
F = v 2 µ = (308 m/s)2 (5.7×10−3 kg/m)
= 540 N
This rather large force is typical for stringed instruments. The necks of such instruments
must be fairly strong, then. Many instruments of the violin family were made more than 200
years ago. The necks of these instruments must be replaced every 80 years or so, because
the wood of a violin neck simply cannot sustain 200 pounds of force over decades of time
without bending.