Homework 4 (Chapter 15)
Homework 4 (Chapter 15)
Wave and Particle Velocity Vector Drawing
A long string is stretched and its left end is oscillated upward and downward. Two points on the string are labeled A
and B.
Part A
⃗ and v ⃗ , to correctly represent the
Points A and B are indicated on the string. Orient the two vectors, v A
B
direction of the wave velocity at points A and B.
Rotate the given vectors to indicate the direction of the wave velocity at the indicated points.
Hint 1. Distinguishing between wave velocity and particle velocity
A wave is a collective disturbance that, typically, travels through some medium, in this case along a
string. The velocity of the individual particles of the medium are quite distinct from the velocity of the
wave as it passes through the medium. In fact, in a transverse wave such as a wave on a string, the
wave velocity and particle velocities are perpendicular.
Hint 2. Wave velocity
A wave on a stretched string travels away from the source of the wave along the length of the string.
ANSWER:
Correct
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Homework 4 (Chapter 15)
Part B
⃗ and v ⃗ to correctly represent the direction of the velocity of
At the instant shown, orient the given vectors v A
B
points A and B.
At each of the points A and B, rotate the given vector to indicate the direction of the velocity.
Hint 1. Distinguishing between wave velocity and particle velocity
A wave is a collective disturbance that, typically, travels through some medium, in this case along a
string. The velocity of the individual particles of the medium are quite distinct from the velocity of the
wave as it passes through the medium. In fact, in a transverse wave such as a wave on a string, the
wave velocity and particle velocities are perpendicular.
Hint 2. Determining velocity from a snapshot
The diagram represents the position of a small portion of the string at a specific instant of time: a
snapshot of the string at this time. Based on only a snapshot, you cannot determine the velocity of an
object, such as a point on the string. However, you also know that the left end of the string is the source
of the wave disturbance. From this information you can deduce what is about to happen to point A’s
position, and from this change in position deduce the direction of point A’s velocity (and similarly for
point B).
Hint 3. Find the change in point A’s position
Based on the location of the source of the wave (the left end of the string), will the wave crest to the
immediate left of point A soon raise or lower point A’s position?
ANSWER:
raise
lower
Hint 4. Find the change in point B’s position
Based on the location of the source of the wave (the left end of the string), will the wave trough to the
immediate left of point B soon raise or lower point B’s position?
ANSWER:
raise
lower
ANSWER:
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Homework 4 (Chapter 15)
Correct
Exercise 15.4
Ultrasound is the name given to frequencies above the human range of hearing, which is about 20000 Hz . Waves
above this frequency can be used to penetrate the body and to produce images by reflecting from surfaces. In a
typical ultrasound scan, the waves travel with a speed of 1500 m/s. For a good detailed image, the wavelength
should be no more than 1.0 mm.
Part A
What frequency is required?
ANSWER:
f
= 1.50×106 Hz Correct
± Speed of Rowing and Swimming
A displacement boat (generally a moving object held up by buoyancy of the water) generates a wave that travels
alongside at the speed of the boat with the first wave crest at the bow of the boat. Because the wavelength
increases with the boat's speed, above a certain speed the second crest appears astern of the boat, so that the
boat must travel uphill on the wave it creates. If it lacks the power to get over the top of this wave, its speed is
limited to roughly the speed of a wave with the length of the boat. Hence we have the following rule of thumb:
The maximum speed of a modestly powered boat is approximately equal to (actually, a little faster than) the wave
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Homework 4 (Chapter 15)
whose wavelength equals the length of the boat measured at the waterline.
Empirically, the propagation velocity of water waves is
−
−
v= √
g
k
,
where k is the wavenumber and g is the gravitational constant. In parts that ask for numerical answers, use 2
g = 9.81m/s .
Part A
Find the velocity of water waves v in terms of the wavelength λ .
Express the velocity in terms of λ and constants like g and π.
Hint 1. Find k in terms of λ
Express the wavenumber k in terms of λ .
Your answer may also include constants like π.
ANSWER:
k
2π
= λ
ANSWER:
−−
−
v
= √
gλ
2π
Correct
An eight­person shell (or "eight") is a light rowing craft approximately 17.6­m long with eight rowers and a coxswain
who steers and calls the stroke. Assume that the speed of the shell is limited according to the rule given in the
introduction.
Part B
Estimate the time T 8 it takes an "eight" to travel 2000 m.
Express the time numerically, in seconds, to three significant figures.
Hint 1. Estimate the speed of the boat
Find v 8 , the maximum speed of the eight.
Express your answer numerically, in meters/second, to three significant figures.
ANSWER:
v8
= 5.24 m/s 4/33
Homework 4 (Chapter 15)
ANSWER:
T8
= 382 s Correct
A good time for an eight over this distance (one typically used in US college crew races) is 390 s.
Part C
Estimate the time T 100 it takes a good human swimmer, 1.80­m tall, to swim 100 m. Assume that the speed of
the swimmer is limited by his or her height.
Express the time numerically, in seconds, to three significant figures.
Hint 1. Estimate the speed of the swimmer
Find the maximum speed v 100 of the swimmer, according to the rule given in the problem introduction.
Express your answer numerically, in meters/second, accurate to three significant figures.
ANSWER:
v 100
= 1.68 m/s ANSWER:
T 100
= 59.7 s Correct
As of 2002, the world record for a swimmer who kicks off the walls of a 25­m pool four times is 46.74 s for
the freestyle and 57.47 s for the breaststroke.
These examples show the power of simple rules. They give estimates valid at the 10% or 20% level, but
these estimates apply over a wide range of situations that are fundamentally the same (from the point of
view of the underlying physics).
Standard Expression for a Traveling Wave
Learning Goal:
To understand the standard formula for a sinusoidal traveling wave.
One formula for a wave with a y displacement (e.g., of a string) traveling in the x direction is
y(x, t) = A sin(kx − ωt)
.
All the questions in this problem refer to this formula and to the wave it describes.
Part A
Which of the following are independent variables?
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Homework 4 (Chapter 15)
Hint 1. What are independent variables?
Independent variables are those that are freely varied to control the value of the function. The
independent variables typically appear on the horizontal axis of a plot of the function.
ANSWER:
x
t
only
only
A
k
only
ω
x
only
only
and t ω
and t A
and k and ω
Correct
Part B
Which of the following are parameters that determine the characteristics of the wave?
Hint 1. What are parameters?
Parameters are constants that determine the characteristics of a particular function. For a wave these
include the amplitude, frequency, wavelength, and period of the wave.
ANSWER:
x
t
only
only
A
k
only
ω
x
only
only
and t ω
and t A
and k and ω
Correct
Part C
What is the phase ϕ(x, t) of the wave?
Express the phase in terms of one or more given variables ( A, k , x , t , and ω ) and any needed
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Homework 4 (Chapter 15)
constants like π.
Hint 1. Definition of phase
The phase is the argument of the trig function, which is expressed in radians.
ANSWER:
ϕ(x, t)
= kx − ωt
Correct
Part D
What is the wavelength λ of the wave?
Express the wavelength in terms of one or more given variables ( A, k , x, t, and ω) and any needed
constants like π.
Hint 1. Finding the wavelength
Consider the form of the wave at time t = 0 . The wave crosses the y axis, sloping upward at x = 0.
The wavelength is the x position at which the wave next crosses the y axis, sloping upward (i.e., the
length of one complete cycle of oscillation).
ANSWER:
λ
= 2π
k
Correct
Part E
What is the period T of this wave?
Express the period in terms of one or more given variables ( A, k , x, t, and ω) and any needed
constants like π.
ANSWER:
T
= 2π
ω
Correct
Part F
What is the speed of propagation v of this wave?
Express the speed of propagation in terms of one or more given variables ( A, k , x , t , and ω ) and any
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Homework 4 (Chapter 15)
needed constants like π.
Hint 1. How to find v
If you've done the previous parts of this problem, you have found the wavelength and the period of this
wave. The speed of propagation is a function of these two quantities: v = λ/T .
ANSWER:
v
= ω
k
Correct
Ant on a Tightrope
A large ant is standing on the middle of a circus tightrope that is stretched with tension T s . The rope has mass per
unit length μ . Wanting to shake the ant off the rope, a tightrope walker moves her foot up and down near the end of
the tightrope, generating a sinusoidal transverse wave of wavelength λ and amplitude A. Assume that the
magnitude of the acceleration due to gravity is g .
Part A
What is the minimum wave amplitude Amin such that the ant will become momentarily "weightless" at some
point as the wave passes underneath it? Assume that the mass of the ant is too small to have any effect on
the wave propagation.
Express the minimum wave amplitude in terms of T s , μ , λ , and g .
Hint 1. Weight and weightless
Weight is generally defined as being the equal and opposite force to the normal force. On a flat surface
in a static situation, the weight is equal to the force due to gravity acting on a mass.
"Weightless" is a more colloquial term meaning that if you stepped on a scale (e.g., in a falling elevator)
it would read zero. Think about what happens to the normal force in this situation.
Note that the force due to gravity does not change and would still be the same as when the elevator was
static.
Hint 2. How to approach the problem
In the context of this problem, when will the ant become "weightless"?
ANSWER:
When it has no net force acting on it
When the normal force of the string equals its weight
When the normal force of the string equals twice its weight
When the string has a downward acceleration of magnitude g
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Homework 4 (Chapter 15)
Correct
The ant will become "weightless" when the normal force between the string and the ant becomes
zero. Once this happens, gravity is the only force acting on the ant; the ant will become
"weightless" and will accelerate downward under the influence of gravity alone.
This same effect can be observed, to a lesser degree, in an elevator. When an elevator accelerates
downward, we feel lighter. If this downward acceleration is equal to g then we feel effectively
"weightless", since the normal force on us from the elevator floor goes to zero.
To solve this problem, you must determine the amplitude for which the maximum acceleration of a
point on the string is equal to −g .
Hint 3. Find the maximum acceleration of the string
Assume that the wave propagates as y(x, t)
acceleration a max of a point on the string?
= A sin(ωt − kx)
. What is the maximum downward
Express the maximum downward acceleration in terms of π and any quantities given in the
problem introduction.
Hint 1. How to approach the problem
Use the formula given for the displacement of the string to find the acceleration of the string as a
function of position and time. Then determine what the maximum value of this acceleration is. (At
some time, the bit of rope underneath the ant will have this maximum downward acceleration.)
Hint 2. Acceleration of a point on the string
Find the vertical acceleration a y (x, t) of an arbitrary point on the string as a function of time.
Express your answer in terms of A, ω, t, k , and x.
Hint 1. How to find the acceleration
Differentiate the expression given for y(x, t), the displacement of a point on the string,
twice.
Hint 2. The first derivative
Differentiate the given equation for the displacement of the string y(x, t) to find the
vertical velocity v y (x, t) of the rope.
Express your answer in terms of A, ω, t, k , and x.
ANSWER:
v y (x, t)
= ωAcos(ωt − kx)
ANSWER:
a y (x, t)
= 2
ω Asin(kx − ωt)
9/33
Homework 4 (Chapter 15)
Correct
Hint 3. Find the maximum downward acceleration
The maximum downward acceleration a max is the most negative possible value of a y (x, t) . As
the wave passes beneath the ant, at some time or another the ant will be at a point where the
acceleration of the string has this most negative value.
What is a max ?
Express your answer in terms of ω and quantities given in the problem introduction.
Hint 1. When will the acceleration reach its most negative value?
The most negative acceleration occurs when sin(ωt − kx)
= 1
.
ANSWER:
a max
= 2
−ω A
Hint 4. Determine ω in terms of given quantities
The angular frequency ω of the wave in the string was not given in the problem introduction. To
solve the problem, find an expression for ω in terms of given quantities.
Express the angular frequency in terms of T s , μ , λ , and π.
Hint 1. How to approach this question
Combine a general formula for ω, a relationship among frequency, wavelength, and
velocity, and a formula for the velocity of a wave on a string to find an expression for ω in
terms of quantities given in the problem introduction.
Hint 2. General formula for ω
The angular frequency of a wave is equal to 2π times the normal frequency: ω
= 2πf
.
Hint 3. Relationship among frequency, wavelength, and velocity
The frequency, wavelength, and velocity of a wave are related by v
= λf
.
Hint 4. Speed of a wave on a string
What is the speed v of any wave on the string described in the problem introduction?
ANSWER:
−−
−
v
= √
Ts
μ
ANSWER:
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Homework 4 (Chapter 15)
−−
−
ω
= √
Ts
2π
μ
λ
ANSWER:
a max
= −
Ts
μ
2
(
2π
λ
) A
Incorrect; Try Again; 9 attempts remaining
Hint 4. Putting it all together
Once you have an expression for the maximum acceleration of a point on the string a max , determine
what amplitude is required such that a max = −g. This will be the minimum amplitude Amin for which
the ant becomes weightless.
ANSWER:
2
A min
= μg(
λ
)
2π
Ts
Correct
Exercise 15.7
Transverse waves on a string have wave speed v = 8.00 m/s, amplitude A = 0.0700 m , and wavelength λ = 0.320 m . The waves travel in the ­x direction, and at t = 0 thex = 0 end of the string has its maximum
upward displacement.
Part A
Find the frequency of these waves.
ANSWER:
f
= 25.0 Hz Correct
Part B
Find the period of these waves.
ANSWER:
11/33
Homework 4 (Chapter 15)
4.00×10−2 s Correct
Part C
Find the wave number of these waves.
ANSWER:
k
= 19.6 rad/m Correct
Part D
Write a wave function describing the wave.
Express your answer in terms of the variables x and t.
ANSWER:
y(x, t)
= 0.07cos(19.6x + 157t)
m Correct
Part E
Find the transverse displacement of a particle at x
= 0.360 m
at time t
= 0.150 s
.
ANSWER:
y
= 4.95×10−2 m Correct
Part F
How much time must elapse from the instant in part E until the particle at x
upward displacement?
= 0.360 m
next has maximum
Express your answer using three significant figures.
ANSWER:
t
= 5.00×10−3 s 12/33
Homework 4 (Chapter 15)
Correct
Wave Propagation in a String of Varying Density
Consider a string of total length L , made up of three segments of equal length. The mass per unit length of the first
segment is μ , that of the second is 2μ , and that of the third μ/4 . The third segment is tied to a wall, and the string
is stretched by a force of magnitude T s applied to the first segment; T s is much greater than the total weight of the
string.
Part A
How long will it take a transverse wave to propagate from one end of the string to the other?
Express the time t in terms of L, μ , and T s .
Hint 1. How do the segments differ?
Consider each segment of the string seperately. Which quantities are the same and which are different
in each of the three segments?
Hint 1. General formula for the speed of a transverse wave on a string
Consider a section of string with mass per unit length μ under tension T s . The speed v with
which a transverse wave travels along this segment of the string is given by
−−
−
v= √
Ts
μ
.
ANSWER:
Both the wave velocity and the tension in the string are the same in all three segments.
Both the wave velocity and the tension are different in the three segments.
The wave velocity is the same in each segment but the tension is different.
The tension is the same in each segment but the wave velocity is different.
Hint 2. Example: speed in the second segment
Find v 2 , the speed of propagation of the wave in the second segment of the string described in the
problem introduction.
Express the wave speed on the second segment in terms of T s , μ , and any constants.
Hint 1. General formula for the speed of a transverse wave on a string
Consider a section of string with mass per unit length μ under tension T s . The speed v with
which a transverse wave travels along this segment of the string is given by
−−
−
v= √
Ts
μ
.
13/33
Homework 4 (Chapter 15)
ANSWER:
−−
−
v2
= √
Ts
2μ
Hint 3. Some math help
Recall that
a
b/c
=
a⋅c
b
and
−
−
−
−
−
−−−−
√m + √n ≠ √m + n
.
ANSWER:
t
= √
L
L
3
3
+
Ts
μ
√
Ts
2μ
L
3
+
√
4T s
μ
Correct
The changes in density along the string are sudden, and the wave will experience them as boundaries.
This will cause a fraction of the wave (energy, amplitude) to be reflected, while the rest is transmitted at
each boundary. Although this will not affect the time it takes for the wave to reach the end of the string
(thus it is not directly relevant to this question), the wave's amplitude will be reduced. Also, after the main
wave has arrived, we may observe later arrivals of waves that have reflected back and forth between the
boundaries before finally reaching the end of the string.
Two Velocities in a Traveling Wave
Wave motion is characterized by two velocities: the velocity with which the wave moves in the medium (e.g., air or
a string) and the velocity of the medium (the air or the string itself).
Consider a transverse wave traveling in a string. The mathematical form of the wave is
y(x, t) = A sin(kx − ωt)
.
Part A
Find the speed of propagation v p of this wave.
Express the velocity of propagation in terms of some or all of the variables A, k , and ω.
Hint 1. Perform an intermediate step
Note that the phase of the wave (kx − ωt), and therefore the displacement of the string, is equal to
zero at (x, t) = (0, 0) .
At what position x
= Δx
is the phase equal to zero a short time t
= Δt
later?
Express your answer in terms of Δt, ω, and k .
ANSWER:
14/33
Homework 4 (Chapter 15)
Δx
= ωΔt
k
ANSWER:
vp
= ω
k
Correct
Part B
Find the y velocity v y (x, t) of a point on the string as a function of x and t.
Express the y velocity in terms of ω, A, k , x, and t.
Hint 1. How to approach the problem
In the problem introduction, you are given an expression for y(x, t), the displacement of the string as a
function of x and t. To find the y velocity, take the partial derivative of y(x, t) with respect to time. That
is, take the time derivative of y(x, t) while treating x as a constant.
Hint 2. A helpful derivative
d
dt
sin(at + b) = a cos(at + b)
ANSWER:
v y (x, t)
= −ωAcos(kx − ωt)
Correct
Part C
Which of the following statements about v x (x, t) , the x component of the velocity of the string, is true?
Hint 1. How to approach this question
You are given a form for y(x, t). You are not given any information about Δx(x, t), but it is assumed
that each point on the string only moves in the y direction, i.e. Δx(x, t) = 0.
ANSWER:
15/33
Homework 4 (Chapter 15)
v x (x, t) = v p
v x (x, t) = v y (x, t)
v x (x, t)
has the same mathematical form as v y (x, t) but is 180 ∘ out of phase.
v x (x, t) = 0
Correct
So the wave moves in the x direction, even though the string does not. What this means is that even
though individual points on the string only move up and down, a given shape or pattern of points (in this
sinusoidal) will move to the right as time progresses.
Part D
Find the slope of the string ∂y(x,t)
∂x
as a function of position x and time t.
Express your answer in terms of A, k , ω, x, and t.
Hint 1. A helpful derivative
d
dx
cos(ax + b) = −a sin(ax + b)
ANSWER:
∂y(x,t)
∂x
= Akcos(kx − ωt)
Correct
Part E
Find the ratio of the y velocity of the string to the slope of the string calculated in the previous part.
Express your answer as a suitable combination of some of the variables ω, k , and v p .
ANSWER:
v y (x,t)
∂y(x,t)
= −
ω
k
∂x
16/33
Homework 4 (Chapter 15)
Correct
To understand why the ratio of the y velocity of the string to its slope is constant, draw the string with a
wave running along it at time t = 0 . In the vicinity of x = 0, the string is sloped upward. The bit of string
at position x = 0 moves downward as the wave moves forward. One­half cycle later, the string in the
vicinity of x = 0 will be sloped downward, and the string at position x = 0 will move upward as the wave
moves forward.
In general, if at some particular (x, t) the slope of the string is positive (∂y(x, t)/∂x > 0), that bit of
string will be moving downward (v y (x, t) < 0 ). If the slope at (x, t) is negative, that bit of string will be
moving upward. This explains why the sign of the ratio of string velocity to slope is always negative.
One way of understanding why the ratio has a constant magnitude is to observe that the more steeply the
string is sloped, the more quickly it will move up or down.
Exercise 15.25
A jet plane at take­off can produce sound of intensity 10.0 W/m2 at 30.6 m away. But you prefer the tranquil
2
sound of normal conversation, which is 1.0 μW/m . Assume that the plane behaves like a point source of sound.
Part A
What is the closest distance you should live from the airport runway to preserve your peace of mind?
Express your answer using two significant figures.
ANSWER:
r
= 97 km Correct
Part B
What intensity from the jet does your friend experience if she lives twice as far from the runway as you do?
Express your answer using two significant figures.
ANSWER:
I
= 0.25 μW/m2 Correct
Part C
What power of sound does the jet produce at take­off?
Express your answer using two significant figures.
ANSWER:
17/33
Homework 4 (Chapter 15)
P
= 1.2×105 W Correct
Exercise 15.22
A piano wire with mass 2.55 g and length 80.0 cm is stretched with a tension of 32.0 N . A wave with frequency
120 Hz and amplitude 1.50 mm travels along the wire.
Part A
Calculate the average power carried by the wave.
ANSWER:
P
= 0.204 W Correct
Part B
What happens to the average power if the wave amplitude is halved?
ANSWER:
P
= 5.11×10−2 W Correct
Exercise 15.34
Two pulses are moving in opposite directions at 1.0 cm/s on a taut string, as shown in the figure . Each square is
1.0 cm.
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Homework 4 (Chapter 15)
Part A
Sketch the shape of the string at the end of 6.0 s ;
ANSWER:
Part B
Sketch the shape of the string at the end of 7.0 s ;
ANSWER:
19/33
Homework 4 (Chapter 15)
Part C
Sketch the shape of the string at the end of 8.0 s .
ANSWER:
20/33
Homework 4 (Chapter 15)
Correct
Creating a Standing Wave
Learning Goal:
To see how two traveling waves of the same frequency create a standing wave.
Consider a traveling wave described by the formula
y (x, t) = A sin(kx − ωt)
1
.
This function might represent the lateral displacement of a string, a local electric field, the position of the surface of
a body of water, or any of a number of other physical manifestations of waves.
Part A
Which one of the following statements about the wave described in the problem introduction is correct?
ANSWER:
The wave is traveling in the +x direction.
The wave is traveling in the −x direction.
The wave is oscillating but not traveling.
The wave is traveling but not oscillating.
Correct
Part B
Which of the expressions given is a mathematical expression for a wave of the same amplitude that is traveling
in the opposite direction? At time t = 0 this new wave should have the same displacement as y1 (x, t) , the
wave described in the problem introduction.
ANSWER:
A cos(kx − ωt)
A cos(kx + ωt)
A sin(kx − ωt)
A sin(kx + ωt)
Correct
The principle of superposition states that if two functions each separately satisfy the wave equation, then the sum
(or difference) also satisfies the wave equation. This principle follows from the fact that every term in the wave
equation is linear in the amplitude of the wave.
Consider the sum of two waves y1 (x, t) + y2 (x, t) , where y1 (x, t) is the wave described in Part A and y2 (x, t) is
the wave described in Part B. These waves have been chosen so that their sum can be written as follows:
(x, t) =
(x)
(t)
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Homework 4 (Chapter 15)
y (x, t) = y (x)y (t)
s
e
t
.
This form is significant because ye (x) , called the envelope, depends only on position, and yt (t) depends only on
time. Traditionally, the time function is taken to be a trigonometric function with unit amplitude; that is, the overall
amplitude of the wave is written as part of ye (x) .
Part C
Find ye (x) and yt (t). Keep in mind that yt (t) should be a trigonometric function of unit amplitude.
Express your answers in terms of A, k , x, ω, and t. Separate the two functions with a comma.
Hint 1. A useful identity
A useful trigonometric identity for this problem is
sin(A + B) = sin(A) cos(B) + cos(A) sin(B)
.
Hint 2. Applying the identity
Since you really need an identity for sin(A − B), simply replace B by −B in the identity from Hint C.1,
keeping in mind that sin(−x) = − sin(x).
ANSWER:
y (x)
e
, yt (t) = 2Asin(kx), cos(ωt)
Correct
Part D
Which one of the following statements about the superposition wave ys (x, t) is correct?
ANSWER:
This wave is traveling in the +x direction.
This wave is traveling in the −x direction.
This wave is oscillating but not traveling.
This wave is traveling but not oscillating.
Correct
A wave that oscillates in place is called a standing wave. Because each part of the string oscillates with
the same phase, the wave does not appear to move left or right; rather, it oscillates up and down only.
Part E
At the position x
present)?
= 0
, what is the displacement of the string (assuming that the standing wave ys (x, t) is
Express your answer in terms of parameters given in the problem introduction.
ANSWER:
22/33
Homework 4 (Chapter 15)
y s (x = 0, t)
= 0
Correct
This could be a useful property of this standing wave, since it could represent a string tied to a post or
otherwise constrained at position x = 0. Such solutions will be important in treating normal modes that
arise when there are two such constraints.
Part F
At certain times, the string will be perfectly straight. Find the first time t1
> 0
when this is true.
Express t1 in terms of ω, k , and necessary constants.
Hint 1. How to approach the problem
The string can be straight only when cos(ωt) = 0 , for then ys (x, t) = 0 also (for all x). For any other
value of cos(ωt) , ys (x, t) will be a sinusoidal function of position x.
ANSWER:
t1
= π
2ω
Correct
Part G
From Part F we know that the string is perfectly straight at time t =
π
2ω
. Which of the following statements
does the string's being straight imply about the energy stored in the string?
a. There is no energy stored in the string: The string will remain straight for all subsequent times.
b. Energy will flow into the string, causing the standing wave to form at a later time.
π
c. Although the string is straight at time t =
, parts of the string have nonzero velocity.
2ω
Therefore, there is energy stored in the string.
d. The total mechanical energy in the string oscillates but is constant if averaged over a complete
cycle.
ANSWER:
a
b
c
d
Correct
± Why the Highest Piano Notes Have Short Strings
23/33
Homework 4 (Chapter 15)
The steel used for piano wire has a breaking (tensile) strength pT of about 3 × 10 9
3
7800 kg/m .
N/m
2
and a density ρ of Part A
What is the speed v of a wave traveling down such a wire if the wire is stretched to its breaking point?
Express the speed of the wave numerically, in meters per second, to the nearest integer.
Hint 1. Find the traveling wave speed in a stretched string
What is the speed v of a traveling wave in a stretched string which has mass per unit length μ and is
under tension FT ?
Answer in terms of FT and μ .
Hint 1. Use dimensional analysis
One way to answer this type of problem is to consider the dimensions of the quantities involved.
Force and mass per unit length have dimensions kg ⋅ m ⋅ s −2 and kg ⋅ m−1 respectively,
whereas we require an answer with dimensions of m ⋅ s −1 . There is only one combination of FT
and μ that has the correct dimensions.
Note that this method does not give the value of any constant multiplier of the relation.
ANSWER:
−
−
−
v
= √
FT
μ
Correct
Hint 2. Express v in terms of tensile strength and density
What is the speed v of a traveling wave in terms of the tensile strength pT and wire density ρ ?
Hint 1. How to approach the problem
Assume that the wire has cross­sectional area A. The tensile strength pT and tension in the wire
at the breaking point FT are related by pT A = FT . The density ρ and mass per unit length μ
are related by ρA = μ . Use these expressions to convert the speed in terms of FT and μ into a
speed in terms of pT and ρ .
ANSWER:
v
= −−
−
√
pT
ρ
ANSWER:
24/33
Homework 4 (Chapter 15)
v
= 620 m/s Correct
This is much less than the speed of sound in steel (5941
tensile strength is much less than the Young's modulus.
m/s
) because, unlike steel, in piano wire the
Part B
Imagine that the wire described in the problem introduction is used for the highest C on a piano (
C8 ≈ 4000 Hz ). If the wire is in tune when stretched to its breaking point, what must the vibrating length of
the wire be?
Express the length numerically, in centimeters, using three significant figures.
Hint 1. A length constraint given f and v
In Part A of this problem, you found the speed of traveling waves in the wire ( v ). You also know the
frequency of oscillation ( f ). What is the wavelength ( λ ) of the wave in the C8 piano string, in terms of
these quantities?
Express your answer in terms of v and f .
ANSWER:
λ
v
= f
Hint 2. Relationship between wavelength and string length
Consider the boundary conditions for a stretched piano wire: Both ends are fixed. If such a wire is
oscillating at its fundamental frequency (its first normal mode), the wavelength will not be equal to the
wire length. What is the wavelength of the first normal mode of a string of length L that is fixed at both
ends?
Express the wavelength in terms of L.
ANSWER:
λ
= 2L
ANSWER:
L
= 7.75 cm Correct
Normal Modes and Resonance Frequencies
Learning Goal:
25/33
Homework 4 (Chapter 15)
To understand the concept of normal modes of oscillation and to derive some properties of normal modes of waves
on a string.
A normal mode of a closed system is an oscillation of the system in which all parts oscillate at a single frequency.
In general there are an infinite number of such modes, each one with a distinctive frequency f i and associated
pattern of oscillation.
Consider an example of a system with normal modes: a string of length L held fixed at both ends, located at x = 0
and x = L. Assume that waves on this string propagate with speed v . The string extends in the x direction, and the
waves are transverse with displacement along the y direction.
In this problem, you will investigate the shape of the normal modes and then their frequency.
The normal modes of this system are products of trigonometric functions. (For linear systems, the time dependance
of a normal mode is always sinusoidal, but the spatial dependence need not be.) Specifically, for this system a
normal mode is described by
y i (x, t) = A i sin(2π
x
λi
) sin(2πf i t).
Part A
The string described in the problem introduction is oscillating in one of its normal modes. Which of the following
statements about the wave in the string is correct?
Hint 1. Normal mode constraints
The key constraint with normal modes is that there are two spatial boundary conditions, yi (0, t)
and yi (L, t) = 0, which correspond to the string being fixed at its two ends.
= 0
ANSWER:
The wave is traveling in the +x direction.
The wave is traveling in the ­x direction.
The wave will satisfy the given boundary conditions for any arbitrary wavelength λ i .
The wavelength λ i can have only certain specific values if the boundary conditions are to be
satisfied.
The wave does not satisfy the boundary condition yi (0; t)
= 0
.
Correct
Part B
Which of the following statements are true?
ANSWER:
The system can resonate at only certain resonance frequencies f i and the wavelength λ i must be
such that yi (0; t) = yi (L; t) = 0.
Ai
must be chosen so that the wave fits exactly on the string.
Any one of Ai or λ i or f i can be chosen to make the solution a normal mode.
26/33
Homework 4 (Chapter 15)
Correct
The key factor producing the normal modes is that there are two spatial boundary conditions, yi (0, t)
and yi (L, t) = 0, that are satisfied only for particular values of λ i .
= 0
Part C
Find the three longest wavelengths (call them λ 1 , λ 2 , and λ 3 ) that "fit" on the string, that is, those that satisfy
the boundary conditions at x = 0 and x = L. These longest wavelengths have the lowest frequencies.
Express the three wavelengths in terms of L. List them in decreasing order of length, separated by
commas.
Hint 1. How to approach the problem
The nodes of the wave occur where
sin(2π
x
λi
) = 0.
This equation is trivially satisfied at one end of the string (with x
= 0
), since sin(0)
= 0
.
The three largest wavelengths that satisfy this equation at the other end of the string (with x = L) are
given by 2πL/λ i = z i , where the z i are the three smallest, nonzero values of z that satisfy the
equation sin(z) = 0 .
Hint 2. Values of z that satisfy sin(z) = 0
The spatial part of the normal mode solution is a sine wave. Find the three smallest (nonzero) values of z (call them z 1 , z 2 , and z 3 ) that satisfy sin(z) = 0 .
Express the three nonzero values of z as multiples of π. List them in increasing order, separated
by commas.
ANSWER:
z1
, z 2 , z 3 = 1,2,3
Correct
Hint 3. Picture of the normal modes
Consider the lowest four modes of the string as shown in the figure.
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Homework 4 (Chapter 15)
The letter N is written over each of the nodes defined as places where the string does not move. (Note
that there are nodes in addition to those at the end of the string.) The letter A is written over the
antinodes, which are where the oscillation amplitude is maximum.
ANSWER:
λ1
, λ 2 , λ 3 = 2L, L,
2L
3
Correct
The procedure described here contains the same mathematics that leads to quantization in quantum
mechanics.
Part D
The frequency of each normal mode depends on the spatial part of the wave function, which is characterized by
its wavelength λ i .
Find the frequency f i of the ith normal mode.
Express f i in terms of its particular wavelength λ i and the speed of propagation of the wave v .
Hint 1. Propagation speed for standing waves
Your expression will involve v , the speed of propagation of a wave on the string. Of course, the normal
modes are standing waves and do not travel along the string the way that traveling waves do.
Nevertheless, the speed of wave propagation is a physical property that has a well­defined value that
happens to appear in the relationship between frequency and wavelength of normal modes.
Hint 2. Use what you know about traveling waves
The relationship between the wavelength and the frequency for standing waves is the same as that for
traveling waves and involves the speed of propagation v .
ANSWER:
f
i
= v
λi
28/33
Homework 4 (Chapter 15)
Correct
The frequencies f i are the only frequencies at which the system can oscillate. If the string is excited at
one of these resonance frequencies it will respond by oscillating in the pattern given by yi (x, t) , that is,
with wavelength λ i associated with the f i at which it is excited. In quantum mechanics these frequencies
are called the eigenfrequencies, which are equal to the energy of that mode divided by Planck's constant −34
h. In SI units, Planck's constant has the value h = 6.63 × 10
J ⋅ s.
Part E
Find the three lowest normal mode frequencies f 1 , f 2 , and f 3 .
Express the frequencies in terms of L, v , and any constants. List them in increasing order, separated by
commas.
ANSWER:
f
1
, f 2 , f 3 = v
2L
,
v
L
,
3v
2L
Correct
Note that, for the string, these frequencies are multiples of the lowest frequency. For this reason the
lowest frequency is called the fundamental and the higher frequencies are called harmonics of the
fundamental. When other physical approximations (for example, the stiffness of the string) are not valid,
the normal mode frequencies are not exactly harmonic, and they are called partials. In an acoustic piano,
the highest audible normal frequencies for a given string can be a significant fraction of a semitone sharper
than a simple integer multiple of the fundamental. Consequently, the fundamental frequencies of the lower
notes are deliberately tuned a bit flat so that their higher partials are closer in frequency to the higher
notes.
Exercise 15.40
A piano tuner stretches a steel piano wire with a tension of 800 N. The steel wire is 0.400 m long and has a mass
of 3.00 g.
Part A
What is the frequency of its fundamental mode of vibration?
ANSWER:
f
= 408 Hz Correct
Part B
What is the nurmber of the highest hamonic that could be heard by a person who is capable of hearing
frequencies up to 10000 Hz?
ANSWER:
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Homework 4 (Chapter 15)
N
= 24
Correct
Exercise 15.41
A thin, taut string tied at both ends and oscillating in its third harmonic has its shape described by the equation y(x, t) = (5.60 cm) sin[(0.0340 rad/cm)x] sin[(50.0 rad/s)t], where the origin is at the left end of the string,
the x­axis is along the string and the y­axis is perpendicular to the string.
Part A
Find the amplitude of the two traveling waves that make up this standing wave.
ANSWER:
A
= 2.80 cm Correct
Part B
What is the length of the string?
ANSWER:
L
= 277 cm Correct
Part C
Find the wavelength of the traveling waves.
ANSWER:
λ
= 185 cm Correct
Part D
Find the frequency of the traveling waves.
ANSWER:
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Homework 4 (Chapter 15)
f
= 7.96 Hz Correct
Part E
Find the period of the traveling waves.
ANSWER:
T
= 0.126 s Correct
Part F
Find the speed of the traveling waves.
ANSWER:
v
= 14.7 m/s Correct
Part G
Find the maximum transverse speed of a point on the string.
ANSWER:
v max
= 2.80 m/s Correct
Video Tutor: Out­of­Phase Speakers
First, launch the video below. You will be asked to use your knowledge of physics to predict the outcome of an
experiment. Then, close the video window and answer the question on the right. You can watch the video again at
any point.
31/33
Homework 4 (Chapter 15)
Part A
Two speakers face each other, and they each emit a sound of wavelength λ . One speaker is 180∘ out of phase
with respect to the other. If we separate the speakers by a distance 1.5λ , how far from the left­most speaker
should we place a microphone in order to pick up the loudest sound? Ignore reflections from nearby surfaces.
Select all that apply.
Hint 1. How to approach the problem.
Make a sketch of the situation. Note that the speakers represent antinodes of the resulting standing
wave pattern.
Will the sound be loudest at the nodes or antinodes of the standing wave pattern? Where will there be no
sound?
ANSWER:
0λ
λ
3
4
1
2
1
4
λ
λ
λ
Correct
Problem 15.50
A 1740­N irregular beam is hanging horizontally by its ends from the ceiling by two vertical wires (A and B), each
1.21 m long and weighing 0.370 N . The center of gravity of this beam is one­third of the way along the beam from
the end where wire A is attached. Ignore the effect of the weight of the wires on the tension in the wires.
Part A
If you pluck both strings at the same time at the beam, what is the time delay between the arrival of the two
pulses at the ceiling?
Express your answer with the appropriate units.
32/33
Homework 4 (Chapter 15)
ANSWER:
tB − tA
= 2.60 ms
Correct
Part B
Which pulse arrives first?
ANSWER:
in wire A
in wire B
Correct
The wave pulse travels faster in wire A, since that wire has the greater tension, so the pulse in wire A
arrives first.
Problem 15.71
A large rock that weighs 164.0 N is suspended from the lower end of a thin wire that is 3.00 m long. The density of
the rock is 3200 kg/m3 . The mass of the wire is small enough that its effect on the tension in the wire can be
neglected. The upper end of the wire is held fixed. When the rock is in air, the fundamental frequency for transverse
standing waves on the wire is 39.0 Hz . When the rock is totally submerged in a liquid, with the top of the rock just
below the surface, the fundamental frequency for the wire is 25.0 Hz .
Part A
What is the density of the liquid?
Express your answer with the appropriate units.
ANSWER:
ρ
= 1890 kg
m3
Correct
33/33