Chapter 13: Oscillations and Waves
Answers and Solutions
1.
Picture the Problem: A bird flaps its wings with a constant frequency.
Strategy: The period is the time for one cycle. The frequency is the inverse of the period, or the number of cycles per
second. In this problem the number of flaps per second is given, so that is the frequency.
Solution: 1. Identify the flaps per second as frequency:
f =
5 flaps 5 cycles
=
= 5 Hz
second second
1
1
=
= 0.2 s
f 5 Hz
Insight: The period and frequency are inverses of each other; when the period is less than 1 second, the frequency will
be greater than 1 hertz.
T=
2. Invert the frequency to determine the period:
2.
Picture the Problem: A bird flaps its wings with a constant frequency.
Strategy: The period is the time for one cycle. The frequency is the inverse of the period, or the number of cycles per
second. In this problem the frequency is doubled.
f old
Tnew 1 f new
f
1
=
= old =
=
⇒ Tnew = 12 Told
Told
1 f old
f new 2 f old
2
Solution: 1. Use a ratio to show the period is halved:
1
1
=
= 0.1 s
f 10 Hz
which is half of answer to problem 1.
Insight: Likewise, when the frequency is cut in half, the period is doubled.
T=
2. Invert the frequency to determine the period:
3.
Picture the Problem: A violin string oscillates 440 times per second.
Strategy: The period is the time for one cycle. The frequency is the inverse of the period, or the number of cycles per
second. In this problem the frequency is increased.
Solution: Use a ratio to show the period is decreased by a
factor of n if the frequency is increased by a factor of n:
f old
Tnew 1 f new
f
1
=
= old =
=
Told
1 f old
f new n f old
n
⇒ Tnew = Told n
Insight: Likewise, when the frequency is decreased, the period is increased.
4.
Picture the Problem: A tennis ball is hit back and forth between two players with a constant period.
Strategy: The period is the time for one cycle. In this case, a cycle is complete when the ball goes to the second player
and then returns to the first player. The frequency is the inverse of the period.
T = 2 Δt = 2 (2.3 s ) = 4.6 s
Solution: 1. The period is twice the time required to
go to the other player:
1
1
=
= 0.22 Hz
T 4.6 s
Insight: Likewise, when the frequency is decreased, the period is increased.
f =
2. Invert the period to determine the frequency:
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13 – 1
Chapter 13: Oscillations and Waves
5.
Pearson Physics by James S. Walker
Picture the Problem: This is a follow-up question to Guided Example 13.2. An air-track cart attached to a spring
oscillates once every 2.4 s. At t = 0 the cart is released from rest at a distance of 0.10 m from its equilibrium position.
Strategy: The amplitude of the motion is the maximum distance the cart travels away from the equilibrium point.
The cart cannot travel farther from equilibrium than its initial release point, so the initial distance equals the amplitude.
Solution: The amplitude is the initial distance from equilibrium or 0.10 m.
Insight: If the cart were pushed from its initial position, rather than released from rest, the amplitude of its motion
would be greater than 0.10 m.
6.
Picture the Problem: This is a follow-up question to Guided Example 13.2. An air-track cart attached to a spring
oscillates once every 2.4 s. At t = 0 the cart is released from rest at a distance of 0.10 m from its equilibrium position.
Strategy: The amplitude of the motion is the maximum distance the cart travels away from the equilibrium point.
The cart cannot travel farther from equilibrium than its initial release point, so the initial distance equals the amplitude.
Solution: Because the amplitude is 0.10 m, when the cart arrives at −0.10 m, it will be at its maximum
distance from equilibrium and its velocity will be zero.
Insight: If the cart were pushed from its initial position, rather than released from rest, the amplitude of its motion
would be greater than 0.10 m, and it would have a nonzero speed at a position of −0.10 m.
7.
Picture the Problem: A mass moves back and forth in simple harmonic motion with an amplitude of 0.25 m and a
period of 1.2 s
Strategy: The amplitude of the motion is the maximum distance the mass travels away from the equilibrium point.
In a single cycle the object travels a distance of four amplitudes: two amplitudes from maximum to minimum, and
two amplitudes during its return.
Solution: Because the period is 1.2 s, after 2.4 s two periods have elapsed. The mass will therefore have
traveled eight amplitudes during that time. The total distance is thus 8×0.25 m = 2.0 m.
Insight: It will take another three periods or 3.6 s for the cart to travel 3.0 meters more.
8.
Picture the Problem: A mass oscillates on a spring, and then its mass is increased by a factor of 4.
Strategy: Use the formula for the period of a mass on a spring to set up a ratio to predict the new period.
Tnew
=
Told
Solution: Use a ratio to show the period is doubled:
2π
2π
mnew
k
mold
k
=
mnew
=
mold
4 mold
mold
= 2 ⇒ Tnew = 2Told
Insight: A larger mass has more inertia, so that the spring force produces a smaller acceleration, and the mass oscillates
with a longer period.
9.
Picture the Problem: A mass oscillates on a spring, and then its mass and the spring constant are both doubled.
Strategy: Use the formula for the period of a mass on a spring to set up a ratio to predict the new period.
Solution: Use a ratio to show
the period will stay the same:
Tnew
=
Told
2π
2π
mnew
knew
mold
kold
=
mnew kold
=
mold knew
2 mold
kold
mold 2 kold
= 1 ⇒ Tnew = Told
Insight: The larger mass has more inertia, but the stiffer spring produces a larger force, so that the two changes cancel
each other and the period of oscillation remains unchanged.
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13 – 2
Chapter 13: Oscillations and Waves
Pearson Physics by James S. Walker
10. Picture the Problem: A mass oscillates on a spring.
Strategy: Use the formula for the period of a mass on a spring together with the given mass and frequency to find the
spring constant.
f =
Solution: 1. Rearrange the formula:
1
1
1
=
=
T 2π m k 2π
k
m
1 k
4π 2 m
2
2
4π m f = k
f2 =
2. Solve for k:
k = 4π 2 (0.32 kg )(1.6 Hz ) = 32 N/m
2
3. Substitute the numerical values:
Insight: A spring constant that is 4 times larger (128 N/m) would be required to double the frequency to 3.2 Hz.
11. Picture the Problem: A mass oscillates on a spring.
Strategy: Use the formula for the period of a mass on a spring together with the given spring constant and period to
find the mass.
m
k
Solution: 1. Rearrange the formula:
T = 2π
2. Solve for m:
4π 2 m = k T 2
m=
⇒ T 2 = 4π 2
m
k
kT2
4π 2
k T 2 (22 N/m )(0.95 s )
m=
=
= 0.50 kg
4π 2
4π 2
2
3. Substitute the numerical values:
Insight: A mass that is four times larger (2.0 kg) would be required to double the period to 1.9 s.
12. Picture the Problem: A mass oscillates on a spring, and then its mass is increased while the spring constant is
decreased.
Strategy: Use the formula for the period of a mass on a spring to set up a ratio that can be used to predict the new
period.
Solution: Suppose the mass
will double and the spring
constant will be cut in half, and
use a ratio to show the period
will increase:
Tnew
=
Told
2π
2π
mnew
knew
mold
kold
=
mnew kold
=
mold knew
2 mold
mold
kold
1
2
kold
= 4 ⇒ Tnew = 2Told
Insight: The larger mass has more inertia, and the weaker spring produces a smaller force, so that the two changes each
result in a smaller acceleration and a longer period of oscillation.
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13 – 3
Chapter 13: Oscillations and Waves
Pearson Physics by James S. Walker
13. Picture the Problem: A mass hangs on a vertical spring, stretching it a distance d while it is at rest. The mass is then
allowed to oscillate about equilibrium.
Strategy: The weight of the mass stretches the spring the distance d. We can use that information to determine the
spring constant. We can also use the given oscillation and time information to determine the period of oscillation. Once
we know these two things we can use the formula for the period of a mass on a spring to find the stretch distance d.

Solution: 1. The weight of the
∑ F = kd − mg = ma = 0
mass stretches the spring:
kd = mg ⇒ k = mg d
56.7 s
Δt
=
= 0.556 s
N 102 oscillations
2. Find the oscillation period T:
T=
3. Now use the period formula to find d:
T = 2π
m
= 2π
k
m
d
= 2π
g
mg d
T 2 = 4π 2
)(
(
d
g
9.81 m s 2 0.556 s
gT 2
=
d=
4π 2
4π 2
)
2
= 0.0768 m = 7.68 cm
Insight: A longer period would imply a weaker spring constant because the force that produces the acceleration would
be smaller. A weaker spring would thus stretch a longer distance d when the mass is attached to it.
14. Picture the Problem: A mass moves back and forth in simple harmonic motion.
Strategy: The frequency and period are inversely proportional.
Solution: The frequency and period of an oscillator are related by f = 1 T . Therefore, if the period T is
increased, the frequency f will decrease.
Insight: Likewise, an increase in frequency (faster oscillation) results in a decrease in the period of oscillation.
15. Picture the Problem: A mass moves back and forth in simple harmonic motion.
Strategy: There must always be a restoring force in order for oscillation to occur.
Solution: The restoring force always pushes the mass back toward the equilibrium point. The farther the
mass is located away from the equilibrium point, the harder the force pushes. These are the characteristics
of a restoring force that lead to simple harmonic motion.
Insight: Suppose the force pushed back toward the equilibrium point, but was constant in magnitude on either side
of the equilibrium point. In this case the object would still oscillate back and forth, but its motion would be not be
described by a simple sine or cosine function.
16. Picture the Problem: A mass oscillates on a spring, and then the spring constant is increased by a factor of 4.
Strategy: Use the formula for the period of a mass on a spring to set up a ratio that can be used to predict the new
period.
Solution: Use a ratio to show the period is cut in half:
Tnew
=
Told
2π
2π
m
knew
m
kold
=
kold
=
knew
kold
4 kold
=
1
⇒ Tnew = 12 Told
2
Insight: The stiffer spring produces a larger force and a larger acceleration, resulting in more rapid change in motion
and a shorter period.
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13 – 4
Chapter 13: Oscillations and Waves
Pearson Physics by James S. Walker
17. Picture the Problem: A mass moves back and forth in simple harmonic motion.
Strategy: The frequency and period are inversely proportional. Use the formula for the period of a mass on a spring to
set up a ratio between the frequency of oscillation and the mass of the object.
Solution: Suppose the mass is increased
by a factor of 4, and use a ratio to show
the frequency will decrease:
f new 1 Tnew
=
=
f old
1 Told
1
2π
k
mnew
1
2π
k
mold
=
mold
=
mnew
mold
4 mold
=
1
⇒ f new =
2
1
2
f old
Insight: The heavier mass will accelerate at a lower rate given the spring force remains the same. The slower
acceleration results in a longer period and a lower frequency of oscillation.
18. The frequency of oscillation for a mass-spring system is inversely proportional to the square root of the mass.
Therefore, in order to increase the frequency of oscillation you should decrease the mass. A smaller mass will
experience a greater acceleration for the same amount of spring force, resulting in faster oscillation.
19. Picture the Problem: Four masses move back and forth in simple harmonic motion.
Strategy: The formula for the period of a mass on a spring can be used to predict the periods of the four masses using
the given spring constant and mass values. Then the periods can be ranked.
Solution: 1. Use the formula
to find the four periods:
TA = 2π
mA
0.1 kg
= 2π
= 1.3 s
10 N/m
kA
TB = 2π
mB
0.4 kg
= 2π
= 1.3 s
40 N/m
kB
TC = 2π
mC
0.4 kg
= 2π
= 2.5 s
kC
10 N/m
TD = 2π
mD
0.1 kg
= 2π
= 0.63 s
kD
40 N/m
2. Rank the periods: TD < TA = TB < TC
Insight: Increasing both the mass and the spring constant by the same factor produces no change in the period.
20. Picture the Problem: A spring is stretched a known distance with a known force. A mass is then attached to the spring
and allowed to oscillate about equilibrium.
Strategy: The force required to stretch the spring a known distance can be used to determine the spring constant. Once
we know the spring constant we can use the formula for the period of a mass on a spring to find the frequency.
Solution: 1. Find the spring constant from
the force required to stretch the spring:
F = − kx ⇒ k =
T = 2π
2. Now use the spring constant to find T:
F
12 N
=
= 75 N/m
x
0.16 m
m
2.2 kg
= 2π
= 1.076 s = 1.1 s
k
75 N/m
1
1
=
= 0.93 Hz
T 1.076 s
Insight: A stiffer spring would require a greater force to stretch the spring the same distance. The stiffer spring would
produce a larger force and a larger acceleration of the mass, resulting in a shorter period and higher frequency.
f =
3. Invert T to find f:
21. Picture the Problem: The processing clock speed of a computer is described in terms of a frequency.
Strategy: The period is the time required to complete one cycle, and in this case it is the time elapsed as a computer
completes one binary operation. The period is also the inverse of the frequency.
1
1
1
=
=
= 5.56 × 10 −10 s = 0.556 ns
f 1.80 GHz 1.80 × 109 Hz
Insight: A higher clock speed implies the computer can perform operations in a shorter period of time.
Solution: Invert f to find T:
T=
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13 – 5
Chapter 13: Oscillations and Waves
Pearson Physics by James S. Walker
22. Picture the Problem: A pendulum swings back and forth in simple harmonic motion.
Strategy: Use the formula for the period of a pendulum to find the relationship between period and length.
Solution: The period of a pendulum is given by T = 2π L g . By examining this formula we can see that
if the length of the pendulum is decreased, the period will also decrease.
Insight: A shorter pendulum will travel a shorter distance along the arc of its path for the same angular displacement.
Because the angles are the same as for a longer pendulum, the force and therefore the acceleration will be the same for
the shorter pendulum, but the distance it travels is shorter. It will therefore complete a cycle in a shorter period of time.
23. Picture the Problem: A pendulum swings back and forth in simple harmonic motion.
Strategy: Use the formula for the period of a pendulum to find the relationship between period and mass.
Solution: The period of a pendulum is given by T = 2π L g . By examining this formula we can see that
the period is independent of mass. Therefore, if the mass of the pendulum bob is increased (and nothing
else is changed) the period will stay the same.
Insight: The period of a pendulum is independent of mass for the same reason that all masses are accelerated at the
same rate by gravity (in a vacuum).
24. Picture the Problem: A pendulum swings back and forth in simple harmonic motion.
Strategy: Use the formula for the period of a pendulum to find its length when given the period.
L
g
L
g
Solution: 1. Solve the formula for L by squaring
both sides of the equation and rearranging:
T = 2π
2. Substitute the numerical values:
9.81 m/ s 2 2.00 s
gT 2
L=
=
4π 2
4π 2
⇒ T 2 = 4π 2
(
⇒L=
)(
)
gT 2
4π 2
2
= 0.994 m
Insight: In order to double the period to 4.00 s, the length must be quadrupled to 3.98 m = 13.0 ft.
25. Picture the Problem: This is a follow-up question to Guided Example 13.7. Distance and time measurements for a
falling mass are recorded, and the values are used to predict the period of a pendulum.
Strategy: Use the distance and time data to determine the local acceleration of gravity g, and then use that value of g to
predict the period of a pendulum by using the given formula.
y = 12 gt 2 for a mass falling from rest, taking
Solution: 1. (a) Determine the acceleration of
gravity g by using the distance and time data:
the positive y direction to be downward
g=
2. (b) Use the formula for the period of a
pendulum to predict the period:
2 y 2 (1.00 m )
=
= 9.83 m/s 2
t2
(0.451 s )2
T = 2π
L
0.500 m
= 2π
= 1.42 s
g
9.83 m /s 2
Insight: A shorter fall time would indicate an even stronger acceleration of gravity, which would result in a shorter
period of oscillation for the pendulum. In fact, you could combine the expressions from parts (a) and (b) to directly
predict the period given L, y, and t: T = 2π L t 2 2 y .
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13 – 6
Chapter 13: Oscillations and Waves
Pearson Physics by James S. Walker
26. Picture the Problem: A pendulum swings back and forth in simple harmonic motion.
Strategy: Use formula for the period of a pendulum, and then recall that f = 1 T to find the frequency.
Solution: 1. Use the formula for the
period of a pendulum:
T = 2π
2. Find the frequency from the period:
f =
L
1.25 m
= 2π
= 2.24 s
g
9.82 m /s 2
1 1 cycle
=
= 0.446 Hz
T 2.24 s
Insight: A stronger acceleration of gravity would result in a shorter period and a higher frequency. In fact, some
instruments can accurately determine g by carefully measuring the frequency of a pendulum-like oscillator.
27. Picture the Problem: Two pendulums swing back and forth in simple harmonic motion.
Strategy: Use the formula for the period of a pendulum to form a ratio of the periods of pendulum A and pendulum B,
given that they are located in places with different accelerations of gravity.
Solution: Use a ratio to show the period of the
pendulum at location A is less than the period at
location B if the acceleration of gravity at location
A is greater than that at location B by a factor of
n, where n > 1:
TA
=
TB
2π
2π
L
gA
L
gB
=
gB
=
gA
gB
=
ng B
1
n
⇒ TA = 1 n TB
Insight: Some instruments turn this problem around and can accurately determine g by carefully measuring the period
of a pendulum-like oscillator.
28. The period of a pendulum is proportional to the square root of its length. Therefore, the period of a pendulum increases
if the pendulum is lengthened. In this way you can adjust the speed of a grandfather clock by adjusting the length of its
pendulum.
29. The resonance condition occurs when the pendulum is driven at its natural frequency.
30. Picture the Problem: Two pendulums swing back and forth in simple harmonic motion. Pendulum A has twice the
amplitude of pendulum B.
Strategy: Use the concept of amplitude to draw a conclusion about the average speeds of the two pendulums.
Solution: The greater the amplitude of any mechanical oscillator, the greater the energy of its motion. Because
the kinetic energy is given by 12 mv 2 , we conclude that an oscillator with a larger amplitude has a larger
maximum kinetic energy and therefore a larger average speed. We conclude that the average speed of the mass
on pendulum A is greater than the average speed of the mass on pendulum B.
Insight: It turns out that the maximum kinetic energy of a pendulum is proportional to the square of the amplitude A,
K ∝ A2 . Because K is also proportional to v 2 we conclude that the maximum speed of the mass on the pendulum is
directly proportional to the amplitude of its motion, and we see that in this case the maximum speed of the mass on
pendulum A would be twice the maximum speed of the mass on pendulum B.
31. Picture the Problem: A playground swing oscillates in simple harmonic motion.
Strategy: Use the formula for the period of a pendulum together with the length of the swing to find the period.
Solution: Use the formula for the period of a pendulum:
T = 2π
L
3.0 m
= 2π
= 3.5 s
g
9.81 m s 2
Insight: The natural frequency of the swing’s motion would therefore be f = 1 T = 1 3.5 s = 0.29 Hz.
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13 – 7
Chapter 13: Oscillations and Waves
Pearson Physics by James S. Walker
32. Picture the Problem: The length of a pendulum is adjusted until its period is 1.0 s.
Strategy: Solve the pendulum period formula for its length.
L
g
Solution: 1. Solve the formula for the length of a pendulum:
T = 2π
⇒ T 2 = 4π 2
2. Substitute the numerical values:
(9.81 m s )(1.0 s )
L=
L
g
⇒L=
gT 2
4π 2
2
2
4π 2
= 0.25 m
Insight: The largest grandfather clocks often have a pendulum whose 0.994-m length gives the clock mechanism an
oscillation period of 2.00 s.
33. Picture the Problem: The measured period of a pendulum is used to determine the local acceleration of gravity.
Strategy: Solve the formula for the period of a pendulum for the acceleration of gravity g.
Solution: 1. Solve the formula for the length of a pendulum:
T = 2π
2. Find the period from the given data:
T=
3. Substitute the numerical values:
g=
L
g
⇒ T 2 = 4π 2
L
g
⇒g=
4π 2 L
T2
16 s
= 3.2 s
5.0 cycles
4π 2 (2.5 m )
(3.2 s)2
= 9.6 m/s 2
Insight: Some instruments can accurately determine g by carefully measuring the period of a pendulum-like oscillator.
34. Picture the Problem: A car brakes suddenly in heavy traffic, causing the cars behind it to bunch up. This bunching
moves through the traffic like a wave.
Strategy: Determine whether the wave is transverse or longitudinal by examining the relative directions of the
oscillations and of the wave propagation.
Solution: The oscillations of the cars (that is, the changes in their speeds) occur along the forward and backward
directions, parallel to the road. The wave also propagates along the direction of the road. Because the oscillations are
parallel to the wave propagation, we conclude that the traffic bunching is a longitudinal wave.
Insight: An analogous situation occurs in spiral galaxies, where gravitational forces cause stars to bunch up during their
orbits around the galactic core. These stellar bunches form the arms of the spiral galaxy.
35. Picture the Problem: Students line up in a row and move their bodies to simulate transverse, longitudinal, and water
waves.
Strategy: Consider the students as the masses through which the waves are propagating. Require that the students move
in a manner that simulates the motions of the masses involved in the propagation of transverse, longitudinal, and water
waves.
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13 – 8
Chapter 13: Oscillations and Waves
Pearson Physics by James S. Walker
Solution: 1. The students could alternately squat and stand up straight in synchronized fashion in order to demonstrate
a transverse wave, similar to the image above. The vertical oscillatory motion of each student would be perpendicular
to the horizontal direction of wave propagation.
2. The students could move closer together and spread further apart in synchronized fashion in order to demonstrate a
longitudinal wave, similar to the image above. The horizontal oscillatory motion of each student would be parallel to
the horizontal direction of wave propagation.
3. In order to simulate a water wave the students would have to move in a circular pattern whose plane of motion is
vertical and oriented along the direction of wave propagation. This combination of transverse and longitudinal motions
would resemble the motion of water molecules as a wave passes through them.
Insight: Each student oscillates near his or her original position, demonstrating that waves transport energy but they do
not transport mass (or students, in this case).
36. Picture the Problem: A wave travels at constant speed through a medium.
Strategy: Consider the oscillatory motion of an individual mass in a medium through which a periodic wave is
propagating.
Solution: Suppose a mass is at its maximum displacement while it is oscillating due to a wave that propagates through
the medium in which it is located. Its maximum displacement indicates that it is immersed in a wave crest. A little after
that it returns to the equilibrium position and then moves to its minimum displacement. At that point it is immersed
in a wave trough. Then it returns to the equilibrium position and on to its maximum displacement, at which time it is
again immersed in a wave crest. The time it took to return to its maximum displacement is one oscillation period, and
during that time one wavelength (crest to crest) passed by. We conclude that a wave travels one wavelength during
the time interval of one oscillation period.
Insight: A stiffer wave medium implies a stronger restoring force and a greater acceleration of the mass, which results
in a smaller oscillation period and a faster wave propagation speed.
37. Picture the Problem: The frequency and wavelength of a periodic wave are measured.
Strategy: Use the definitions of the frequency, period, and speed of a wave to answer the questions.
Solution: 1. (a) The frequency is the number of wave
cycles per second, which is also the number of
oscillations per second for individual masses:
4.0 oscillations 4.0 wave cycles
=
= 4.0 Hz
1 second
1 second
f =
2. (b) The period is the inverse of the frequency:
T=
3. (c) The speed of the wave is one wavelength per period:
v=
1
1 second
=
= 0.25 s
f 4.0 cycles
λ
T
=
3.0 m
= 12 m/s
0.25 s
Insight: A softer wave medium implies a weaker restoring force and a smaller acceleration of the masses, which results
in a longer oscillation period and a slower wave propagation speed.
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13 – 9
Chapter 13: Oscillations and Waves
Pearson Physics by James S. Walker
38. Picture the Problem: You produce a transverse wave by oscillating one end of a rope up and down.
Strategy: Use the definition of frequency, period, and speed of a wave to answer the questions.
Solution: 1. (a) The frequency is the number of wave
cycles per second, which is also the number of
oscillations per second for individual masses:
2.0 oscillations 2.0 wave cycles
=
= 2.0 Hz
1 second
1 second
f =
2. (b) The period is the inverse of the frequency:
T=
3. (c) The speed of the wave is one wavelength per period:
v=
1
1 second
=
= 0.50 s
f 2.0 cycles
λ
T
⇒ λ = vT = (5.0 m s )(0.50 s ) = 2.5 m
Insight: A softer wave medium implies a weaker restoring force and a smaller acceleration of the masses, which results
in a longer oscillation period and a slower wave propagation speed.
39. Picture the Problem: A crowd produces a “wave” at a sports stadium.
Strategy: Determine whether the wave is transverse or longitudinal by examining the relative directions of the
oscillations and of the wave propagation.
Solution: The motion of each fan in the crowd is vertical, perpendicular to the horizontal direction of wave propagation.
We conclude that the wave is transverse.
Insight: A longitudinal wave could propagate through the crowd, but it would cause many fans to bump into each other
and likely result in the spilling of many refreshments!
40. Picture the Problem: A wave propagates through a chocolate candy bar and a chocolate cake.
Strategy: Note that the wave propagation speed depends upon the restoring forces that allow the masses in the medium
to oscillate about their equilibrium positions. Use this fact to predict the relative wave speeds.
Solution: The cake is a softer medium than a solid chocolate bar. A softer wave medium implies a weaker restoring
force and a smaller acceleration of the masses, which results in a longer oscillation period and a slower wave propagation speed. We therefore expect the speed of sound to be greater in a chocolate candy bar than in a chocolate cake.
Insight: The speed of sound in steel (6100 m/s) is much faster than the speed of sound in hard wood (3960 m/s)
because steel is a much stiffer medium than wood. Likewise, each of these speeds is much greater than the speed of
sound in air (340 m/s).
41. Picture the Problem: Waves transport energy as they travel through various media.
Strategy: Each wave type is distinguished by the type of energy that it transports.
Solution: 1. Light waves carry electromagnetic energy. An example of this is the energy we receive from sunlight.
2. Radio waves carry electromagnetic energy. Examples of this are the energy a car stereo receives from a radio station,
the energy received by a cordless telephone base, or the energy received by an automatic garage door opener.
3. Water waves carry mechanical energy. When you see a boat oscillating up and down in response to the water waves
passing by, you can see that the boat gains mechanical energy from the wave.
4. Sound waves carry mechanical energy in the form of pressure variations. Your eardrum vibrates in response to these
pressure variations, and your brain interprets those vibrations as sound.
Insight: A lithotripter is a medical device that uses concentrated ultrasound waves to deliver so much sound energy to a
kidney stone that it is broken into tiny pieces.
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13 – 10
Chapter 13: Oscillations and Waves
Pearson Physics by James S. Walker
42. Picture the Problem: The frequency and speed of a wave are measured.
Strategy: Use the definition of frequency, period, and speed of a wave to answer the questions.
Solution: 1. (a) The frequency is the number of wave
cycles per second, which is also the number of
oscillations per second for individual masses:
f =
5.0 oscillations 5.0 wave cycles
=
= 5.0 Hz
1 second
1 second
2. (b) The period is the inverse of the frequency:
T=
1
1 second
=
= 0.20 s
f 5.0 cycles
λ
⇒ λ = vT = (6.0 m s )(0.20 s ) = 1.2 m
T
Insight: A faster wave speed would correspond to a longer wavelength because the wave must travel one wavelength
during one oscillation period, and in this example the oscillation period is fixed.
v=
3. (c) Find the wavelength from the speed and period:
43. Picture the Problem: The frequency and wavelength of waves that pass a fishing boat are measured.
Strategy: Use the relationship between wave speed, frequency, and wavelength to answer the question.
Solution: 1. The frequency is the number of wave
cycles per second, which is also the number of
oscillations per second for individual masses:
f =
12 oscillations 12 wave cycles
=
= 0.2667 Hz
45 seconds
45 seconds
λ
= λ f = (7.5 m )(0.2667 Hz ) = 2.0 m s
T
Insight: If the wave crests were closer together but the frequency was the same, we would conclude that the wavelength
was shorter than 7.5 m and that the wave speed was slower than 2.0 m/s.
v=
2. The speed of the wave is one wavelength per period:
44. Picture the Problem: The speed and wavelength of a passing periodic wave are measured.
Strategy: Use the relationship among wave speed, frequency, and wavelength to answer the question.
Solution: 1. The wavelength is the distance between
adjacent crests, so the distance between a crest and an
adjacent trough is only half a wavelength:
1
2
2. The frequency of the wave is
related to wavelength and speed:
v=λf
λ = 2.4 m ⇒ λ = 4.8 m
⇒ f =
v
λ
=
5.6 m/s
= 1.2 Hz
4.8 m
Insight: In the laboratory it is easier to measure the period of a wave with a stopwatch than it is to measure the distance
between adjacent crests because the wave is continuously moving.
45. Picture the Problem: Two waves are superimposed.
Strategy: Use the principle of wave superposition to draw the resultant waves.
Solution: 1. (a) The waves are superimposed:
(b) The waves are superimposed:
Insight: When two waves produce destructive interference the energy carried by each wave remains, but it is
temporarily stored in potential energy (due to the interactions between the oscillating atoms) rather than kinetic energy
and the amplitude of the resultant wave is zero.
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13 – 11
Chapter 13: Oscillations and Waves
Pearson Physics by James S. Walker
46. Picture the Problem: Two wave pulses travel toward each
other at time t = 0. We will make sketches of the wave
pulses at times t = 1.0 s, 2.0 s, 2.5 s, 3.0 s, and 4.0 s.
Strategy: Move the pulses one meter forward for each second. When the pulses overlap, add their amplitudes.
Solution: Sketches for times t = 1.0 s, 2.0 s, 2.5 s, 3.0 s, and 4.0 s are shown below.
t = 1.0 s:
t = 2.0 s:
Add the amplitudes when the waves overlap at t = 2.5 s:
t = 3.0 s:
t = 4.0 s:
Insight: The pulses pass right through each other and continue their motion unchanged.
47. Picture the Problem: This is a follow-up question to Guided Example 13.10. A string 1.30 m in length and fixed at
both ends is oscillating in its first harmonic mode. The wave speed is 22.0 m/s.
Strategy: Find the oscillation frequency given the string length and wave propagation speed, noting that only half a
wavelength fits on a string in its first harmonic mode.
Solution: 1. Only half a wavelength
spans the length of the string:
1
2
2. The frequency of the wave is
related to wavelength and speed:
λ = 1.3 m ⇒ λ = 2.6 m
v=λf
⇒ f =
v
λ
=
22.0 m/s
= 8.46 Hz
2.6 m
Insight: The speed of the waves on the string depends only upon the properties of the string – its tension and mass per
unit length. In this problem the higher first-harmonic frequency is a consequence of the higher wave speed, which in
turn was caused by some adjustment to the string itself, such as increasing its tension. It is not correct to say the higher
frequency caused the waves to travel faster.
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13 – 12
Chapter 13: Oscillations and Waves
Pearson Physics by James S. Walker
48. Picture the Problem: A guitar string is vibrating at a lower frequency than desired.
Strategy: Note that the frequency of vibration of a string fixed at both ends (such as that found on a guitar) is given by
f = v λ . Use this relationship to answer the question.
Solution: The first harmonic frequency of the guitar string is given by f = v λ = v 2 L , where λ = 2L because only
half a wavelength spans the length of the string that vibrates in the first harmonic. Because the length L of the string is
fixed, we conclude that in order to get the string in tune, the speed of waves on the string should be increased. This is
normally accomplished by tightening the string (increasing its tension).
Insight: Guitars are tuned in this way by turning knobs at the end of the guitar neck (on the headstock) that adjust the
tension in the string and therefore change the frequency of vibration.
49. Picture the Problem: Two identical wave pulses with positive displacement head toward each other from opposite
directions and then pass through each other.
Strategy: Apply the principle of wave superposition to answer the question.
Solution: When the two wave pulses meet they will pass through each other. The principle of superposition states that
while they are passing through each other their amplitudes will add. Therefore, at the moment the pulses are fully
overlapped the string will be displaced with twice the amplitude of a single wave pulse. After passing through each
other the two pulses will continue onward as if nothing had happened.
Insight: If waves did not pass through each other in this manner it would be impossible to enjoy an orchestra concert.
The waves from the flutes would collide with and scatter the waves from the violins and other instruments, creating an
incomprehensible noise instead of a lovely melody.
50. A standing wave is formed when two waves of the same frequency pass through the same medium but in opposite
directions. At some times the crests will overlap troughs and produce destructive interference. At other times the crests
will overlap crests and create crests with double amplitude (constructive interference). Therefore your classmate is
correct to say that standing waves involve both constructive and destructive interference.
51. Every musical instrument creates standing waves of sound. A wind instrument creates a standing wave along an air
column that is composed of the generated wave (produced by the musician’s lips or by a reed) propagating in one
direction and a reflected wave traveling in the other direction. A string instrument creates standing waves on a string
that is fixed at both ends. Lasers create standing waves of light, and a radio tuner can be thought of as an electrical
oscillator caused by counter-propagating electrical waves.
52. Yes. A higher harmonic of a standing wave always consists of a greater number of nodes and antinodes along the same
length of a medium than a lower harmonic. That fact implies that the wavelength of a higher harmonic is always shorter
than the wavelength of a lower harmonic. Because the wave speed is the same for both harmonics, we conclude that the
frequency f = v λ of a higher harmonic must always be greater than the frequency of a lower harmonic.
53. Picture the Problem: A guitar string of known length that is fixed at both ends vibrates with a standing wave with a
known number of antinodes.
Strategy: The harmonic number for a standing wave on a string that is fixed at both ends is equal to the number of
antinodes. Determine the harmonic number and then find the corresponding wavelength of the standing wave using the
given length of the string.
Solution: 1. The standing wave has two antinodes so it is the second harmonic standing wave.
2. The second harmonic standing wave has a wavelength equal to the length of the string, which is 66 cm or 0.66 m.
Insight: If we knew the frequency of this standing wave, say, 440 Hz, we could determine the speed of the waves on the
guitar string v = f λ = (440 Hz )(0.66 m ) = 290 m/s.
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13 – 13
Chapter 13: Oscillations and Waves
Pearson Physics by James S. Walker
54. Picture the Problem: A string of known length that is fixed at both ends vibrates in standing wave patterns.
Strategy: The wavelengths that correspond to standing wave patterns on a string that is fixed at both ends are
2L L
2L
, , ...
λ = 2 L, L,
, where L is the length of the string and n is the harmonic number. Use this series
n
3 4
to determine the allowable wavelengths of standing waves.
Solution: The string has length 0.33 m, so the allowable wavelengths for standing waves are 0.66 m, 0.33 m, 0.22 m,
0.0825 m, … We conclude the wavelengths that produce standing waves are wavelengths 1, 2, and 4. Wavelength
3 (0.44 m) is 4/3 the length of the string, and does not correspond with the wavelength of any standing wave.
Insight: Using the formula λ = 2L n we conclude that wavelength 1 is the third (n = 3) harmonic, wavelength 2 is the
second harmonic, and wavelength 4 is the first harmonic (fundamental frequency).
55. Picture the Problem: A string that is fixed at both ends vibrates in its first harmonic standing wave pattern.
Strategy: The wavelength that corresponds to the first harmonic standing wave pattern on a string that is fixed at both
ends is λ = 2 L , where L is the length of the string. Use this relationship to determine the length of the string.
Solution: The wavelength is measured to be 0.65 m, so the string length is L =
λ
2
=
0.65 m
= 0.325 m = 0.33 m
2
Insight: The wavelength λ and the string length L are the same when the string vibrates in its second harmonic
standing wave pattern.
56. Picture the Problem: A mass-spring system oscillates on Earth and its period is measured. The system is then
transported to the Moon and is set into oscillation again.
Strategy: Use the formula that predicts the period of a mass on a spring to answer the question.
Solution: No. The primary difference between the surface of the Moon and the surface of Earth is the acceleration of
gravity. The period of a mass on a spring T = 2π m k depends only on the mass and the spring constant k, not the
acceleration of gravity. We conclude that the mass-spring system will oscillate with the same period on the Moon as it
had on Earth.
Insight: The period of a pendulum on the Moon will be almost 2.5 times longer than the period of the same pendulum
on Earth:
2π
TMoon
=
TEarth
2π
L g Moon
L g Earth
=
g Earth
9.81 m s 2
=
= 2.46
g Moon
1.62 m s 2
57. Picture the Problem: A mass moves back and forth in simple harmonic motion with amplitude A and period T.
Strategy: Use the principles of simple harmonic motion to answer the conceptual question.
Solution: Suppose the mass begins at one extreme of its displacement at t = 0. In the first half of its period T it will
move through the equilibrium position and all the way to the other extreme of its displacement, a distance of 2A. It
therefore requires a time 12 T for the mass to travel a distance of 2A. Thus the mass will move a distance 2A in half a
cycle and a distance A in another quarter of a cycle, for a total time elapsed of
3
4
T to move distance 3A.
Insight: The distance traveled in each case is the same regardless of where the mass starts in its cycle.
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13 – 14
Chapter 13: Oscillations and Waves
Pearson Physics by James S. Walker
58. Picture the Problem: Four mass-spring systems have different masses and spring force constants.
Strategy: Use the relationship T = 2π m k to determine the ranking of the periods.
Solution: 1. Find T
for system A:
TA = 2π
mA
m
= 2π
kA
k
2. Repeat to find TB :
TB = 2π
mB
2m
= 2π
= 2 TA
kB
k
3. Repeat to find TC :
TC = 2π
3m
1
TA
=
6k
2
4. Repeat to find TD :
TD = 2π
mD
m 1
= 2π
= TA
kD
4k 2
5. By comparing the resulting periods we arrive at the ranking D < C < A < B.
Insight: Measuring the period of oscillation is often the most accurate way of measuring a spring force constant.
59. Picture the Problem: An old car with worn-out shock absorbers oscillates with a given frequency when it hits a speed
bump. The driver adds a couple of passengers to the car and hits another speed bump.
Strategy: Note the relationship between frequency of a mass on a spring and the mass in order to answer the conceptual
question.
Solution: 1. (a) Write the period in terms of frequency: f = 1 T = (1 2π ) k m . From this expression we can see that
increasing m will decrease f. We conclude that the resulting frequency of oscillation is less than it was before.
2. (b) The best explanation is A. Increasing the mass on a spring increases its period and thus decreases its frequency.
Statements B and C are both false.
Insight: The shock absorbers are designed to damp out such oscillations for the comfort of the passengers.
60. Picture the Problem: A rocking chair completes one full oscillation each time it returns back to its original position.
Strategy: The period is the time to complete one cycle, and the frequency is the number of cycles per second.
Solution: 1. Divide the total time by the
number of cycles to determine the period:
T=
t
21 s
=
= 1.75 s
n 12 cycles
1
1
=
= 0.57 Hz
T 1.75 s
Insight: Because period and frequency are inverses of each other, whenever the period is greater than one second, the
frequency will be less than one hertz.
f =
2. Invert the period to determine the frequency:
61. Picture the Problem: A fishing bobber moves up and down with periodic motion. One period is the amount of time for
the bob to drop down and rise back up to its original position.
Strategy: We can find the period by taking the inverse of the frequency.
T=
Solution: Invert the frequency to obtain the period:
1
1
=
= 0.38 s
f 2.6 Hz
Insight: Because frequency is the inverse of period, whenever the frequency is greater than one hertz, the period will be
less than one second.
62. Picture the Problem: The ends of a medical tuning fork vibrate rapidly to help diagnose a nervous system disorder.
Strategy: The period of an oscillator is the inverse of its frequency.
Solution: Find the inverse of the frequency:
T=
1
1
=
= 0.00781 s = 7.81 ms
f 128 Hz
Insight: Tuning forks typically have two tines which oscillate in opposite directions. The net acceleration of the entire
fork is thus zero even though the tips are accelerating at a large rate.
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13 – 15
Chapter 13: Oscillations and Waves
Pearson Physics by James S. Walker
63. Picture the Problem: As a basketball is dribbled it moves up and down in periodic motion. One period is the time for
the ball to drop and then return to the player’s hand.
Strategy: The time for one dribble is the period, or the inverse of the frequency. Multiplying the period by the number
of dribbles will give the total time.
Solution: Multiply the inverse of the frequency by
the number of dribbles to obtain the total time:
t = 12 T =
12
12
=
= 6.7 s
f 1.8 Hz
Insight: The greater the frequency, the more dribbles that will occur over a given time.
64. Picture the Problem: A heart beats in a regular periodic pattern. To measure your heart rate you typically count the
number of pulses in one minute.
Strategy: Convert the time from minutes to seconds to obtain the frequency in hertz. The period is obtained from the
inverse of the frequency.
Solution: 1. Multiply the heart rate by the correct
conversion factor to get the frequency in Hz.
beats ⎞ ⎛ 1 min ⎞
⎛
f = ⎜ 74
⎟⎜
⎟ = 1.2 Hz
⎝
min ⎠ ⎝ 60 s ⎠
2. Invert the frequency to obtain the period:
T=
1
60 s
=
= 0.81 s
f 74 beats
Insight: Typical resting heartbeats have frequencies around one hertz and periods of about one second.
65. Picture the Problem: A force-versus-stretch distance plot for a spring is
shown in the figure.
Strategy: Determine the spring constant of the spring by finding the slope of
the F vs. x plot. Use the spring constant together with the given mass to find the
period of oscillation.
Solution: 1. (a) The magnitude of the slope of the F-versus-x graph is the
spring constant. The line extends from the origin to the point (25 cm, 5 N), so
rise 5.0 N 100 cm
=
×
= 20 N/m
the slope is
run 25 cm
m
2. (b) Now use the spring constant to find T:
T = 2π
m
0.10 kg
= 2π
= 0.44 s
k
20 N/m
Insight: A stiffer spring would require a larger force for the same amount of stretch, so its F vs. x plot would have a
steeper slope.
66. Picture the Problem: A mass attached to a spring is pulled away from equilibrium and released. The mass then
oscillates about the equilibrium position at a frequency determined by the stiffness of the spring.
Strategy: We can determine the spring force constant by using the given period and mass together with the equation for
the period of a mass on a spring.
2
m
⎛ 2π ⎞
⇒ k=⎜ ⎟ m
⎝ T ⎠
k
Solution: 1. Solve the period
equation for the spring force constant:
T = 2π
2. Substitute the numerical values:
⎛ 2π ⎞
⎛ 2π ⎞
k=⎜ ⎟ m=⎜
⎝ T ⎠
⎝ 0.77 s ⎟⎠
2
2
(0.46 kg ) =
31 N/m
Insight: Measuring the period of oscillation is often the most accurate way to measure a spring force constant.
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13 – 16
Chapter 13: Oscillations and Waves
Pearson Physics by James S. Walker
67. Picture the Problem: When you measure your pulse you typically count the number of pulses per minute. The number
of beats per minute, or frequency, can change depending on whether you are resting or exercising.
Strategy: Convert the time rate from per second to per minute to obtain the frequency in beats/min.
beats ⎞ ⎛ 60 s ⎞
⎛
⎜⎝1.45
⎟⎜
⎟ = 87.0 beats min
s ⎠ ⎝ min ⎠
Solution: 1. (a) Multiply the frequency by the
conversion factor to obtain the heart rate:
2. (b) The number of beats per minute will increase, because more beats per second means more beats per minute.
beats ⎞ ⎛ 60 s ⎞
⎛
⎜⎝1.55
⎟⎜
⎟ = 93.0 beats min
s ⎠ ⎝ min ⎠
3. (c) Multiply the new frequency by the
conversion factor to obtain the heart rate:
Insight: Increasing the frequency in any set of units (such as beat/s or beat/min) will increase the frequency in any other
set of units. The period, however, will decrease.
68. Picture the Problem: When a 0.50-kg mass is attached to a vertical spring, the
spring stretches by 15 cm.
Strategy: We can use the displacement of the spring to calculate the spring force
constant. The spring force constant, together with the given period, can then be
inserted into the period equation to solve for the necessary mass.
Solution: 1. Use the spring force equation
to solve for the spring force constant:
F = ky ⇒ k =
2. Insert the numeric values to obtain k:
k=
3. Solve the period equation for the mass:
T = 2π
4. Substitute the numerical values:
⎛ 0.75 s ⎞
m=⎜
⎝ 2π ⎟⎠
F mg
=
y
y
(0.50 kg )(9.81 m/s 2 )
0.15 m
= 32.7 N/m
2
m
⎛ T ⎞
⇒ m=⎜ ⎟ k
⎝ 2π ⎠
k
2
(32.7 N/m ) =
0.47 kg
Insight: Increasing the mass will increase the period because the period is proportional to the square root of the mass.
69. If the pendulum begins its cycle at its farthest left position, it will swing through the vertical (equilibrium) position,
reach its farthest right position, swing back through vertical, and return to its farthest left position to complete one cycle.
We conclude the pendulum passes through the vertical position two times each cycle.
70. Picture the Problem: A pendulum of length L oscillates with a period T.
Strategy: Consider the relationship between the period and the length of a pendulum in order to answer the conceptual
question.
Solution: The period of a pendulum is directly proportional to the square root of the length. In order to triple the period
you must therefore change the length by a factor of nine. We conclude that the length of the pendulum must be
increased to 9L.
2
2
2
g Tnew
4π 2 ⎛ Tnew ⎞
⎛ 3 T old ⎞
L
Insight: You can also use a ratio: new =
= 9 ⇒ Lnew = 9 Lold
=⎜
=⎜
⎟
2
2
Lold
⎝ Told ⎠
⎝ T old ⎟⎠
g Told 4π
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13 – 17
Chapter 13: Oscillations and Waves
Pearson Physics by James S. Walker
71. Picture the Problem: Pendulum A has a length L and a mass m. Pendulum B has a length L and a mass 2m.
Strategy: Consider the relationship between the period and the mass of a pendulum in order to answer the conceptual
question.
Solution: The period of a pendulum is independent of the mass of the pendulum bob. We conclude that the extra mass
on pendulum B will not affect its period, and the period of pendulum A is equal to the period of pendulum B.
Insight: The fact that each pendulum has the same period is related to the fact that all masses are accelerated at the
same rate by gravity.
72. Picture the Problem: One pendulum has a length L and is released from an angle of 10°. A second pendulum also has
length L but is released from an angle of 20°.
Strategy: Consider the relationship between the period and the amplitude of a pendulum in order to answer the
conceptual question.
Solution: The period of a pendulum is independent of the amplitude of its motion. We conclude that the extra amplitude
of Bobby’s pendulum will not affect its period, and the periods of the two pendulums are identical. Therefore, Billie’s
bob will reach the vertical position at the same time as Bobby’s bob.
Insight: If the amplitude of the pendulum is very large (greater than 45°) the small angle approximation breaks down
and its movement is not accurately described by simple harmonic motion.
73. Picture the Problem: The picture shows a metronome that oscillates with a mass (the bow tie)
attached to a thin metal rod that pivots about a point near the belly of the penguin.
Strategy: The device can be considered a physical pendulum whose moment of inertia about the
pivot point can be adjusted by moving the bow tie up and down the thin metal rod. Use the equation
for the relationship between the period of a pendulum and its length to answer the conceptual
question.
Solution: The period of a pendulum is proportional to the square root of its length. In order to reduce
the period and increase the frequency of oscillation pendulum length should be decreased. We
conclude that the penguin’s bow tie should be moved downward in order to increase the frequency.
Insight: The reverse would be true if this were a standard pendulum with the pivot point at the top and the mass below.
In such a case the mass would have to be moved upward in order to decrease the pendulum length.
74. Picture the Problem: A grandfather clock keeps correct time at sea level but is taken to the top of a nearby mountain.
Strategy: The acceleration of gravity will be slightly smaller at high altitude than it is at sea level because the
mountaintop location is farther from the center of Earth. Use this fact together with the relationship between
the period of a pendulum and the acceleration of gravity to answer the conceptual question.
Solution: 1. (a) The period of a pendulum is inversely proportional to the square root of the acceleration of gravity.
Decreasing g will therefore increase T. We conclude that if the clock is taken to the top of a nearby mountain the period
will increase and the clock will run slow.
2. (b) The best explanation is A. Gravity is weaker at the top of the mountain, leading to a greater period of oscillation.
Statement B is partly true but ignores the change in the acceleration of gravity. Statement C could only be true if the
mountain were filled with material with a higher than average density, so that the nearby location of extra mass more
than compensates for the increased distance from the center of Earth.
Insight: Some gravity meters operate on this principle, precisely determining the local acceleration of gravity by
accurately measuring the period of a pendulum.
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13 – 18
Chapter 13: Oscillations and Waves
Pearson Physics by James S. Walker
75. Picture the Problem: The frequency of a pendulum is measured.
Strategy: First invert the frequency in order to find the period of the pendulum. Then solve the pendulum period
formula for its length.
1
1
=
= 4.17 s
f 0.24 Hz
Solution: 1. Invert the frequency to find the period:
T=
2. Now solve the formula for the length of a pendulum:
T = 2π
L
g
L
g
⇒ T 2 = 4π 2
⇒L=
gT 2
4π 2
(9.81 m s )(4.17 s )
L=
2
2
= 4.3 m
4π 2
Insight: If the frequency is to be increased by a factor of 4, the length must be decreased by a factor of 16.
3. Substitute the numerical values:
76. Picture the Problem: The period of a pendulum in the service shaft of a building measured.
Strategy: Find the average period of the pendulum using the measurement data. Then solve the pendulum period
formula for its length.
110 s
= 8.46 s
13 oscillations
Solution: 1. Find the period from the data:
T=
2. Now solve the formula for the length of a pendulum:
T = 2π
L
g
L
g
⇒ T 2 = 4π 2
⇒L=
gT 2
4π 2
(9.81 m s )(8.46 s )
L=
2
2
= 17.8 m = 58.4 ft
4π 2
Insight: A very long pendulum like this one could be used as a dramatic example of a Foucault pendulum, which was
the first easily accessible experiment that definitively proved that Earth rotates on its axis.
3. Substitute the numerical values:
77. Picture the Problem: The length and period of a simple pendulum are measured in order to determine the acceleration
of gravity.
Strategy: Use the period of the pendulum and its length to calculate the local acceleration of gravity.
2
L
⎛ 2π ⎞
⇒ g=⎜ ⎟ L
⎝ T ⎠
g
Solution: 1. Solve the period equation for g:
T = 2π
2. Substitute the numerical values:
⎡ 2π ⎤
2
g = ⎢1
⎥ (2.5 m ) = 9.6 m/s
16
s
(
)
⎣5
⎦
2
Insight: The small variations in gravity around the surface of Earth can be determined by carefully measuring the
period of a pendulum with a precisely known length.
78. Picture the Problem: The period of a pendulum is measured.
Strategy: Calculate the length of the pendulum from its period.
2
L
⎛ T ⎞
⇒ L=⎜ ⎟ g
⎝ 2π ⎠
g
Solution: 1. Solve the period equation for length:
T = 2π
2. Insert the numeric values:
⎛ 2.00 s ⎞
L=⎜
⎝ 2π ⎟⎠
2
(9.81 m/s ) = 0.994 m = 99.4 cm
2
Insight: This is the length of the pendulum in many large mechanical clocks such as grandfather clocks. Smaller clocks,
such as wall-mounted mechanical clocks, have a pendulum about 0.248 m long with a period of 1.00 second.
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13 – 19
Chapter 13: Oscillations and Waves
Pearson Physics by James S. Walker
79. Picture the Problem: A pendulum is carefully constructed so that it passes the equilibrium position every 1.000 s.
Strategy: A pendulum will pass the equilibrium point twice each cycle, once while moving to the left and once while
moving to the right. Use this fact to find the period of the “seconds pendulum,” and then use the given length to
determine the local acceleration of gravity.
1.000 s
= 2.000 s
0.5 oscillations
Solution: 1. (a) Find the period from the data:
T=
2. (b) Solve the period equation for g:
T = 2π
3. Substitute the numerical values:
⎡ 2π ⎤
g=⎢
(0.9933 m ) = 9.803 m/s 2
⎣ 2.000 s ⎥⎦
2
L
⎛ 2π ⎞
⇒ g=⎜ ⎟ L
⎝ T ⎠
g
2
Insight: If this pendulum were made very long, it could be used as a dramatic example of a Foucault pendulum, which
was the first easily accessible experiment that definitively proved that Earth rotates on its axis.
80. Picture the Problem: A pendulum in the United Nations building oscillates back and forth through the vertical.
Strategy: The time the pendulum takes to move from maximum displacement to equilibrium position is one-quarter of
a period. Use the relationship between period and length to determine the time.
T 1⎛
L⎞ π
22.9 m
= ⎜ 2π
=
= 2.40 s
4 4⎝
g ⎟⎠ 2 9.81 m/s 2
Solution: Substitute the numerical values:
Insight: The full period of this pendulum is 4(2.4 s) = 9.6 seconds. A pendulum with only half this length would have a
period of 6.8 s.
81. Picture the Problem: A microphone that dangles from a radio booth behaves as a simple pendulum.
Strategy: We can use the period of the pendulum to determine its length.
2
L
⎛ T ⎞
⇒ L=⎜ ⎟ g
⎝ 2π ⎠
g
Solution: 1. Solve the period equation for L:
T = 2π
2. Substitute the numerical values:
⎛ 1 60.0 s ⎞
L = ⎜ 10
⎝ 2π ⎟⎠
2
(9.81 m/s ) = 8.95 m = 29.4 ft
2
Insight: A long pendulum will oscillate slowly. This problem shows that it takes 6 s for one oscillation of a pendulum
about 9 m long. A shorter pendulum, such as that on a grandfather clock, is only about a 1 m long and has a period of
about 2 s.
82. A wave travels a distance of one wavelength during one oscillation period. Therefore, a time corresponding to two
oscillation periods is required for a wave to travel two wavelengths.
83. Picture the Problem: Light is a periodic wave that travels at a constant speed, and the different colors of light have
different wavelengths.
Strategy: Recall the relationships among wavelength, frequency, and speed of a periodic wave to answer the conceptual
question.
Solution: The product of the wavelength and frequency is equal to the wave speed, v = λ f . Because red light has
the longer wavelength, we conclude that it also has the lower frequency, so that the product of its wavelength and
its frequency is equal to its speed as well as the speed of the blue light. Hence blue light has the greater frequency.
Insight: The speed of all electromagnetic waves in vacuum is exactly the same, 3.00×108 m/s, for all wavelengths,
from the longest radio waves, through the infrared and visible waves, and all the way down to x-rays and gamma rays.
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13 – 20
Chapter 13: Oscillations and Waves
Pearson Physics by James S. Walker
84. Picture the Problem: Fans at a stadium often create a transverse “wave” by standing and sitting in synchronized
fashion. In this problem we consider what motion would be required to make a longitudinal wave.
Strategy: The fans must move in a manner that simulates the motions of the masses involved in the propagation of a
longitudinal wave.
Solution: The fans would have to move closer together and spread further apart in synchronized fashion in order to
demonstrate a longitudinal wave. The horizontal oscillatory motion of each student would be parallel to the horizontal
direction of wave propagation.
Insight: Each fan oscillates about his or her original position, demonstrating that waves transport energy but they do not
transport mass (or in this case, people).
85. Picture the Problem: In a classic TV commercial, a group of cats feed
from bowls of cat food that are lined up side by side. Initially there is one
cat for each bowl. When an additional cat is added to the scene, it runs to a
bowl at the end of the line and begins to eat. The cat that was there
originally moves to the next bowl, displacing that cat, which moves to the
next bowl, and so on down the line.
Strategy: Consider the relative motions of the individual cats and the
propagation of the “wave.” If the motions are perpendicular, it is a
transverse wave, but if they are parallel, it is a longitudinal wave.
Solution: The cats move horizontally, and so does the “wave”. Because the motion of an individual cat is parallel to
the direction of wave propagation, we conclude this is a longitudinal wave.
Insight: A key difference between this motion and an actual wave is that each cat does not oscillate about an
equilibrium position, but moves to the next bowl of food. In this way mass is transported along the line of cats, and not
energy, as is the case for a wave.
86. Picture the Problem: The three waves, A, B, and C, shown
in the figure at right propagate on strings with equal tensions
and equal mass per length.
Strategy: Use v = λ f and the fact that each wave propagates
with the same speed to determine the requested rankings.
Solution: 1. (a) The waves on the three strings have the same speed, v = λ f . The wave with the longest wavelength
therefore has the lowest frequency. The ranking for frequency is B < C < A.
2. (b) Wavelength is simply the distance from crest to crest. Therefore the ranking for wavelength is A < C < B.
Insight: If these were sound waves, the ranking for sound pitch would be the same as for frequency: B < C < A.
87. Picture the Problem: Ripples on a pond expand outward from where you are dipping
your finger in a periodic fashion.
Strategy: Use v = λ f and the given frequency and wavelength to find the various values.
Solution: 1. Each time you dip your finger you create a wave. Therefore the frequency of
the wave is f = 2 dips second = 2 Hz
2. The period is the inverse of the frequency, T = 1 (2 Hz ) = 0.50 s
3. The wavelength is the distance between adjacent crests, which is given in the problem
statement as λ = 0.18 m
4. The speed of the waves is v = f λ = (2 Hz )(0.18 m ) = 0.36 m/s
Insight: Dipping your finger even more frequently will not make the waves go any faster;
it will only shorten the wavelength. This is because the speed of any wave is determined
by the properties of the medium, not by wavelength or frequency.
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13 – 21
Chapter 13: Oscillations and Waves
Pearson Physics by James S. Walker
88. Picture the Problem: The dimensions of a wave are labeled in the
image shown at the right.
Strategy: Set the wavelength equal to the horizontal crest-to-crest
distance, or double the horizontal crest-to-trough distance. Set the
amplitude equal to the vertical crest-to-midline distance, or half the
vertical crest-to-trough distance.
Solution: 1. (a) Double the horizontal crest-to-trough distance:
λ = 2 (28 cm ) = 56 cm
2. (b) Halve the vertical crest-to-trough distance:
A=
1
2
(13 cm ) =
6.5 cm
Insight: Note the difference between how wavelength and amplitude are measured. The wavelength is the entire
distance from crest to crest, but the amplitude is measured from the equilibrium level to the crest.
89. Picture the Problem: A surfer determines the speed of the water waves that pass her by measuring their frequency and
wavelength.
Strategy: Use the relationship between frequency, wavelength, and speed to find the speed.
Solution: Multiply wavelength by frequency:
waves ⎞ ⎛ 1 min ⎞
⎛
v = λ f = (34 m ) ⎜ 14
⎟⎜
⎟ = 7.9 m/s
⎝
min ⎠ ⎝ 60 sec ⎠
Insight: The wave speed is determined by the properties of the water itself. An increase in the wave frequency would be
accompanied by a decrease in wavelength so that their product (which equals the wave speed) remains the same.
90. Picture the Problem: The speed and wavelength of a tsunami (tidal wave) are measured.
Strategy: Use the relationship between frequency, wavelength, and speed to find the frequency.
Solution: Calculate the frequency:
f =
v
λ
=
(750 km/h ) ⎛
1h ⎞
−4
⎜⎝
⎟ = 6.7 × 10 Hz
310 km 3600 s ⎠
Insight: Although the tsunami has a very high speed, the long wavelength gives it a low frequency.
91. Picture the Problem: The image shows water
waves passing to a shallow region where the speed
decreases. The wavelength in the shallow area
decreases because the frequency remains the same.
Strategy: Use the relationships among frequency, wavelength, and speed to find the frequency in the deep water.
Then use the constant frequency and the speed in the shallow water to calculate the new wavelength.
Solution: 1. Calculate the frequency:
f =
2. Calculate the shallow water wavelength:
λ2 =
v1
λ1
=
2.0 m/s
= 1.333 Hz
1.5 m
v2
1.6 m/s
=
= 1.2 m
f 1.333 Hz
Insight: Note that increasing the speed, while holding the frequency constant, will increase the wavelength. This would
happen if the water wave travels from a shallow region (say, a coral reef) into a deeper region.
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13 – 22
Chapter 13: Oscillations and Waves
Pearson Physics by James S. Walker
92. Picture the Problem: An observer sitting on a dock measures the frequency and wavelength of the water waves that
pass by.
Strategy: Convert the measurements of number of waves per measured time interval into period and frequency. Then
convert the crest-to-trough measurement into wavelength. Finally, use the relationships among frequency, wavelength,
and speed to find the speed of the waves.
Solution: 1. (a) :Calculate the period of the
waves from the measurements:
T=
45 s
= 4.1 s
11 waves
2. (b) Invert the period to find the frequency:
f =
1 11 waves
=
= 0.24 Hz
T
45 s
3. (c) The wavelength is the distance
between adjacent crests, or twice the
distance between a crest and a trough:
λ = 2 × Lcrest-to-trough = 2 (3.0 m ) = 6.0 m
4. (d) Calculate the wave speed:
v = f λ = (0.24 Hz )(6.0 m ) = 1.4 m/s
Insight: It is pretty easy to measure the frequency of waves by counting crests while tracking time with a stopwatch.
Measuring the wavelength is much more difficult because the waves are moving.
93. Picture the Problem: Several waves are depicted in the provided
figure. The waves have different wavelengths and amplitudes.
Strategy: To determine which wave is the sum of the others, recall that
wave amplitudes add together. At some point two crests will add
together and make a wave amplitude that is larger than either of the
other amplitudes. Identify the resultant wave as the one that has an
amplitude that is larger than the others.
Solution: The blue wave has an amplitude that is greater than the
amplitude of either the red wave or the green wave, so we conclude that
it is the resultant wave.:
Insight: There is an exception to the strategy outlined above. If the red wave and the green wave had the same
wavelength and were out of phase with each other (crests overlap troughs) the resultant wave would be zero
everywhere. That would be an example in which a resultant wave does not have an amplitude that is larger than
either of the two waves that are summed. If the waves have different wavelengths the above strategy always works.
94. Picture the Problem: Two waves (depicted with a brown color)
are added together.
Strategy: The resultant wave has an amplitude that is the sum of
the amplitudes of the two brown waves.
Solution: Where the two brown waves are both zero, the sum of
their amplitudes is also zero. For the leftmost pulses, the positive
pulse from the upper brown wave is cancelled by the negative
pulse of the lower; the resultant is zero. For the rightmost pulses,
the negative pulse of the lower brown wave half cancels the
double-amplitude of the upper brown wave. But the lower brown
wave amplitude goes to zero halfway through the upper brown
wave pulse, so the remainder of the resultant wave mimics the
upper brown wave. The resultant of the two brown waves is the
lower wave drawn at the right in blue.
Insight: One of the characteristics that distinguish waves from particles is that waves pass through each other but
particles collide. When waves pass through each other their amplitudes add by the principle of wave superposition.
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13 – 23
Chapter 13: Oscillations and Waves
Pearson Physics by James S. Walker
95. Picture the Problem: A string fixed at both ends vibrates. The three lowest-frequency
standing waves have frequencies 100 Hz, 200 Hz, and 300 Hz.
Strategy: Note the principles regarding standing waves on a string when answering the
conceptual questions.
Solution: 1. (a) The fundamental frequency is the lowest frequency standing wave that
is possible on a string that is fixed at both ends. We conclude that the fundamental
frequency of this string is 100 Hz.
2. (b) The three lowest harmonic standing waves on a string that is fixed at both ends
are depicted in the figure at the right. You can see from the first harmonic wave that
half a wavelength spans the length of the string. We conclude that the wavelength of
the first harmonic is twice the length of the string, or 2.0 m.
Insight: The 300-Hz standing wave is the third harmonic for this string, and its
wavelength is two-thirds the length of the string, or 0.67 m.
96. Picture the Problem: The wave speed and fundamental frequency are measured for a string that is fixed at both ends.
Strategy: Use the relationships among frequency, wavelength, and speed to find the wavelength of the waves. Then use
the relationship between string length and wavelength for the fundamental standing wave to find the string length.
Solution 1. Calculate the wavelength:
λ=
2. The string length is half the wavelength
of the fundamental standing wave:
L=
v 32 m/s
=
= 0.492 m
f
65 Hz
λ
2
=
0.492 m
= 0.246 m = 0.25 m
2
Insight: If instead the frequency of the second harmonic standing wave had been measured as 65 Hz, the string length
would have been equal to the wavelength of 0.49 m.
97. Picture the Problem: The string length and fundamental frequency are measured for a string that is fixed at both ends.
Strategy: Use the relationships among string length and wavelength for the fundamental standing wave to find the
wavelength of the standing wave. Then use the relationship between frequency, wavelength, and speed to find the speed
of the waves.
Solution 1. The wavelength is twice the string
length for the fundamental standing wave:
λ = 2 L = 2 (0.83 m ) = 1.66 m
2. Calculate the wave speed:
v = f λ = (26 Hz )(1.66 m ) = 43 m/s
Insight: If instead the frequency of the second harmonic standing wave had been measured as 26 Hz, the wavelength
would have been equal to the string length 0.83 m and the wave speed would have been 22 m/s.
98. Picture the Problem: The string length and fundamental frequency are measured for a string that is fixed at both ends.
Strategy: Use the relationships among frequency, wavelength, and speed to find the speed of the waves. Then use the
relationship between string length and wavelength for the first harmonic (fundamental) standing wave to predict the
change in the fundamental frequency, noting that the speed of the waves will not change.
Solution 1. The wavelength is twice the string
length for the fundamental standing wave:
λ = 2 L = 2 (0.64 m ) = 1.28 m
2. Calculate the wave speed:
v = f λ = (41 Hz )(1.28 m ) = 52.5 m/s
3. (a) The wave speed on the string depends on the properties of the string such as tension and mass per unit length.
Because these do not change, the wave speed will not change when the string is made longer. Noting that v = f λ , we
can see that increasing the string length will increase the wavelength, but because v remains constant, the first harmonic
frequency f must decrease:
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13 – 24
Chapter 13: Oscillations and Waves
Pearson Physics by James S. Walker
4. (b) Calculate the frequency
given the new string length:
f =
v
λ
=
v
52.5 m/s
=
= 30 Hz
2 L 2 (0.88 m )
Insight: Because the frequency and wavelength are inverses of each other if the wave speed remains constant, the
percentage changes are not the same. In this example, a 37.5% increase in string length (0.64 m to 0.88 m) resulted
in a 27% decrease in the frequency (41 Hz to 30 Hz).
99. Picture the Problem: Two wave pulses travel toward each
other at time t = 0 with a propagation speed of
1.0 m/s. We will make sketches of the wave pulses at times
t = 1.0 s, 2.0 s, 2.5 s, 3.0 s, and 4.0 s.
Strategy: Move the pulses one meter forward for each second. When the pulses overlap, add their amplitudes.
Solution: Sketches for times t = 1.0 s, 2.0 s, 2.5 s, 3.0 s, and 4.0 s are shown below.
t = 2.0 s:
t = 1.0 s:
Add the amplitudes when the waves overlap at t = 2.5 s:
t = 4.0 s:
t = 3.0 s:
Insight: The pulses pass right through each other and continue their motion unchanged.
100. Picture the Problem: Two wave pulses travel toward
each other at time t = 0 with a propagation speed of
1.0 m/s. We will make sketches of the wave pulses
at times t = 1.0 s, 2.0 s, 2.5 s, 3.0 s, and 4.0 s.
Strategy: Move the pulses one meter forward for each second. When the pulses overlap, add their amplitudes.
Solution: Sketches for times t = 1.0 s, 2.0 s, 2.5 s, 3.0 s, and 4.0 s are shown below.
t = 1.0 s:
t = 2.0 s:
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13 – 25
Chapter 13: Oscillations and Waves
Pearson Physics by James S. Walker
Add the amplitudes when the waves overlap at t = 2.5 s:
t = 4.0 s:
t = 3.0 s:
Insight: The pulses pass right through each other and continue their motion unchanged.
101. Picture the Problem: A string of known length that is fixed at both
ends vibrates with a standing wave that has three antinodes.
Strategy: Because there are three antinodes this wave is oscillating in
the third harmonic. Use the relationship between wavelength and
string length to find the wavelength of the waves.
Solution: 1. (a) The number of antinodes equals the number of the
harmonic, so this standing wave is the third harmonic.
λ3 =
2. (b) The wavelength is two-thirds the length of the string:
2 L 2 (66 cm )
=
= 44 cm
n
3
Insight: The distance between each antinode corresponds to a half wavelength. Because there are three antinodes in the
full length, the wavelength is 2/3 of the string length.
102. Picture the Problem: An object undergoes simple harmonic motion with a period T. In the time T the object moves
through a distance 4A, where A is the amplitude.
Strategy: Use the principles of simple harmonic motion to answer the conceptual question.
Solution: An object that undergoes simple harmonic motion moves through a distance of 4A during each cycle, where A
is the amplitude of the motion. It therefore travels a distance 6A in the time 3T 2. We conclude that the distance
6A = 24 cm and that A = 4 cm.
Insight: Another approach is to note that the object travels a distance
24 cm
= 16 cm in one period of oscillation.
32
We can then set 16 cm = 4A and find that A = 4 cm.
103. Picture the Problem: An astronaut uses a Body Mass Measurement Device (BMMD) while in orbit.
Strategy: The BMMD can be treated as a mass on a spring. Use the equation for the period of a mass on a spring in
order to find the mass of the astronaut.
Solution: 1. Solve the period equation for the mass:
T = 2π
2. Substitute the numerical values:
m=
m
kT 2
⇒ m=
k
4π 2
(2600 N/m )(0.85 s)2
4π 2
= 48 kg
Insight: When using a BMMD an astronaut must remain as motionless as possible. Moving arms and legs at a different
frequency from the natural frequency of the spring can alter the reading. This astronaut has a small stature; while she is
weightless in orbit, on Earth’s surface she would weigh 48 kg × 2.2 lb/kg = 106 lb.
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13 – 26
Chapter 13: Oscillations and Waves
Pearson Physics by James S. Walker
104. Picture the Problem: Sunspots are dark spots on the surface of the Sun caused by magnetic fields in the Sun.
The density of sunspots varies in time with a regular period.
Strategy: The frequency is the number of cycles divided by the time.
Solution: Take the inverse of the period
and convert the units to hertz:
f =
1 ⎛ 1 ⎞⎛
yr
=⎜
⎜⎝
⎟
T ⎝ 11 yr ⎠ 3.156 × 107
⎞
−9
⎟ = 2.9 × 10 Hz
s⎠
Insight: Frequencies can range from very small (such as the frequency of the sunspot cycle) to very high frequencies
such as the frequency of gamma radiation (1025 Hz).
105. Picture the Problem: The Sun vibrates in a complicated three-dimensional pattern but with a known period.
Strategy: Calculate the inverse of the period to find the frequency.
Solution: Calculate the frequency:
f =
1
1
⎛ 1 min ⎞
=
⎜
⎟ = 0.0029 Hz
T 5.7 min ⎝ 60 s ⎠
Insight: By tracking the oscillation patterns in the Sun, helioseismologists can study the solar dynamics.
106. Picture the Problem: Two springs are attached to a mass m as shown in the figure
at the right. The springs attached to block 1 are connected in parallel while the
springs attached to block 2 are in a push/pull arrangement.
Strategy: Determine the effective force constants of the two springs connected in
the arrangements shown in the figure. Then use the relationship between the period
and the force constant to predict the effect upon T.
Solution: 1. (a) Twice as much force is required to stretch two springs connected in parallel (as in block 1) than is
required to stretch a single spring the same distance. However, block 2 experiences the very same restoring force as
block 1 because whenever one spring is stretched, the other is compressed, and the two forces add to make a double
force. This means that the effective force constants of the arrangements are identical. We conclude that the period of
block 1 is equal to the period of block 2.
2. (b) The best explanation is B. The two blocks experience the same restoring force for a given displacement from
equilibrium, and thus they have equal periods of oscillation. Statement A is true, but irrelevant because the springs for
block 2 are not connected in series because they are not connected to each other. Statement C is false because the forces
add, they don’t cancel.
Insight: When two springs are truly connected in series they will stretch twice as far as a single spring when the same
force is applied. This means their force constant is effectively half that of a single spring.
107. Picture the Problem: The length of the string, frequency of oscillation, and wave speed are measured for a standing
wave on a string that is fixed at both ends.
Strategy: The number of antinodes is equal to the harmonic number for a wave on a string. Use the relationship
between string length and wavelength for the fundamental frequency, together with the relationships among frequency,
wavelength, and speed to find the fundamental frequency. The harmonic number is the ratio of the oscillation frequency
and the fundamental frequency.
Solution: 1. Find the wavelength of
the fundamental standing wave:
λ = 2 L = 2 (1.33 m ) = 2.66 m
2. Find the frequency of the
fundamental standing wave:
f =
3. The harmonic number is the ratio of the frequency
to the fundamental frequency. It is also equal to the
number of antinodes in the standing wave pattern:
n=
v
λ
=
402 m/s
= 151 Hz
2.66 m
f n 603 Hz
=
= 4
f1 151 Hz
Insight: To increase the number of antinodes to 5, the frequency of oscillation would have to be increased by a factor of
5/4, from 603 Hz to 754 Hz.
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13 – 27
Chapter 13: Oscillations and Waves
Pearson Physics by James S. Walker
108. Picture the Problem: The figure shows the x-versus-t graph for a
mass that oscillates on a spring.
Strategy: Determine the period of motion, as well as the amplitude,
directly from the graph. Then use the expression for the period of a
mass on a spring to find the spring constant.
Solution: 1. (a) From the graph we can determine that the period, or time to complete one cycle, is 4.0 seconds.
2. (b) From the graph we can determine that the amplitude, or maximum distance from the equilibrium position,
is 0.50 meters.
3. (c) Use the expression for the period of a
mass on a spring to find the spring constant:
T = 2π
m
k
⇒ k = 4π 2
m
3.8 kg
= 4π 2
= 9.4 N/m
2
T
(4.0 s)2
Insight: A stiffer spring would exert a larger force on the mass, accelerating it at a higher rate, and producing a period
that is shorter than 4.0 s.
109. Picture the Problem: A mass on a spring oscillates with the
amplitude and period shown in the figure.
Strategy: The maximum force occurs at points of maximum
displacement. Calculate the maximum force from Hooke’s law and the
known maximum displacement. The spring constant was determined
in the previous question.
Solution: 1. (a) From the graph find the times of maximum displacement: t = 1.0 s, 3.0 s, 5.0 s. The
maximum force is exerted at these same times because the force is given by Hooke’s law F = − k x .
4. (b) From the graph find the times when the
displacement (and therefore the force) is zero:
t = 0, 2.0 s, 4.0 s, 6.0 s
5. (c) Use the maximum displacement together
with Hooke’s law to find the maximum force:
Fmax = − k xmax = − (9.38 N/m )(0.50 m ) = 4.7 N
Insight: At the endpoints of the motion the speed of the mass is zero, but the acceleration of the mass ( a = F m ) is a
maximum at that location because the force is a maximum.
110. Picture the Problem: A crow lands on a branch and it bobs up and down like a mass on a spring. When an eagle lands
on the same branch the period of the motion will be longer because the eagle is more massive.
Strategy: Use the mass of the crow and the period of oscillation to determine the spring force constant of the branch.
Calculate the mass of the eagle from the spring force constant of the branch and the period of the eagle’s oscillation.
2
m
⎛ 2π ⎞
⇒ k=⎜ ⎟ m
⎝ T ⎠
k
Solution: 1. (a) Solve the period equation
for the spring force constant:
T = 2π
2. Insert the mass and period:
⎛ 2π ⎞
k=⎜
⎝ 1.4 s ⎟⎠
3. (b) Solve the period equation for the mass:
T = 2π
4. Insert the spring force constant and the period:
⎛ 3.6 s ⎞
m=⎜
⎝ 2π ⎟⎠
2
(0.45 kg ) = 9.06 N/m =
9.1 N/m
2
m
⎛ T ⎞
⇒ m=⎜ ⎟ k
⎝ 2π ⎠
k
2
(9.06 N/m) =
3.0 kg
Insight: Even though the amplitudes of oscillation for the crow and the eagle could have been different, they do not
affect the period of motion. Therefore it is possible to use the oscillation of the branch to measure the mass of the eagle.
This eagle is a little on the small side, as the average mass of the American bald eagle is 5.5 kg (about 12 lb).
Copyright © 2014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13 – 28
Chapter 13: Oscillations and Waves
Pearson Physics by James S. Walker
111. Picture the Problem: A pendulum bob is free to swing using the full length of the
string on the right side of its oscillation. However, when the bob moves to the left
side of the oscillation a peg acts as a pivot point, decreasing the effective length of
the string.
Strategy: The period is the time for the pendulum to complete one full oscillation.
With the peg in place consider the oscillation to be in two parts: Half of the
oscillation will occur with a shorter string and half with the longer string. Since
the period is proportional to the square root of the string length, it takes a shorter
time for the oscillation with the peg.
Solution 1. (a) The period is less than the period of the same pendulum without the
peg because the peg shortens the pendulum length, and hence the period, for half
the cycle.
1⎛
⎞

2. (b) Write out the time for the left half of the oscillation:
=π
T = ⎜ 2π
2⎝
g ⎟⎠
g
1⎛
L⎞
L
=π
2π
⎜
⎟
g⎠
g
2⎝
3. Write out the time for the right half of the oscillation:
TL =
4. Add the two times to get the full period:
⎛ 
T = π⎜
+
⎝ g
5. (c) Evaluate the period for the specified lengths:
⎛ 0.25 m
1.0 m ⎞
+
= 1.5 s
T =π⎜
2
9.81 m/s 2 ⎟⎠
⎝ 9.81 m/s
Insight: The period without the peg would be T = 2π
L⎞
g ⎟⎠
1.0 m
= 2.0 s which is longer than the time with the peg.
9.81 m/s 2
112. Reports will vary. A typical resonator assembly is about 4 mm in length and oscillates at 32,768 Hz. Quartz watches are
accurate because they produce a very precise frequency. The first quartz clock was invented in 1927 at Bell Telephone
Laboratories.
113. Some fine crystal wine glasses oscillate at a natural frequency within the audible range produced by a human voice.
If a person sings this note loudly, the sound wave in air drives the standing wave in the glass, and the amplitude of the
standing wave in the glass increases until the glass shatters.
114. Picture the Problem: A cricket chirps at a rate that is governed by the air temperature (Dolbear’s law).
Strategy: Solve Dolbear’s law (N = T – 39) for T using the given number of chirps N.
35 + 39 chirps 13 s
T = N + 39 =
= 74 °F
Solution: Solve Dolbear’s law for T. The calculated
1 chirp/13 s/ °F
answer corresponds to choice C.
Insight: You can use Dolbear’s law to verify that a temperature of 84°F will produce 45 chirps in 13 s.
115. Picture the Problem: A cricket chirps at a rate that is governed by the air temperature (Dolbear’s law).
Strategy: The frequency at any temperature can be obtained by dividing the number of chirps (N = T − 39) by
13 seconds.
Solution: Write the frequency
as a function of temperature.
The calculated answer
corresponds to choice B.
f =
N T − 39 (68 °F)(1 chirp/ °F) − 39 chirps
=
=
= 2.2 Hz
t
13 s
13 s
Insight: Many wind-up alarm clocks tick at 2 Hz, about the same frequency as this rapidly chirping cricket.
Copyright © 2014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13 – 29