Chapter 15 Traveling Waves Conceptual Pr

1451
Chapter 15
Traveling Waves
Conceptual Problems
1 • [SSM] A rope hangs vertically from the ceiling. A pulse is sent up
the rope. Does the pulse travel faster, slower, or at a constant speed as it moves
toward the ceiling? Explain your answer.
Determine the Concept The speed of a transverse wave on a uniform rope
increases with increasing tension. The waves on the rope move faster as they
move toward the ceiling because the tension increases due to the weight of the
rope below the pulse.
2 • A pulse on a horizontal taut string travels to the right. If the rope’s
mass per unit length decreases to the right, what happens to the speed of the pulse
as it travels to the right? (a) It slows down. (b) It speeds up. (c) Its speed is
constant. (d) You cannot tell from the information given.
Determine the Concept The speed v of a pulse on the string varies with the
tension FT in the string and its mass per unit length μ according to μ
T
Fv =.
Because the rope’s mass per unit length decreases to the right, the speed of the
pulse increases.
()
bis correct.
3 • As a sinusoidal wave travels past a point on a taut string, the arrival
time between successive crests is measured to be 0.20 s. Which of the following is
true? (a) The wavelength of the wave is 5.0 m. (b) The frequency of the wave is
5.0 Hz. (c) The velocity of propagation of the wave is 5.0 m/s. (d) The
wavelength of the wave is 0.20 m. (e) There is not enough information to justify
any of these statements.
Determine the Concept The distance between successive crests is one
wavelength and the time between successive crests is the period of the wave
motion. Thus, T = 0.20 s and f = 1/T = 5.0 Hz. )(bis correct.
4 • Two harmonic waves on identical strings differ only in amplitude.
Wave A has an amplitude that is twice that of wave B’s. How do the energies of
these waves compare? (a)
E
A
=
E
B. (b)
E
A
=
2
E
B. (c)
E
A
=
4
E
B. (d) There is
not enough information to compare their energies.
Picture the Problem The average energy transmitted by a wave on a string is
proportional to the square of its amplitude and is given by
(
)
xAE ΔΔ 22
2
1
av
μω =
where A is the amplitude of the wave,
μ is the linear density (mass per unit length)

Chapter 15
1452
of the string,
ω is the angular frequency of the wave, and Δx is the length of the
string.
Because the waves on the strings
differ only in amplitude, the energy
of the wave on string A is given by:
2
AA cAE =
Express the energy of the wave on
string B:
2
BB cAE =
Divide the first of these equations by
the second and simplify to obtain:
2
B
A
2
B
2
A
B
A⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
== A
A
cA
cA
E
E
Because AA = 2AB:
4
22
B
B
B
A=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
=A
A
E
E⇒
(
)
cis correct.
5 • [SSM] To keep all of the lengths of the treble strings (unwrapped
steel wires) in a piano all about the same order of magnitude, wires of different
linear mass densities are employed. Explain how this allows a piano manufacturer
to use wires with lengths that are the same order of magnitude.
Determine the Concept The resonant (standing wave) frequencies on a string are
inversely proportional to the square root of the linear density of the string
(
)
λμ
T
Tf =. Thus extremely high frequencies (which might otherwise require
very long strings) can be accommodated on relatively short strings if the strings
are linearly less dense than the high frequency strings. High frequencies are not a
problem as they use short strings anyway.
6 • Musical instruments produce sounds of widely varying frequencies.
Which sounds waves have the longer wavelengths? (a) The lower frequencies.
(b) The higher frequencies. (c) All frequencies have the same wavelength.
(d) There is not enough information to compare the wavelengths of the different
frequency sounds.
Determine the Concept Once the sound has been produced by a vibrating string,
membrane, or air column, the speed with which it propagates is the product of its
frequency and wavelength
()
.
λ
fv
=
For a given medium and temperature, v is
constant. Hence the wavelength and frequency for a given sound are inversely
proportional and
()
ais correct.

Traveling Waves
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7 • In Problem 6, which sound waves have the higher speeds? (a) The
lower frequency sounds. (b) The higher frequency sounds. (c) All frequencies
have the same wave speed. (d) There is not enough information to compare their
speeds.
Determine the Concept Once the sound has been produced by a vibrating string,
membrane, or air column, the wave speed with which it propagates depends on
the properties of the medium in which it is propagating and is independent of the
frequency and wavelength of the sound.
(
)
cis correct.
8 • Sound travels at 343 m/s in air and 1500 m/s in water. A sound of
256 Hz is made under water, but you hear the sound while walking along the side
of the pool. In the air, the frequency is (a) the same, but the wavelength of the
sound is shorter, (b) higher, but the wavelength of the sound stays the same,
(c) lower, but the wavelength of the sound is longer, (d) lower, and the
wavelength of the sound is shorter, (e) the same, and the wavelength of the sound
stays the same.
Determine the Concept In any medium, the wavelength, frequency, and speed of
a sound wave are related through
λ = v/f. Because the frequency of a wave is
determined by its source and is independent of the nature of the medium, if v is
greater in water than in air,
λ will be shorter in air than in water. )(ais correct.
9 • While out on patrol, the battleship Rodger Young hits a mine, begins to
burn, and ultimately explodes. Sailor Abel jumps into the water and begins
swimming away from the doomed ship, while Sailor Baker gets into a life raft.
Comparing their experiences later, Abel tells Baker, ″I was swimming
underwater, and heard a big explosion from the ship. When I surfaced, I heard a
second explosion. What do you think it could be?″ Baker says, ″I think it was
your imagination—I only heard one explosion. ″ Explain why Baker only heard
one explosion, while Abel heard two.
Determine the Concept There was only one explosion. Sound travels faster in
water than air. Abel heard the sound wave in the water first, then, surfacing,
heard the sound wave traveling through the air, which took longer to reach him.
10 • True or false: A 60-dB sound has twice the intensity of a 30-dB sound.
Picture the Problem The intensity level β , measured in decibels, is given by
()()
0
logdB10 II=
β
where I0 = 10−12 W/m2 is defined to be the threshold of
hearing.

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1454
Express the intensity level of the 60-
dB sound:
()
0
60
logdB10dB60 I
I
= ⇒ 0
6
60 10 II =
Express the intensity level of the 30-
dB sound:
()
0
30
logdB10dB30 I
I
= ⇒ 0
3
30 10 II =
Because I60 = 103I30, the statement is false.
11 • [SSM] At a given location, two harmonic sound waves have the
same displacement amplitude, but the frequency of sound A is twice the
frequency of sound B. How do their average energy densities compare? (a) The
average energy density of A is twice the average energy density of B. (b) The
average energy density of A is four times the average energy density of B. (c) The
average energy density of A is 16 times the average energy density of B. (d) You
cannot compare the average energy densities from the data given.
Determine the Concept The average energy density of a sound wave is given by
2
0
2
2
1
av s
ρωη =where ρ is the average density of the medium, s0 is the
displacement amplitude of the molecules making up the medium, and
ω is the
angular frequency of the sound waves.
Express the average energy density
of sound A:
2
A,0
2
AA
2
1
A av, s
ωρη =
The average energy density of sound
B is given by:
2
B,0
2
BB
2
1
B av, s
ωρη =
Dividing the first of these equation
by the second yields: 2
B,0
2
BB
2
1
2
A,0
2
AA
2
1
B av,
A av,
s
s
ωρ
ωρ
η
η
=
Because the sound waves are
identical except for their frequencies:
2
B
A
2
B
A
2
B
2
A
B av,
A av,
2
2⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
== f
f
f
f
π
π
ω
ω
η
η
Because fA = 2fB:
4
22
B
B
B av,
A av, =
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
=f
f
η
η
⇒
(
)
bis correct.
12 • At a given location, two harmonic sound waves have the same
frequency but the amplitude of sound A is twice the amplitude of sound B. How
do their average energy densities compare? (a) The average energy density of A is
twice the average energy density of B. (b) The average energy density of A is four
times the average energy density of B (c) The average energy density of A is 16

Traveling Waves
1455
times the average energy density of B (d) You cannot compare the average energy
densities from the data given.
Determine the Concept The average energy density of a sound wave is given by
2
0
2
2
1
av s
ρωη =where ρ is the average density of the medium, s0 is the
displacement amplitude of the molecules making up the medium, and
ω is the
angular frequency of the sound waves.
Express the average energy density
of sound A:
2
A,0
2
AA
2
1
A av, s
ωρη =
The average energy density of sound
B is given by:
2
B,0
2
BB
2
1
B av, s
ωρη =
Dividing the first of these equation
by the second yields: 2
B,0
2
BB
2
1
2
A,0
2
AA
2
1
B av,
A av,
s
s
ωρ
ωρ
η
η
=
Because the sound waves are
identical except for their
displacement amplitudes:
2
B,0
A,0
2
B,0
2
A,0
B av,
A av, ⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
== s
s
s
s
η
η
Because B,0A,0 2ss =
4
22
B,0
B,0
B av,
A av, =
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
=s
s
η
η
⇒
()
bis correct.
13 • What is the ratio of the intensity of normal conversation to the sound
intensity of a soft whisper (at a distance of 5.0 m)? (a) 103, (b) 2, (c) 10−3, (d) 1/2.
Hint: See Table 15-1.
Determine the Concept The problem is asking us for the ratio of the intensities
corresponding to normal conversation and a soft whisper. Table 15-1 includes the
quantities we need in order to find this ratio.
Use Table 15-1 to find the ratio of
I/I0, where I0 is the threshold of
hearing, for normal conversation:
6
onconversati
normal
0
10=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
I
I
Use Table 15-1 to find the ratio of
I/I0 for a soft whisper:
3
whisper
soft
0
10=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
I
I