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# Chapter 15 Traveling Waves Conceptual Pr

publicité ```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
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 v = FT μ .
Because the rope’s mass per unit length decreases to the right, the speed of the
pulse increases. (b ) is 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. (b) is 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) EA = EB . (b) EA = 2EB . (c) EA = 4EB . (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 (ΔE )av = 12 μω 2 A 2 Δx
where A is the amplitude of the wave, μ is the linear density (mass per unit length)
1451
1452 Chapter 15
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:
E A = cAA2
Express the energy of the wave on
string B:
E B = cAB2
Divide the first of these equations by
the second and simplify to obtain:
EA cAA2 ⎛ AA
=
=⎜
EB cAB2 ⎜⎝ AB
Because AA = 2AB:
E A ⎛ 2 AB ⎞
⎟ = 4 ⇒ (c ) is correct.
=⎜
E B ⎜⎝ AB ⎟⎠
⎞
⎟⎟
⎠
2
2
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
f = TT μ λ . 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 (v = fλ ). For a given medium and temperature, v is
constant. Hence the wavelength and frequency for a given sound are inversely
proportional and (a ) is correct.
Traveling Waves 1453
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. (c ) is 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. (a ) is 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
β = (10 dB) log(I I 0 ) where I0 = 10−12 W/m2 is defined to be the threshold of
hearing.
1454 Chapter 15
Express the intensity level of the 60dB sound:
60dB = (10 dB) log
I 60
⇒ I 60 = 106 I 0
I0
Express the intensity level of the 30dB sound:
30 dB = (10 dB) log
I 30
⇒ I 30 = 103 I 0
I0
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
η av = 12 ρω 2 s02 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:
η av, A = 12 ρ Aω A2 s02, A
The average energy density of sound
B is given by:
η av, B = 12 ρ Bω B2 s02, B
Dividing the first of these equation
by the second yields:
η av, A 12 ρ Aω A2 s02, A
=
η av, B 12 ρ Bω B2 s02, B
Because the sound waves are
identical except for their frequencies:
η av, A ω A2 ⎛ 2πf A ⎞ ⎛ f A ⎞
⎟ =⎜ ⎟
=
=⎜
η av, B ω B2 ⎜⎝ 2πf B ⎟⎠ ⎜⎝ f B ⎟⎠
Because fA = 2fB:
η av, A ⎛ 2 f B ⎞
⎟ = 4 ⇒ (b ) is correct.
=⎜
η av, B ⎜⎝ f B ⎟⎠
2
2
2
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
η av = 12 ρω 2 s02 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:
η av, A = 12 ρ Aω A2 s02, A
The average energy density of sound
B is given by:
η av, B = 12 ρ Bω B2 s02, B
Dividing the first of these equation
by the second yields:
η av, A 12 ρ Aω A2 s02, A
=
η av, B 12 ρ Bω B2 s02, B
Because the sound waves are
identical except for their
displacement amplitudes:
η av, A s02, A ⎛ s0, A ⎞
⎟
=
=⎜
η av, B s02, B ⎜⎝ s0, B ⎟⎠
Because s0, A = 2s0, B
η av, A ⎛ 2s0, B ⎞
⎟ = 4 ⇒ (b ) is correct.
=⎜
η av, B ⎜⎝ s0, B ⎟⎠
2
2
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:
⎛ I ⎞
⎜⎜ ⎟⎟
= 10 6
normal
⎝ I 0 ⎠ conversati
on
Use Table 15-1 to find the ratio of
I/I0 for a soft whisper:
⎛I ⎞
⎜⎜ ⎟⎟
= 103
⎝ I 0 ⎠soft
whisper
1456 Chapter 15
Dividing the first of these equation
by the second yields:
⎛ I ⎞
⎜⎜ ⎟⎟
normal
⎝ I 0 ⎠ conversati
on
⎛ I ⎞
⎜⎜ ⎟⎟
⎝ I 0 ⎠ soft
whisper
(a )
=
10 6
= 10 3
3
10
is correct.
14 •
What is the ratio of the intensity level of normal conversation to the
sound intensity level 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 sound
intensity levels (decibel levels) corresponding to normal conversation and a soft
whisper. Table 15-1 includes the quantities we need in order to find this ratio.
From Table 15-1, the sound intensity
level corresponding to normal
conversation is:
β normal
From Table 15-1, the sound intensity
level corresponding to a soft whisper
is:
β soft
Dividing the first of these equation
by the second yields:
β normal
= 60 dB
conversation
= 30 dB
whisper
conversation
β soft
=
60 dB
=2
30 dB
whisper
(b )
is correct.
15 •
To increase the sound intensity level by 20 dB requires the sound
intensity to increase by what factor? (a) 10, (b) 100, (c) 1000, (d) 2.
Determine the Concept The sound intensity level is given by
β = (10 dB) log(I I 0 ) where I is the intensity of a given sound and I0 is the
intensity of the threshold of hearing.
⎛ I1 ⎞
⎟⎟
⎝ I0 ⎠
The sound intensity level for a sound
whose intensity is I1 is given by:
β1 = (10 dB) log⎜⎜
The sound intensity level for a sound
whose intensity is I2 is given by:
β 2 = (10 dB) log⎜⎜
⎛ I2 ⎞
⎟⎟
I
⎝ 0⎠
Traveling Waves 1457
Subtract the second equation from
the first and simplify to obtain:
⎛
⎛ I1 ⎞ ⎞
⎟⎟ ⎟
⎟
⎝ I2 ⎠⎠
β1 − β 2 = (10 dB) ⎜⎜ log⎜⎜
⎝
β −β
Solving for I1/I2 yields:
1
2
I1
= 10 10 dB
I2
For a 20-dB increase in the sound
intensity level:
20 dB
I1
10 dB
= 10
= 10 2 ⇒ (b ) is correct.
I2
16 •
You are using a hand-held sound level meter to measure the intensity
level of the roars produced by a lion prowling in the high grass. To decrease the
measured sound intensity level by 20 dB requires the lion move away from you
until its distance from you has increased by what factor? (a) 10, (b) 100, (c) 1000,
(d) You cannot tell the required distance from the data given.
Determine the Concept The sound intensity level is given by
β = (10 dB) log(I I 0 ) where I is the intensity of a given sound and I0 is the
intensity of the threshold of hearing. The intensity of the a given sound varies
with distance from its source according to I = Pav (4π r 2 ) .
⎛ I1 ⎞
⎟⎟
I
⎝ 0⎠
The sound intensity level for a sound
whose intensity is I1 is given by:
β1 = (10 dB) log⎜⎜
The sound intensity level for a sound
whose intensity is I2 is given by:
β 2 = (10 dB) log⎜⎜
Subtract the second equation from
the first and simplify to obtain:
β1 − β 2 = (10 dB) ⎜⎜ log⎜⎜
⎛ I2 ⎞
⎟⎟
⎝ I0 ⎠
⎛
⎝
⎛ I1 ⎞ ⎞
⎟⎟ ⎟
⎟
⎝ I2 ⎠⎠
β −β
Solving for I1/I2 yields:
1
2
I1
= 10 10 dB
I2
Letting r1 be the lion’s initial
distance from you and r2 be the
lion’s final distance from you,
substitute for I1 and I2 and simplify
to obtain:
Pav
β1 − β 2
β1 − β 2
r
4π r12 r22
= 2 = 10 10 dB ⇒ 2 = 10 10 dB
Pav
r1
r1
4π r22
1458 Chapter 15
For a 20-dB decrease in the sound
intensity level:
20 dB
r2
= 10 10 dB = 10 ⇒ (a ) is correct.
r1
17 •
One end of a very light (but strong) thread is attached to an end of a
thicker and denser cord. The other end of the thread is fastened to a sturdy post
and you pull the other end of the cord so the thread and cord are taut. A pulse is
sent down the thicker, denser cord. True or false:
(a) The pulse that is reflected back from the thread-cord attachment point is
inverted compared to the initial incoming pulse.
(b) The pulse that continues past the thread-cord attachment point is not inverted
compared to the initial incoming pulse.
(c) The pulse that continues past the thread-cord attachment point has an
amplitude that is smaller than the pulse that is reflected.
(a) False. Because the reflection medium (cord) in which the pulse travels initially
has a greater linear density than the transmission medium (thread), there is no
phase shift in the reflected pulse.
(b) True. Because the thread-cord attachment point of the media is more like a
loose end than a fixed end, the pulse transmitted into the light thread is in phase
with the incoming pulse in the thicker, denser cord.
(c) True. Because the string is attached to a thicker, denser cord, the reflected
pulse behaves almost as though it was reflected from a fixed end and is, therefore,
inverted. The pulse in the light thread is in phase with the incoming pulse in the
thicker, denser cord. Because energy is conserved and there is some energy
transmitted and some reflected, the transmitted wave cannot have an amplitude as
large as the amplitude of the pulse in the thicker, denser cord.
18 •
Light traveling in air strikes a glass surface at an incident angle of 45&deg;.
True or false: (a) The angle between the reflected light ray and the incident ray is
90&deg;. (b) The angle between the reflected light ray and the refracted light ray is less
than 90&deg;.
Determine the Concept The ray diagram shows the relationship between the
incident, reflected, and refracted rays.
Traveling Waves 1459
vair &gt; vglass
Incident ray
air
glass
45&deg;
θ1
Refracted ray
Reflected ray
(a) True. Because the angle of incidence is equal to the angle of reflection, the
sum of these angles in this example is 90&deg;. Note: this is a special case and is not
generally true.
(b) False. Because light travels slower in glass than in air, the refracted ray makes
an angle of less than 45&deg; with the normal. Hence the angle it makes with the
reflected ray is greater than 90&deg;.
19 •
[SSM] Sound waves in air encounter a 1.0-m wide door into a
classroom. Due to the effects of diffraction, the sound of which frequency is least
likely to be heard by all the students in the room, assuming the room is full?
(a) 600 Hz, (b) 300 Hz, (c) 100 Hz, (d) All the sounds are equally likely to be
heard in the room. (e) Diffraction depends on wavelength not frequency, so you
cannot tell from the data given.
Determine the Concept If the wavelength is large relative to the door, the
diffraction effects are large and the waves spread out as they pass through the
door. Because we’re interested in sounds that are least likely to be heard
everywhere in the room, we want the wavelength to be short and the frequency to
be high. Hence (a ) is correct.
20 •
Microwave radiation in modern microwave ovens has a wavelength on
the order of centimeters. Would you expect significant diffraction if this radiation
was aimed at a 1.0-m wide door? Explain.
Determine the Concept No. Because the wavelength of the radiation is small
relative to the door, the diffraction effects are small and the waves do not spread
out significantly as they pass through the door.
1460 Chapter 15
21 ••
[SSM] Stars often occur in pairs revolving around their common
center of mass. If one of the stars is a black hole, it is invisible. Explain how the
existence of such a black hole might be inferred by measuring the Doppler
frequency shift of the light observed from the other, visible star.
Determine the Concept The light from the visible star will be shifted about its
mean frequency periodically due to the relative approach toward and recession
away from Earth as the star revolves about the common center of mass.
22 ••
Figure 15-30 shows a wave pulse at time t = 0 moving to the right.
(a) At this particular time, which segments of the string are moving up?
(b) Which segments are moving down? (c) Is there any segment of the string at
the pulse that is instantaneously at rest? Answer these questions by sketching the
pulse at a slightly later time and a slightly earlier time to see how the segments of
the string are moving.
Determine the Concept The graph shown below shows the pulse at an earlier
time (t &lt; 0) and later time (t &gt; 0). (a) One can see that at t = 0, the portion of the
string between 1 cm and 2 cm is moving down, (b) the portion between 2 cm and
3 cm is moving up, and (c) the string at x = 2 cm is instantaneously at rest.
t=0
t&gt;0
y
t&lt;0
0
1
2
3
4
5
6
7
8
9
10
x , cm
23 ••
Make a sketch of the velocity of each string segment versus position
for the pulse shown in Figure 15-30.
Determine the Concept The velocity of the string at t = 0 is shown. Note that the
velocity is negative for 1 cm &lt; x &lt; 2 cm and is positive for 2 cm &lt; x &lt; 3 cm.
vy
Traveling Waves 1461
0
1
2
3
4
5
6
7
8
9
10
x , cm
24 ••
An object of mass m hangs on a very light rope that is connected to the
ceiling. You pluck the rope just above the object, and a wave pulse travels up to
the ceiling and back. Compare the round-trip time for such a wave pulse to the
round-trip time of a wave pulse on the same rope if an object of mass 9m is hung
on the rope instead. (Assume the rope does not stretch, that is, the mass-to-ceiling
distance is the same in each case.)
Determine the Concept If the mass providing the tension in the rope is increased
by a factor of n, then the tension in the rope increases by the same factor and the
speed of the pulse on the rope increases by a factor of n and the round trip time
for the pulse is decreased by a factor of 1 n .
Let l represent the distance the pulse
travels. Then the time for the round
trip is given by:
t=
l
v
Express the speed of the wave when
the object of mass m provides the
tension in the rope:
vm =
Express the speed of the wave when
the object of mass 9m provides the
tension in the rope:
v9 m =
Substituting for v9 m in equation (1)
yields:
t9 m =
(1)
FT, m
μ
FT,9 m
μ
l
FT,9 m
=l
μ
FT,9 m
(2)
μ
Similarly, the travel time when the
object whose mass is m is providing
the tension is given by:
tm =
l
FT, m
μ
=l
μ
FT, m
(3)
1462 Chapter 15
Divide equation (2) by equation (3)
and simplify to obtain:
t9m
=
tm
Because FT,9 m = 9 FT, m :
FT, m
t9 m
=
= 1 ⇒ t9 m =
tm
9 FT, m 3
FT, m
FT,9 m
t
1
3 m
The explosion of a depth charge beneath the surface of the water is
25 ••
recorded by a helicopter hovering above its surface, as shown in Figure 15-31.
Along which path⎯A, B, or C⎯will the sound wave take the least time to reach
the helicopter? Explain why you chose the path you did.
Determine the Concept Path C. Because the wave speed is highest in the water,
and more of path C is underwater than are paths A or B, the sound wave will
spend the least time on path C.
26 ••
Does a speed of Mach 2 at an altitude of 60,000 feet mean the same as
a speed of Mach 2 near ground level? Explain clearly.
Determine the Concept No. The term Mach 2 means that the speed is twice the
speed of sound at a given altitude. Because the speed of sound is lower at high
altitudes than at ground level (due to lower density and colder temperature), Mach
2 means less than twice the speed of sound at sea level.
Estimation and Approximation
27 ••
Many years ago, Olympic 100 m dashes were started by the sound
from a starter’s pistol, with the starter positioned several meters down the track,
just on the inside of the track. (Today, the pistol that is used is often only a
trigger, which is used to electronically activate speakers behind each sprinter’s
starting blocks. This method avoids the problem of one runner hearing sound
before the other runners.) Estimate the time advantage the inside lane (relative to
the runner at the outside lane of 8 runners) would have if all runners started when
they heard the sound from the starter’s pistol.
Picture the Problem Due to the differences in the distances from the starter to
the inside- and outside-lane sprinters, the sound of the starting pistol took more
time to reach the outside-lane sprinter than it did to reach the inside-lane sprinter.
Express the difference in time for the
two sprinters in terms of the
distances the sound must travel and
the speed of sound:
Δt = touside − tinside
lane
=
l outside
lane
v
lane
−
l inside
lane
v
Traveling Waves 1463
Assuming that the separation of the
inside- and outside-lane sprinters is
w:
l outside = l 2inside + w2
lane
lane
Substitute for l outside to obtain:
lane
Assuming that the separation of the
inside- and outside-lane sprinters is
7.0 m and that the starter is 5.0 m
from the inside-lane sprinter yields:
Δt =
Δt =
l 2inside + w2 − l inside
lane
lane
v
(5.0 m )2 + (7.0 m )2 − 5.0 m
343 m/s
= 11 ms
Remarks: Because a 100-m sprint is often decided by one or two hundredths
of a second, this difference could be significant. This difference led to the
invention of the electronic starter’s sound.
28 ••
Estimate the speed of the bullet as it passes through the helium balloon
in Figure 15-32. Hint: A protractor would be beneficial.
Picture the Problem You can use a protractor to measure the angle of the shock
cone and then estimate the speed of the bullet using sin θ = v u . The speed of
sound in helium at room temperature (293 K) is 977 m/s.
Relate the speed of the bullet u to the
speed of sound v in helium and the
angle of the shock cone θ :
sin θ =
Measuring θ yields:
θ ≈ 70&deg;
Substitute numerical values and
evaluate u:
u=
v
v
⇒u =
u
sin θ
977 m/s
= 1.0 km/s
sin 70&deg;
29 ••
The new student townhouses at a local college are in the form of a
semicircle half-enclosing the track field. To estimate the speed of sound in air, an
ambitious physics student stood at the center of the semicircle and clapped his
hands rhythmically at a frequency at which he could not hear the echo of the clap,
because it reached him at the same time as his next clap. This frequency was
about 2.5 claps/s. Once he established this frequency, he paced off the distance to
the townhouses, which was 30 double strides. Assuming that the length of each
stride is equal to half his height (5 ft 11 in), estimate the speed of sound in air
using these data. How far off is your estimation from the commonly accepted
value of 343 m/s?
1464 Chapter 15
Picture the Problem Let d be the distance to the townhouses. We can relate the
speed of sound to the distance to the townhouses to the frequency of the clapping
for which no echo is heard. 5 ft 11 in is equal to 1.8 m.
Relate the speed of sound to the
distance it travels to the townhouses
and back to the elapsed time:
Express d in terms of the number of
strides and distance covered per
stride:
v=
2d
Δt
d = (30 strides )(1.8 m/stride ) = 54 m
Relate the elapsed time Δt to the
frequency f of the clapping:
Δt =
Substitute numerical values and
evaluate v:
v=
The percent difference between
this result and 343 m/s is:
% diff =
1
1
=
= 0.40 s
f 2.5 claps/s
2(54 m )
= 270 m/s = 0.27 km/s
0.40 s
343 m/s − 270 m/s
= 21%
343 m/s
Speed of Waves
30 •
(a) The bulk modulus of water is 2.00 &times; 109 N/m2. Use this value to
find the speed of sound in water. (b) The speed of sound in mercury is 1410 m/s.
What is the bulk modulus of mercury (ρ = 13.6 &times; 103 kg/m3)?
Picture the Problem The speed of sound in a fluid is given by v = B ρ where
B is the bulk modulus of the fluid and ρ is its density.
(a) Express the speed of sound in
water in terms of its bulk modulus:
v=
Substitute numerical values and
evaluate v:
v=
B
ρ
2.00 &times; 10 9 N/m 2
= 1.414 km/s
1.00 &times; 10 3 kg/m 3
= 1.41 km/s
(b) Solving v = B ρ for B yields:
B = ρv 2
Substitute numerical values and
evaluate B:
B = 13.6 &times; 10 3 kg/m 3 (1410 m/s )
(
)
= 2.70 &times; 1010 N/m 2
2
Traveling Waves 1465
31 •
Calculate the speed of sound waves in hydrogen gas (M = 2.00 g/mol
and γ = 1.40) at T = 300 K.
Picture the Problem The speed of sound in a gas is given by v = γRT M where
R is the gas constant, T is the absolute temperature, M is the molecular mass of the
gas, and γ is a constant that is characteristic of the particular molecular structure
of the gas. Because hydrogen gas is diatomic, γ = 1.4.
Express the dependence of the speed
of sound in hydrogen gas on the
absolute temperature:
Substitute numerical values and
evaluate v:
v=
v=
γRT
M
1.4(8.314 J/mol ⋅ K )(300 K )
2.00 &times; 10 −3 kg/mol
= 1.32 km/s
32 •
A 7.00-m-long string has a mass of 100 g and is under a tension of
900 N. What is the speed of a transverse wave pulse on this string?
Picture the Problem The speed of a transverse wave pulse on a string is given by
v = FT μ where FT is the tension in the string, m is its mass, L is its length, and
μ is its mass per unit length.
Express the dependence of the speed
of the pulse on the tension in the
string:
Substitute numerical values and
evaluate v:
v=
FT
μ
where μ is the mass per unit
length of the string.
v=
900 N
= 251 m/s
0.100 kg
7.00 m
33 ••
[SSM] (a) Compute the derivative of the speed of a wave on a string
with respect to the tension dv/dFT, and show that the differentials dv and dFT obey
dv v = 12 dFT FT . (b) A wave moves with a speed of 300 m/s on a string that is
under a tension of 500 N. Using the differential approximation, estimate how
much the tension must be changed to increase the speed to 312 m/s. (c) Calculate
ΔFT exactly and compare it to the differential approximation result in Part (b).
Assume that the string does not stretch with the increase in tension.
1466 Chapter 15
Picture the Problem (a) The speed of a transverse wave on a string is given by
v = FT μ where FT is the tension in the wire and μ is its linear density. We can
differentiate this expression with respect to FT and then separate the variables to
show that the differentials satisfy dv v = 12 dFT FT . (b) We’ll approximate the
differential quantities to determine by how much the tension must be changed to
increase the speed of the wave to 312 m/s. (c) We can use v = FT μ to obtain
an exact expression for ΔFT,
(a) Evaluate dv/dFT:
dv
d ⎡ FT ⎤ 1
1
1 v
= ⋅
=
⎥=
⎢
dF dFT ⎣ μ ⎦ 2 FT μ 2 FT
Separate the variables to obtain:
dv
1 dFT
=
v
2 FT
(b) Solve the equation derived in Part
(a) for dFT:
dFT = 2 FT
dv
v
Approximate dFT with ΔFT and dv
with Δv to obtain:
ΔFT = 2 FT
Δv
v
Substitute numerical values and
evaluate ΔFT:
⎛ 312 m/s − 300 m/s ⎞
⎟⎟
ΔFT = 2(500 N )⎜⎜
300 m/s
⎝
⎠
= 40 N
(c) The exact value for (ΔF)exact is
given by:
Express the wave speeds for the two
tensions:
Dividing the second equation by the
first and simplifying yields:
(ΔF )exact = FT,2 − FT,1
v1 =
FT,1
μ
and v2 =
(1)
FT,2
μ
FT,2
v2
=
v1
μ
FT,1
μ
=
FT,2
⎛v ⎞
⇒ FT,2 = FT,1 ⎜⎜ 2 ⎟⎟
FT,1
⎝ v1 ⎠
2
Traveling Waves 1467
Substituting for FT,2 in equation (1)
yields:
(ΔFT )exact
2
⎛v ⎞
= FF,1 ⎜⎜ 2 ⎟⎟ − FT,1
⎝ v1 ⎠
⎡⎛ v ⎞ 2 ⎤
= FF,1 ⎢⎜⎜ 2 ⎟⎟ − 1⎥
⎢⎣⎝ v1 ⎠
⎥⎦
Substitute numerical values and
evaluate (ΔFT)exact:
(ΔFT )exact
⎡⎛ 312 m/s ⎞ 2 ⎤
= (500 N )⎢⎜
⎟ − 1⎥
⎣⎢⎝ 300 m/s ⎠
⎦⎥
= 40.8 N
The percent error between the exact
and approximate values for ΔFT is:
(ΔFT )exact − ΔFT
(ΔFT )exact
=
40.8 N − 40.0 N
40.8 N
≈ 2%
34 ••
(a) Compute the derivative of the speed of sound in air with respect to
the absolute temperature, and show that the differentials dv and dT obey
dv v = 12 dT T . (b) Use this result to estimate the percentage change in the speed
of sound when the temperature changes from 0 to 27&deg;C. (c) If the speed of sound
is 331 m/s at 0&deg;C, estimate its value at 27&deg;C using the differential approximation.
(d) How does this approximation compare with the result of an exact calculation?
Picture the Problem The speed of sound in a gas is given by v = γRT M where
R is the gas constant, T is the absolute temperature, M is the molecular mass of the
gas, and γ is a constant that is characteristic of the particular molecular structure
of the gas. We can differentiate this expression with respect to T and then separate
the variables to show that the differentials satisfy dv v = 12 dT T . We’ll
approximate the differential quantities to estimate the percentage change in the
speed of sound when the temperature increases from 0 to 27&deg;C. Lacking
information regarding the nature of the gas, we can express the ratio of the speeds
of sound at 300 K and 273 K to obtain an expression that involves just the
temperatures.
(a) Evaluate dv/dT:
dv
d ⎡ γRT
=
⎢
dT dT ⎣ M
1v
=
2T
⎤ 1 M ⎛ γR ⎞
⎜ ⎟
⎥=
⎦ 2 γRT ⎝ M ⎠
1468 Chapter 15
Separate the variables to obtain:
dv
1 dT
=
v
2 T
(b) Approximate dT with ΔT and dv
with Δv and substitute numerical
values to obtain:
Δv 1 ⎛ 300 K − 273 K ⎞
= ⎜
⎟ = 5.0%
v
2⎝
273 K
⎠
(c) Using a differential
approximation, approximate the
speed of sound at 300 K:
v300 K ≈ v273 K + v273 K
Substitute numerical values and
evaluate v300 K:
(d) Use v = γRT M to express
the speed of sound at 300 K:
Use v = γRT M to express the
speed of sound at 273 K:
Divide the first of these equations by
the second and solve for and evaluate
v300 K:
Δv
v
⎛ Δv ⎞
= v273 K ⎜1 +
⎟
v ⎠
⎝
v300 K = (331 m/s )(1 + 0.0495)
= 347 m/s
v300 K =
v273 K =
v300 K
v273 K
γR(300 K )
M
γR(273 K )
M
γR(300 K )
=
300
M
=
273
γR(273 K )
M
and
v300 K = (331m/s )
300
= 347 m/s
273
Note that these two results agree to three significant figures.
35 ••• Derive a convenient formula for the speed of sound in air at
temperature t in Celsius degrees. Begin by writing the temperature as T = T0 +
ΔT, where T0 = 273 K corresponds to 0&deg;C and ΔT = t, the Celsius temperature.
The speed of sound is a function of T, v(T). To a first-order approximation, you
can write v(T ) ≈ v(T0 ) + (dv dT )T0 ΔT , where (dv / dT )T 0 is the derivative
evaluated at T = T0. Compute this derivative, and show that the result leads to
⎛
t ⎞
= (331 + 0.606t ) m/s.
v = (331 m / s)⎜ 1 +
⎝ 2T0 ⎟⎠
Traveling Waves 1469
Picture the Problem We can use the approximate expression for v(T) given in
the problem statement and the result of Problem 34 to show that
v = (331 + 0.606 t ) m/s .
The speed of sound as a function of
T is given by:
⎛ dv ⎞
v(T ) ≈ v(T0 ) + ⎜
⎟ ΔT
⎝ dT ⎠T0
From Problem 44, dv/dT is given by:
dv 1 v(T )
=
dt 2 T
Evaluating
dv ⎤
yields:
dT ⎥⎦ T =T0
dv ⎤
1 v(T0 )
=
⎥
dT ⎦ T =T0 2 T0
Substitute equation (2) in equation (1)
and simplify to obtain:
⎛ 1 v(T0 ) ⎞
⎟⎟ΔT
v(T ) ≈ v(T0 ) + ⎜⎜
⎝ 2 T0 ⎠
⎛ ΔT ⎞
⎟⎟
= v(T0 )⎜⎜1 +
2
T
0 ⎠
⎝
For T0 = 273&deg; K, v(T0 ) = 331 m/s ,
⎛
⎞
t
⎟⎟
v = (331 m/s)⎜⎜1 +
⎝ 2(273 K ) ⎠
and ΔT = t :
(1)
(2)
= (331 + 0.606 t ) m/s
The Wave Equation
36 •
Show explicitly that the following functions satisfy the wave equation
2
∂ y 1 ∂2 y
= 2 2 : (a) y(x,t) = k(x + vt)3, (b) y(x, t) = Aeik(x – vt) , where A and k are
2
∂x
v ∂t
constants and i = − 1 , and (c) y(x,t) = ln k(x – vt).
Picture the Problem To show that each of the functions satisfies the wave
equation, we’ll need to find their first and second derivatives with respect to x and
t and then substitute these derivatives in the wave equation.
(a) Find the first two spatial
3
derivatives of y ( x,t ) = k (x + vt ) :
∂y
2
= 3k (x + vt )
∂x
and
∂2 y
= 6k (x + vt )
∂x 2
(1)
1470 Chapter 15
Find the first two temporal
3
derivatives of y ( x,t ) = k (x + vt ) :
∂y
2
= 3kv(x + vt )
∂t
and
∂2 y
= 6kv 2 (x + vt )
2
∂t
(2)
Express the ratio of equation (1) to
equation (2):
∂2 y
∂x 2 = 6k ( x + vt ) = 1
∂ 2 y 6kv 2 ( x + vt )
v2
∂t 2
3
confirming that y (x, t ) = k (x + vt )
satisfies the general wave equation.
(b) Find the first two spatial
derivatives of y (x,t ) = Aeik ( x−vt ) :
∂2 y
∂y
= ikAeik ( x−vt ) , 2 = i 2 k 2 Aeik ( x−vt )
∂x
∂x
or
∂2 y
= −k 2 Aeik ( x−vt )
(3)
∂x 2
Find the first two temporal
derivatives of y (x,t ) = Aeik ( x−vt ) :
∂y
= −ikvAeik ( x−vt ) ,
∂t
∂ 2 y 2 2 2 ik ( x−vt )
= i k v Ae
∂t 2
or
∂2 y
= −k 2v 2 Aeik ( x−vt )
∂t 2
(4)
Express the ratio of equation (3) to
equation (4):
∂2 y
2
ik ( x − vt )
1
∂x 2 = − k Ae
= 2
2 2
ik ( x − vt )
2
∂ y − k v Ae
v
2
∂t
confirming that y (x, t ) = Aeik ( x−vt )
satisfies the general wave equation.
(c) Find the first two spatial
derivatives of
y ( x, t ) = ln k ( x − vt ) :
k
∂y
=
∂x x − vt
and
∂2 y
k2
=
−
∂x 2
(x − vt )2
(5)
Traveling Waves 1471
Find the first two temporal
derivatives of y ( x, t ) = ln k ( x − vt ) :
Express the ratio of equation (5)
to equation (6):
37
•
vk
∂y
=−
∂t
x − vt
and
∂2 y
v2k 2
=
−
∂t 2
(x − vt )2
(6)
2
∂2 y − k
2
∂x 2 = ( x − vt ) = 1
∂2 y
v2k 2
v2
−
∂t 2
(x − vt )2
confirming that y ( x, t ) = ln k ( x − vt )
satisfies the general wave equation.
Show that the function y = A sin kx cos ωt satisfies the wave equation.
∂2 y 1 ∂2 y
=
. To show that
∂x 2 v 2 ∂ t 2
y = A sin kx cos ω t satisfies this equation, we’ll need to find the first and second
derivatives of y with respect to x and t and then substitute these derivatives in the
wave equation.
Picture the Problem The wave equation is
Find the first two spatial derivatives
of y = A sin kx cos ω t :
Find the first two temporal
derivatives of y = A sin kx cos ω t :
∂y
= Ak cos kx cos ωt
∂x
and
∂2 y
= − Ak 2 sin kx cos ωt
∂x 2
∂y
= −ωA sin kx sin ωt
∂t
and
∂2 y
= −ω 2 A sin kx cos ωt
2
∂t
(1)
(2)
1472 Chapter 15
Express the ratio of equation (1) to
equation (2):
∂2 y
2
2
∂x 2 = − Ak sin kx cos ωt = k
∂ 2 y − Aω 2 sin kx cos ωt ω 2
∂t 2
1
= 2
v
confirming that y = A sin kx cos ω t
satisfies the general wave equation.
Harmonic Waves on a String
38 •
One end of a string 6.0 m long is moved up and down with simple
harmonic motion at a frequency of 60 Hz. If the wave crests travel the length of
the string in 0.50 s, find the wavelength of the waves on the string.
Picture the Problem We can find the wavelength of the waves on the string
using the relationship v = fλ .
v
f
Express the wavelength of the
waves:
fλ = v ⇒ λ =
The speed of the wave is equal to
the distance divided by the time:
v=
Δx 6.0 m
=
= 12 m/s
Δt 0.50 s
Substitute numerical values to
obtain:
λ=
12 m/s
= 20 cm
60 s −1
39 •
[SSM] A harmonic wave on a string with a mass per unit length of
0.050 kg/m and a tension of 80 N, has an amplitude of 5.0 cm. Each point on the
string moves with simple harmonic motion at a frequency of 10 Hz. What is the
power carried by the wave propagating along the string?
Picture the Problem The average power propagated along the string by a
harmonic wave is Pav = 12 μω 2 A2v, where v is the speed of the wave, and μ, ω, and
A are the linear density of the string, the angular frequency of the wave, and the
amplitude of the wave, respectively.
Express and evaluate the power
propagated along the string:
Pav = 12 μω 2 A2v
Traveling Waves 1473
The speed of the wave on the string
is given by:
FT
v=
Substitute for v and simplify to obtain:
μ
Pav = 12 μω 2 A2
FT
= 2(πfA)
FT μ
2
=
μ
1
2
(2πf )2 A2
FT μ
Substitute numerical values and evaluate Pav:
[(
)
]
Pav = 2 π 10 s −1 (0.050 m )
2
(80 N )(0.050 kg/m ) =
9.9 W
40 •
A 2.00-m-long rope has a mass of 0.100 kg. The tension is 60.0 N. An
oscillator at one end sends a harmonic wave with an amplitude of 1.00 cm down
the rope. The other end of the rope is terminated so that all of the energy of the
wave is absorbed and none is reflected. What is the frequency of the oscillator if
the power transmitted is 100 W?
Picture the Problem The average power propagated along the rope by a
harmonic wave is Pav = 12 μω 2 A2v, where v is the velocity of the wave, and μ, ω,
and A are the linear density of the string, the angular frequency of the wave, and
the amplitude of the wave, respectively.
Rewrite the power equation in terms
of the frequency of the wave:
Solve for the frequency and simplify
to obtain:
The wave speed is given by:
Substitute for v and simplify to
obtain:
Pav = 12 μω 2 A2 v = 2π 2 μf 2 A2 v
f =
2 Pav
Pav
1
=
2
2π μA v 2πA μv
v=
FT
f =
1
2πA
2
μ
2 Pav
μ
FT
μ
=
1
2πA
2 Pav
μFT
1474 Chapter 15
Substitute numerical values and evaluate f:
f =
1
2π (0.0100 m )
2(100 W )
⎛ 0.100 kg ⎞
⎟ (60.0 N )
⎜
⎝ 2.00 m ⎠
= 171 Hz
41 ••
The wave function for a harmonic wave on a string is
y(x, t) = (1.00 mm) sin(62.8 m–1 x + 314 s–1 t). (a) In what direction does this
wave travel, and what is the wave’s speed? (b) Find the wavelength, frequency,
and period of this wave. (c) What is the maximum speed of any point on the
string?
Picture the Problem y ( x, t ) = A sin (kx − ωt ) describes a wave traveling in the +x
direction. For a wave traveling in the −x direction, we have y ( x, t ) = A sin (kx + ωt ) .
We can determine A, k, and ω by examination of the wave function. The
wavelength, frequency, and period of the wave can, in turn, be determined from k
and ω.
(a) Because the sign between the kx and ωt terms is positive, the wave is traveling
in the −x direction.
The speed of the wave is:
(b) The coefficient of x is k and:
Substitute numerical values and
evaluate λ:
The coefficient of t is ω and:
The period of the wave motion is the
reciprocal of its frequency:
(c) Express and evaluate the
maximum speed of any string
segment:
v=
k=
ω
k
=
2π
λ
314 s −1
= 5.00 m/s
62.8 m −1
⇒λ =
2π
k
λ=
2π
= 10.0 cm
62.8 m −1
f =
ω 314 s −1
=
= 50.0 Hz
2π
2π
T=
1
1
=
= 0.0200 s
f 50.0 s −1
vmax = Aω = (1.00 mm )(314 rad/s )
= 0.314 m/s
Traveling Waves 1475
42 ••
A harmonic wave on a string with a frequency of 80 Hz and an
amplitude of 0.025 m travels in the +x direction with a speed of 12 m/s. (a) Write
a suitable wave function for this wave. (b) Find the maximum speed of a point on
the string. (c) Find the maximum acceleration of a point on the string.
Picture the Problem y ( x,t ) = A sin (kx − ωt ) describes a wave traveling in the +x
direction. We can find ω and k from the data included in the problem statement
and substitute in the general equation. The maximum speed of a point on the
string can be found from vmax = Aω and the maximum acceleration from
amax = Aω 2 .
(a) Express the general form of the
equation of a harmonic wave
traveling to the right:
y ( x,t ) = A sin (kx − ω t )
ω = 2πf = 2π (80 s −1 ) = 503 s −1
Evaluate ω:
= 5.0 &times; 10 2 s −1
Determine k:
Substitute to obtain: y ( x,t ) =
k=
ω
v
=
503 s −1
= 42 m −1
12 m/s
(0.025 m )sin[(42 m −1 ) x − (5.0 &times;10 2 s −1 )t ]
(
(b) The maximum speed of a point
on the string is:
vmax = Aω = (0.025 m ) 503 s −1
(c) Express the maximum
acceleration of a point on the string:
amax = Aω 2
Substitute numerical values and
evaluate amax:
a max = (0.025 m ) 503 s −1
)
= 13 m/s
(
)
2
= 6.3 km/s 2
A 200-Hz harmonic wave with an amplitude equal to 1.2 cm moves
43 ••
along a 40-m-long string that has a mass of 0.120 kg and a tension of 50 N.
(a) What is the average total energy of the waves on a 20-m-long segment of
string? (b) What is the power transmitted past a given point on the string?
Picture the Problem The average power propagated along the string is
Pav = 12 μω 2 A2v. The average energy on a segment of length L is the average
power multiplied by the time for the wave to travel the distance L
1476 Chapter 15
Pav = 12 μω 2 A2v
(a) Express the power transmitted
past a given point on the string. The
average energy passing a point in
time t is the average power
multiplied by the time:
and
Eav = Pavt
The time for the wave to travel the
length L is L/v:
t=
Combine the previous steps and
solve for the average energy on the
string:
Eav = 12 μvω 2 A 2
L
v
L 1
= μLω 2 A 2
v 2
2
= 12 μL(2πf ) A 2 = 2π 2 μLf 2 A 2
Substitute numerical values and evaluate Eav:
⎛ 0.120 kg ⎞
⎟⎟ 200 s −1
Eav = 2π 2 ⎜⎜
40
m
⎝
⎠
(
) (20 m )(0.012 m)
2
2
(b) Express Pav:
Pav = 12 μvω 2 A 2
To calculate Pav we need an
expression for the wave speed v:
v=
Substitute for v to obtain:
= 6.8 J
FT
μ
FT
Pav = 12 μ (2πf ) A 2
2
= 2π 2 μf 2 A 2
μ
FT
μ
Substitute numerical values and evaluate Pav:
(
⎛ 0.060 kg ⎞
−1
Pav = 2π 2 ⎜
⎟ 200 s
⎝ 20 m ⎠
) (0.012 m )
2
2
50 N
= 44 W
0.060 kg
20 m
44 ••
On a real string, some of the energy of the wave dissipates as the wave
travels down the string. Such a situation can be described by a wave function
whose amplitude A(x) depends on x: y = A(x) sin (kx – ωt) where
A(x) = (A0e–bx). What is the power transported by the wave as a function of x,
where x &gt; 0?
Traveling Waves 1477
Picture the Problem The power propagated along the string by a harmonic wave
is P = 12 μω 2 A2v where v is the velocity of the wave, and μ, ω, and A are the linear
density of the string, the angular frequency of the wave, and the amplitude of the
wave, respectively.
Express the power associated with
the wave at the origin:
P = 12 μω 2 A2v
Express the amplitude of the
wave at x:
A( x ) = A0 e −bx
Substitute in equation (1) to obtain:
P(x ) = 12 μω 2 A0e −bx v
(
=
1
2
(1)
)
2
μω 2 A02ve −2bx
45 ••
[SSM] Power is to be transmitted along a taut string by means of
transverse harmonic waves. The wave speed is 10 m/s and the linear mass density
of the string is 0.010 kg/m. The power source oscillates with an amplitude of
0.50 mm. (a) What average power is transmitted along the string if the frequency
is 400 Hz? (b) The power transmitted can be increased by increasing the tension
in the string, the frequency of the source, or the amplitude of the waves. By how
much would each of these quantities have to increase to cause an increase in
power by a factor of 100 if it is the only quantity changed?
Picture the Problem The average power propagated along a string by a
harmonic wave is Pav = 12 μω 2 A2 v where v is the speed of the wave, and μ, ω, and
A are the linear density of the string, the angular frequency of the wave, and the
amplitude of the wave, respectively.
(a) Express the average power
transmitted along the string:
Pav = 12 μω 2 A2 v = 2π 2 μf 2 A2 v
Substitute numerical values and
evaluate Pav:
Pav = 2π 2 (0.010 kg/m ) 400 s −1
(
&times; 0.50 &times; 10 − 3
(
)
m ) (10 m/s )
2
2
= 79 mW
(b) Because Pav ∝ f 2 , increasing the frequency by a factor of 10 would increase
the power by a factor of 100.
1478 Chapter 15
Because Pav ∝ A2 , increasing the amplitude by a factor of 10 would increase
the power by a factor of 100.
Because Pav ∝ v and v ∝ F , increasing the tension by a factor of 104 would
increase v by a factor of 100 and the power by a factor of 100.
46 ••• Two very long strings are tied together at the point x = 0. In the region
x &lt; 0, the wave speed is v1, while in the region x &gt; 0, the speed is v2. A sinusoidal
wave is incident on the knot from the left (x &lt; 0); part of the wave is reflected and
part is transmitted. For x &lt; 0, the displacement of the wave is describabed by
y(x,t) = A sin (k1x – ωt) + B sin(k1x + ωt), while for x &gt; 0, y(x,t) = C sin (k2x – ωt),
where ω/k1 = v1 and ω/k2 = v2. (a) If we assume that both the wave function y and
its first spatial derivative ∂y/∂x must be continuous at x = 0, show that
C/A = 2v2/( v1 + v2), and that B/A = (v1 – v2)/( v1 + v2). (b) Show that
B2 + (v1/v2)C2 = A2.
Picture the Problem We can use the assumption that both the wave function and
its first spatial derivative are continuous at x = 0 to establish equations relating A,
B, C, k1, and k2. Then, we can solve these simultaneous equations to obtain
expressions for B and C in terms of A, v1, and v2.
(a) Let y1(x, t) represent the wave
function in the region x &lt; 0, and
y2(x, t) represent the wave function
in the region x &gt; 0. Express the
continuity of the two wave functions
at x = 0:
y1 (0, t ) = y2 (0, t )
and
A sin [k1 (0) − ωt ] + B sin[k1 (0) + ωt ]
= C sin[k 2 (0) − ωt ]
or
A sin (− ωt ) + B sin ωt = C sin (− ωt )
Because the sine function is odd; that
is, sin (− θ ) = − sin θ :
− A sin ωt + B sin ωt = −C sin ωt
and
A− B =C
(1)
Differentiate the wave functions with
respect to x to obtain:
∂y1
= Ak1 cos(k1 x − ωt )
∂x
+ Bk1 cos(k1 x + ωt )
and
∂y2
= Ck 2 cos(k 2 x − ωt )
∂x
Traveling Waves 1479
Express the continuity of the slopes
of the two wave functions at x = 0:
∂y 1
∂y
= 2
∂x x =0 ∂x x =0
and
Ak1 cos[k1 (0 ) − ωt ] + Bk1 cos[k1 (0) + ωt ]
= Ck 2 cos[k 2 (0) − ωt ]
or
Ak1 cos(− ωt ) + Bk1 cos ωt
= Ck 2 cos(− ωt )
Because the cosine function is even;
that is, cos(− θ ) = cos θ :
Ak1 cos ωt + Bk1 cos ωt = Ck 2 cos ωt
and
k1 A + k1 B = k 2C
(2)
Multiply equation (1) by k1 and add
it to equation (2) to obtain:
2k1 A = (k1 + k 2 )C
Solving for C yields:
C=
2k1
2
A=
A
k1 + k 2
1 + k 2 k1
Solve for C/A and substitute ω/v1
for k1 and ω/v2 for k2 to obtain:
2v2
C
2
2
=
=
=
A 1 + k 2 k1 1 + v1 v2
v1 + v2
Substitute in equation (1) to obtain:
⎛ 2v2 ⎞
⎟⎟ A
A − B = ⎜⎜
⎝ v2 + v1 ⎠
Solving for B/A yields:
v −v
B
= 1 2
A
v1 + v2
1480 Chapter 15
(b) We wish to show that
B2 + (v1/v2)C2 = A2
Use the results of (a) to obtain the
expressions B = −[(1 − α)/(1 + α)] A
and C = 2A/(1 + α), where α = v1/v2.
B2 +
2
2
⎛1−α ⎞ 2
⎛ 2 ⎞ 2
2
⎜
⎟ A + α⎜
⎟ A =A
1
+
1
+
α
α
⎝
⎠
⎝
⎠
2
and check to see if the resulting
equation is an identity:
2
⎛ 2 ⎞
⎛1−α ⎞
⎟ =1
⎜
⎟ + α⎜
⎝1+ α ⎠
⎝1+ α ⎠
(1 − α )2 + 4α
(1 + α )2
Substitute these expressions into
B2 + (v1/v2)C2 = A2
v1 2
C = A2
v2
=1
1 − 2α + α 2 + 4α
=1
(1 + α )2
1 + 2α + α 2
=1
(1 + α )2
(1 + α )2
(1 + α )2
=1
1=1
The equation is an identity:
Therefore, B 2 +
v1 2
C = A2
v2
Remarks: Our result in (a) can be checked by considering the limit of B/A as
v2/v1 → 0. This limit gives B/A = +1, telling us that the transmitted wave has
zero amplitude and the incident and reflected waves superpose to give a
standing wave with a node at x = 0.
Harmonic Sound Waves
47
•
A sound wave in air produces a pressure variation given by
p( x, t ) = 0.75 cos
π
(x − 343t ) , where p is in pascals, x is in meters, and t is in
2
seconds. Find (a) the pressure amplitude, (b) the wavelength, (c) the frequency,
and (d) the wave speed.
Picture the Problem The pressure variation is of the form
p( x,t ) = p0 cos k ( x − vt ) where k is the wave number, p0 is the pressure amplitude,
and v is the wave speed. We can find λ from k and f from ω and k.
(a) By inspection of the equation:
p0 = 0.75 Pa
Traveling Waves 1481
(b) Because k =
(c) Solve v =
ω
k
2π
λ
=
=
π
2
:
2πf
for f to obtain:
k
λ = 4.00 m
f =
kv
2π
π
Substitute numerical values and
evaluate f:
f = 2
(d) By inspection of the equation:
v = 343 m/s
(343 m/s )
2π
= 85.8 Hz
48 •
(a) Middle C on the musical scale has a frequency of 262 Hz. What is
the wavelength of this note in air? (b) The frequency of the C an octave above
middle C is twice that of middle C. What is the wavelength of this note in air?
Picture the Problem The frequency, wavelength, and speed of the sound waves
are related by v = fλ.
(a) The wavelength of middle C is
given by:
λ=
v 340 m/s
=
= 1.30 m
f
262 s −1
(b) Evaluate λ for a frequency twice
that of middle C:
λ=
v
340 m/s
=
= 0.649 m
f 2 262 s −1
(
)
49 •
[SSM] The density of air is 1.29 kg/m3. (a) What is the displacement
amplitude for a sound wave with a frequency of 100 Hz and a pressure amplitude
of 1.00 &times; 10–4 atm? (b) The displacement amplitude of a sound wave of frequency
300 Hz is 1.00 &times;10–7 m. What is the pressure amplitude of this wave?
Picture the Problem The pressure amplitude depends on the density of the
medium ρ, the angular frequency of the sound wave μ, the speed of the wave v,
and the displacement amplitude s0 according to p0 = ρωvs0 .
(a) Solve p0 = ρωvs0 for s0:
s0 =
p0
ρωv
1482
Chapter 15
Substitute numerical values and evaluate s0:
s0 =
(1.00 &times;10 atm)(1.01325 &times;10 Pa/atm) = 3.64 &times;10
2π (1.29 kg/m )(100 s )(343 m/s )
−4
5
−5
−1
3
m = 36.4 μm
(b) Use p0 = ρωvs0 to find p0:
(
)(
)
(
)
p0 = 2π 1.29 kg/m 3 300 s −1 (343 m/s ) 1.00 &times;10 −7 m = 83.4 mPa
50 •
The density of air is 1.29 kg/m3. (a) What is the displacement
amplitude of a sound wave that has a frequency of 500 Hz at the pain-threshold
pressure amplitude of 29.0 Pa? (b) What is the displacement amplitude of a sound
wave that has the same pressure amplitude as the wave in Part (a), but has a
frequency of 1.00 kHz?
Picture the Problem The pressure amplitude depends on the density of the
medium ρ, the angular frequency of the sound wave μ, the wave speed v, and the
displacement amplitude s0 according to p0 = ρωvs0 .
(a) Solve p0 = ρωvs0 for s0:
Substitute numerical values and
evaluate s0:
s0 =
p0
ρωv
s0 =
29.0 Pa
2π 1.29 kg/m 3 (500 Hz )(343 m/s )
(
)
= 20.9 μm
(b) Proceed as in (a) with
f = 1.00 kHz:
s0 =
29.0 Pa
2π 1.29 kg/m (1.00 kHz )(343 m/s )
(
3
)
= 10.4 μm
51 •
A typical loud sound wave that has a frequency of 1.00 kHz has a
pressure amplitude of about 1.00 &times; 10–4 atm. (a) At t = 0, the pressure is a
maximum at some point x1. What is the displacement at that point at t = 0?
(b) Assuming the density of air is 1.29 kg/m3, what is the maximum value of the
displacement at any time and place?
Picture the Problem The pressure or density wave is 90&deg; out of phase with the
displacement wave. When the displacement is zero, the pressure and density
changes are either a maximum or a minimum. When the displacement is a
Traveling Waves 1483
maximum or minimum, the pressure and density changes are zero. We can use
p0 = ρωvs0 to find the maximum value of the displacement at any time and place.
(a) If the pressure is a maximum at x1 when t = 0, the displacement s is zero.
(b) Solve p0 = ρωvs0 for s0:
s0 =
p0
ρωv
Substitute numerical values and evaluate s0:
(1.00 &times;10 atm )(1.01325 &times;10 Pa/atm) =
=
2π (1.29 kg/m )(1.00 kHz )(343 m/s )
−4
s0
5
3
3.64 μm
52 •
An octave represents a change in frequency by a factor of two. Over
how many octaves can a typical person hear?
Picture the Problem A human can hear sounds between roughly 20 Hz and
20 kHz; a factor of 1000. An octave represents a change in frequency by a factor
of 2. We can evaluate 2N = 1000 to find the number of octaves heard by a person
who can hear this range of frequencies.
Relate the number of octaves to the
difference between 20 kHz and
20 Hz:
2 N = 1000
Take the logarithm of both sides of
the equation to obtain:
log 2 N = log103 ⇒ N log 2 = 3
Solving for N yields:
N=
3
= 9.97 ≈ 10
log 2
53 ••
In the oceans, whales communicate by sound transmission through the
water. A whale emits a sound of 50.0 Hz to tell a wayward calf to catch up to the
pod. The speed of sound in water is about 1500 m/s. (a) How long does it take the
sound to reach the calf if he is 1.20 km away? (b) What is the wavelength of this
sound in the water? (c) If the whales are close to the surface, some of the sound
energy might refract out into the air. What would be the frequency and
wavelength of the sound in the air?
1484 Chapter 15
Picture the Problem (a) We can use the definition of average speed to find the
time required for the sound to travel to the calf. (b) We can use the relationship
between wavelength, frequency, and speed to find the wavelength of the sound in
water. (c) The frequency of the sound does not change as it travels from water to
air, but its wavelength changes because of the difference in the speed of sound in
water and in air.
(a) Relate the time it takes the sound
to reach the calf to the distance from
the whale to the calf and the speed of
sound in water:
d 1.20 km
=
= 0.80 s
v 1500 m/s
Δt =
(b) The wavelength of this sound in
water is the ratio of its speed in water
to its frequency:
λwater =
(c) Because the frequency does not
change as it travels from water to air:
f = 50.0 Hz
The wavelength of this sound in air
is the ratio of its speed in air to its
frequency:
λair =
vwater 1500 m/s
=
= 30 m
f
50.0 Hz
vair 343 m/s
=
= 6.86 m
f
50.0 Hz
Waves in Three Dimensions: Intensity
54 •
A spherical sinusoidal source radiates sound uniformly in all
directions. At a distance of 10.0 m, the sound intensity level is 1.00 &times; 10–4 W/m2.
(a) At what distance from the source is the intensity 1.00 &times; 10–6 W/m2? (b) What
power is radiated by this source?
Picture the Problem The intensity of the sound from the spherical sinusoidal
source varies inversely with the square of the distance from the source. The power
radiated by the source is the product of the intensity of the radiation and the
surface area over which it is distributed.
(a) Relate the intensity I1 at a
distance R1 from the source to the
energy per unit time (power) arriving
at the point of interest:
I1 =
Pav,1
⇒ Pav,1 = 4πR12 I1
4πR12
Traveling Waves 1485
At a distance R2 from the source:
Because Pav,1 = Pav, 2 :
Substituting numerical values and
evaluating R2 gives:
I2 =
Pav, 2
⇒ Pav, 2 = 4πR22 I 2
4πR22
4πR12 I1 = 4πR22 I 2 ⇒ R2 = R1
I1
I2
1.00 &times; 10 −4 W/m 2
R2 = (10.0 m )
1.00 &times; 10 −6 W/m 2
= 100 m
(b) Solve I =
Pav
for Pav:
4π r 2
Substitute numerical values and
evaluate Pav:
Pav = 4π r 2 I
(
Pav = 4π (10.0 m ) 1.00 &times; 10 −4 W/m 2
2
)
= 126 mW
55 •
[SSM] A loudspeaker at a rock concert generates a sound that has an
intensity level equal to 1.00 &times; 10–2 W/m2 at 20.0 m and has a frequency of
1.00 kHz. Assume that the speaker spreads its energy uniformly in three
dimensions. (a) What is the total acoustic power output of the speaker? (b) At
what distance will the sound intensity be at the pain threshold of 1.00 W/m2?
(c) What is the sound intensity at 30.0 m?
Picture the Problem Because the power radiated by the loudspeaker is the
product of the intensity of the sound and the area over which it is distributed, we
can use this relationship to find the average power, the intensity of the radiation,
or the distance to the speaker for a given intensity or average power.
(a) Use Pav = 4πr 2 I to find the total
acoustic power output of the speaker:
(b) Relate the intensity of the sound
at 20 m to the distance from the
speaker:
(
Pav = 4π (20.0 m ) 1.00 &times; 10 −2 W/m 2
2
= 50.27 W = 50.3 W
1.00 &times;10 − 2 W/m 2 =
Pav
4π (20.0 m )
2
)
1486 Chapter 15
Relate the threshold-of-pain
intensity to the distance from the
speaker:
1.00 W/m 2 =
Divide the first of these equations by
the second and solve for r:
r=
(c) Use I =
Pav
to find the intensity
4πr 2
Pav
4πr 2
(1.00 &times;10 )(20.0 m )
2
−2
I (30.0 m ) =
at 30.0 m:
= 2.00 m
50.3 W
2
4π (30.0 m )
= 4.45 &times; 10 −3 W/m 2
56 ••
When a pin of mass 0.100 g is dropped from a height of 1.00 m, 0.050
percent of its energy is converted into a sound pulse that has a duration of 0.100 s.
(a) Estimate how far away the dropped pin can be heard if the minimum audible
intensity is 1.00 &times; 10–11 W/m2. (b) Your result in Part (a) is much too large in
practice because of background noise. If you assume that the intensity must be at
least 1.00 &times; 10–8 W/m2 for the sound to be heard, estimate how far away the
dropped pin can be heard. (In both parts, assume that the intensity is P/4πr2.)
Picture the Problem We can use conservation of energy to find the acoustical
energy resulting from the dropping of the pin. The power developed can then be
found from the given time during which the energy was transformed from
mechanical to acoustical form. We can find the range at which the dropped pin
can be heard from I = P/4π r2.
(a) Assuming that I = P/4π r2,
express the distance at which one can
hear the dropped pin:
Use conservation of energy to
determine the sound energy
generated when the pin falls:
r=
P
4π I
E = ε mgh
(
)
&times; (9.81 m/s )(1.00 m )
= (0.00050) 0.100 &times; 10 −3 kg
2
= 4.905 &times; 10 −7 J
The power of the sound pulse is
given by:
P=
E 4.905 &times;10 −7 J
=
Δt
0.100 s
= 4.905 &times;10 −6 W
Traveling Waves 1487
Substitute numerical values and
evaluate r:
r=
4.905 &times; 10 −6 W
4π 1.00 &times; 10 −11 W/m 2
(
)
= 0.20 km
(b) Repeat the last step in (a) with
I = 1.00 &times; 10–8 W/m2:
4.905 &times; 10 −6 W
r=
= 6.2 m
4π (1.00 &times; 10 −8 W/m 2 )
*Intensity Level
57 •
[SSM] What is the intensity level in decibels of a sound wave that
has an intensity of (a) 1.00 &times; 10–10 W/m2 and (b) 1.00 &times; 10–2 W/m2?
Picture the Problem The intensity level β of a sound wave, measured in
decibels, is given by β = (10 dB) log(I I 0 ) where I0 = 10−12 W/m2 is defined to be
the threshold of hearing.
⎛ 1.00 &times; 10 −10 W/m 2 ⎞
⎟⎟
−12
2
⎝ 10 W/m
⎠
(a) Using its definition, calculate the
intensity level of a sound wave whose
intensity is 1.00 &times; 10–10 W/m2:
β = (10 dB)log⎜⎜
(b) Proceed as in (a) with
I = 1.00 &times; 10–2 W/m2:
⎛ 1.00 &times; 10 −2 W/m 2 ⎞
⎟⎟
β = (10 dB)log⎜⎜
2
−12
10
W/m
⎝
⎠
= 10 log10 2 = 20.0 dB
= 10 log1010 = 100 dB
What is the intensity of a sound wave if, at a particular location, the
58 •
intensity level is (a) β = 10 dB and (b) β = 3 dB?
Picture the Problem (a) and (b) The intensity level β of a sound wave, measured
in decibels, is given by β = (10 dB) log(I I 0 ) where I0 = 10−12 W/m2 is defined to
be the threshold of hearing.
(a) Solve β = (10 dB) log(I I 0 ) for I
I = 10 β
(10 dB )
I0
to obtain:
Evaluate I for β = 10 dB:
I = 10 (10 dB ) (10 dB ) I 0 = 10 I 0
(
)
= 10 10 −12 W/m 2 = 10 −11 W/m 2
1488 Chapter 15
(b) Proceed as in (a) with β = 3 dB:
I = 10 (3 dB ) (10 dB ) I 0 = 2 I 0
(
)
= 2 10 −12 W/m 2 = 2 &times; 10 −12 W/m 2
59 •
At a certain distance, the sound intensity level of a dog’s bark is
50 dB. At that same distance, the sound intensity of a rock concert is 10,000 times
that of the dog’s bark. What is the sound intensity level of the rock concert?
Picture the Problem The intensity level β of a sound wave, measured in
decibels, is given by β = (10 dB) log(I I 0 ) where I0 = 10−12 W/m2 is defined to be
the threshold of hearing.
⎛ I concert
⎝ I0
⎞
⎟⎟
⎠
Express the sound intensity level of
the rock concert:
β concert = (10 dB)log⎜⎜
Express the sound intensity level of
the dog’s bark:
⎛ I dog ⎞
⎟⎟
50 dB = (10 dB) log⎜⎜
⎝ I0 ⎠
Solve for the intensity of the dog’s
bark:
I dog = 10 5 I 0 = 10 5 10 −12 W/m 2
Express the intensity of the rock
concert in terms of the intensity
of the dog’s bark:
I concert = 10 4 I dog = 10 4 10 −7 W/m 2
Substitute in equation (1) and
evaluate βconcert:
β concert = (10 dB)log⎜⎜
(
(1)
)
= 10 −7 W/m 2
(
)
= 10 −3 W/m 2
⎛ 10 −3 W/m 2 ⎞
⎟
−12
2 ⎟
⎝ 10 W/m ⎠
= (10 dB)log10 9 = 90 dB
60 •
What fraction of the acoustic power of a noise would have to be
eliminated to lower its sound intensity level from 90 to 70 dB?
Picture the Problem We can express the intensity levels at both 90 dB and 70 dB
in terms of the intensities of the sound at those levels. By subtracting the two
expressions, we can solve for the ratio of the intensities at the two levels and then
find the fractional change in the intensity that corresponds to a decrease in
intensity level from 90 dB to 70 dB.
Traveling Waves 1489
Express the intensity level at 90 dB:
⎛I ⎞
90 dB = (10 dB) log⎜⎜ 90 ⎟⎟
⎝ I0 ⎠
Express the intensity level at 70 dB:
⎛I ⎞
70 dB = (10 dB)log⎜⎜ 70 ⎟⎟
⎝ I0 ⎠
Δβ = 20 dB
Express Δβ = β90 − β70:
⎛I ⎞
⎛I ⎞
= (10 dB) log⎜⎜ 90 ⎟⎟ − (10 dB) log⎜⎜ 70 ⎟⎟
⎝ I0 ⎠
⎝ I0 ⎠
⎛I ⎞
= (10 dB) log⎜⎜ 90 ⎟⎟
⎝ I 70 ⎠
Solving for I90 yields:
I 90 = 100I 70
Express the fractional change in the
intensity from 90 dB to 70 dB:
I 90 − I 70 100 I 70 − I 70
=
= 99%
I 90
100 I 70
Because P ∝ I , the fractional change in power is 99% .
61 ••
A spherical source radiates sound uniformly in all directions. At a
distance of 10 m, the sound intensity level is 80 dB. (a) At what distance from the
source is the intensity level 60 dB? (b) What power is radiated by this source?
Picture the Problem The intensity at a distance r from a spherical source varies
with distance from the source according to I = Pav 4πr 2 . We can use this
relationship to relate the intensities corresponding to an 80-dB intensity level (I80)
and the intensity corresponding to a 60-dB intensity level (I60) to their distances
from the source. We can relate the intensities to the intensity levels
through β = (10 dB) log(I I 0 ) .
(a) Express the intensity of the sound
where the intensity level is 80 dB:
I 80 =
Pav
4π r802
Express the intensity of the sound
where the intensity level is 60 dB:
I 60 =
Pav
4π r602
1490 Chapter 15
Divide the first of these equations by
the second to obtain:
I 80
I 60
Pav
2
r602
4π (10 m )
=
=
Pav
100 m 2
4π r602
Solving for r60 yields:
r60 = (10 m )
Find the intensity of the 80-dB
⎛I ⎞
80 dB = (10 dB) log⎜⎜ 80 ⎟⎟
⎝ I0 ⎠
and
I 80 = 108 I 0 = 10 −4 W/m 2
Find the intensity of the 60-dB
⎛I ⎞
60 dB = (10 dB) log⎜⎜ 60 ⎟⎟
⎝ I0 ⎠
and
I 60 = 10 6 I 0 = 10 −6 W/m 2
Substitute numerical values for I80 and
I60 and evaluate r60:
r60 = (10 m )
(b) Using the intensity corresponding
to an intensity level of 80 dB,
express and evaluate the power
P = I 80 A = 10 − 4 W/m 2 4π(10 m )
(
I 80
I 60
10 −4 W/m 2
= 0.10 km
10 −6 W/m 2
)[
2
]
= 0.13 W
62 ••
Harry and Sally are sitting on opposite sides of a circus tent when an
elephant trumpets a loud blast. If Harry experiences a sound intensity level of 65
dB and Sally experiences only 55 dB, what is the ratio of the distance between
Sally and the elephant to the distance between Harry and the elephant?
Picture the Problem The intensity of the sound heard by Harry and Sally
depends inversely on the square of the distance between the elephant and each of
them. We can use the definition of sound intensity (decibel) level
( β = (10 dB) log(I I 0 ) ) and the definition of intensity ( I = Pav A ) to find the ratio
of these distances.
Traveling Waves 1491
Express the sound intensity level at
Harry’s location:
⎛ IH
⎝ I0
β H = (10 dB)log⎜⎜
⎞
⎟⎟
⎠
⎛ Pav ⎞
⎟⎟
= (10 dB)log⎜⎜
2
4
π
r
I
H 0 ⎠
⎝
⎛
Pav ⎞
⎟⎟
2
⎝ 4π rS I 0 ⎠
Similarly, the sound intensity level at
Sally’s location is:
β S = (10 dB)log⎜⎜
The difference in decibel level’s at
the two locations is given by:
Δβ = β H − β S
Substituting for βH, βS, and Δβ yields:
⎛ Pav ⎞
⎛ Pav ⎞
⎟
⎜⎜
⎟⎟ = 65 dB − 55 dB = 10 dB
(
)
−
Δβ = (10 dB)log⎜⎜
10
dB
log
2
2
⎟
r
I
r
I
π
π
4
4
H 0 ⎠
S 0 ⎠
⎝
⎝
Simplifying this expression yields:
⎛ r2 ⎞
r2
log⎜⎜ S2 ⎟⎟ = 1 ⇒ S2 = 101
rH
⎝ rH ⎠
Solving for the ratio rS rH yields:
rS
= 10 = 3.2
rH
Three noise sources produce intensity levels of 70, 73, and 80 dB
63 ••
when acting separately. When the sources act together, the resultant intensity is
the sum of the individual intensities. (a) Find the sound intensity level in decibels
when the three sources act at the same time. (b) Discuss the effectiveness of
eliminating the two least intense sources in reducing the intensity level of the
noise.
Picture the Problem We can find the intensities of the three sources from their
intensity levels and, because their intensities are additive, find the intensity level
when all three sources are acting.
(a) Express the sound intensity level
when the three sources act at the
same time:
⎛ I 3 sources ⎞
⎟⎟
I
0
⎝
⎠
β 3 sources = (10 dB)log⎜⎜
⎛I +I +I ⎞
= (10 dB)log⎜⎜ 70 73 80 ⎟⎟
I0
⎠
⎝
1492 Chapter 15
Find the intensities of each of the
three sources:
⎛I ⎞
70 dB = (10 dB)log⎜⎜ 70 ⎟⎟ ⇒ I 70 = 10 7 I 0
⎝ I0 ⎠
⎛I ⎞
73 dB = (10 dB)log⎜⎜ 73 ⎟⎟ ⇒ I 73 = 10 7.3 I 0
⎝ I0 ⎠
and
⎛I ⎞
80 dB = (10 dB)log⎜⎜ 80 ⎟⎟ ⇒ I 80 = 10 8 I 0
⎝ I0 ⎠
Substituting in the expression for β3 sources yields:
⎛ 10 7 I 0 + 10 7.3 I 0 + 10 8 I 0 ⎞
⎟⎟ = (10 dB)log 10 7 + 10 7.3 + 10 8
I
0
⎝
⎠
(
β 3 sources = (10 dB)log⎜⎜
)
= 81dB
(b) Find the intensity level with the
two least intense sources eliminated:
⎛ 10 8 I 0 ⎞
⎟⎟ = 80 dB
⎝ I0 ⎠
β 80 = (10 dB)log⎜⎜
Eliminating the 70-dB and 73-dB sources does not reduce the intensity level
significantly.
64 ••
Show that if two people are different distances away from a sound
source, the difference Δβ between the intensity levels reaching the people, in
decibels, will always be the same, no matter the power radiated by the source.
Picture the Problem Let the two people be identified by the numerals 1 and 2
and use the definition of the intensity level, in decibels, to express Δβ as a
function of the distances r1 and r2 of the two people from the source.
Express the difference in the sound
intensity level heard by the two
people:
Δβ = β 2 − β1
The sound level intensity (decibel
level) heard by the second person is
given by:
β 2 = (10 dB)log⎜⎜
⎛ I2 ⎞
⎟⎟
I
⎝ 0⎠
⎛ P ⎞
= (10 dB)log⎜⎜ av ⎟⎟
⎝ A2 I 0 ⎠
⎛ Pav ⎞
⎟⎟
= (10 dB)log⎜⎜
2
⎝ 4π r2 I 0 ⎠
Traveling Waves 1493
⎛
Pav ⎞
⎟⎟
2
⎝ 4π r1 I 0 ⎠
In like manner:
β1 = (10 dB)log⎜⎜
Substitute in the expression for Δβ to obtain:
⎛ Pav ⎞
⎛ Pav ⎞
⎟⎟ − (10 dB)log⎜⎜
⎟⎟
Δβ = (10 dB)log⎜⎜
2
2
4
π
r
I
4
π
r
I
2 0 ⎠
1 0 ⎠
⎝
⎝
Simplifying yields:
⎛ Pav
⎜
4π r22 I 0
Δβ = (10 dB) log⎜
⎜ Pav
⎜
2
⎝ 4π r1 I 0
⎞
⎟
⎟
⎟
⎟
⎠
⎛ r2 ⎞
= (10 dB) log⎜⎜ 1 2 ⎟⎟
⎝ r2 ⎠
This result shows that the difference in sound level intensities depends only on
the distances to the source and not on the source’s power output.
65 ••• Everyone at a party is talking equally loudly. One person is talking to
you and the sound intensity level at your location is 72 dB. Assuming that all 38
people at the party are at the same distance from you as the person who you are
talking to, find the sound intensity level at your location.
Picture the Problem The sound intensity level can be found from the intensity of
the sound due to the 38 people. When 38 people are talking, the intensities add.
Express the sound level when all
38 people are talking:
⎛ 38 I1 ⎞
⎟⎟
I
⎝ 0 ⎠
β 38 = (10 dB) log⎜⎜
⎛I ⎞
= (10 dB) log 38 + (10 dB) log⎜⎜ 1 ⎟⎟
⎝ I0 ⎠
= (10 dB) log 38 + 72 dB
= 88 dB
An equivalent but longer solution:
Express the sound intensity level
when all 38 people are talking:
⎛ 38 I1 ⎞
⎟⎟
⎝ I0 ⎠
β 38 = (10 dB) log⎜⎜
1494 Chapter 15
⎛ I1 ⎞
⎟⎟
⎝ I0 ⎠
Express the sound intensity level
when only one person is talking:
β1 = 72 dB = (10 dB) log⎜⎜
Solving for I1 yields:
I1 = 10 7.2 I 0 = 107.2 10 −12 W/m 2
(
)
= 1.58 &times; 10 −5 W/m 2
Express the sound intensity when all
38 people are talking:
I 38 = 38I1
The sound intensity level is:
β 38 = (10 dB)log ⎢
(
)
⎡ 38 1.58 &times; 10 −5 W/m 2 ⎤
⎥
10 −12 W/m 2
⎣
⎦
= 88 dB
66 ••• When a violinist pulls the bow across a string, the force with which the
bow is pulled is fairly small, about 0.60 N. Suppose the bow travels across the A
string, which vibrates at 440 Hz, at 0.50 m/s. A listener 35 m from the performer
hears a sound of 60-dB intensity. Assuming that the sound radiates uniformly in
all directions, with what efficiency is the mechanical energy of bowing converted
to sound energy?
Picture the Problem Let η represent the efficiency with which mechanical
energy is converted to sound energy. Because we’re given information regarding
the rate at which mechanical energy is delivered to the string and the rate at which
sound energy arrives at the location of the listener, we’ll take the efficiency to be
the ratio of the sound power delivered to the listener divided by the power
delivered to the string. We can calculate the power input directly from the given
data. We’ll calculate the intensity of the sound at 35 m from its intensity level at
that distance and use this result to find the power output.
Express the efficiency of the
conversion of mechanical energy to
sound energy:
Find the power delivered by the bow
to the string:
Using β = (10 dB) log(I I 0 ) , find the
intensity of the sound at 35 m:
η=
Pout
Pin
Pin = Fv = (0.60 N )(0.50 m/s ) = 0.30 W
⎛ I 35 m ⎞
⎟⎟
60 dB = (10 dB) log⎜⎜
I
⎝ 0 ⎠
and
I 35 m = 10 6 I 0 = 1.00 &times; 10 −6 W/m 2
Traveling Waves 1495
(
)
The rate at which sound energy is
emitted is given by:
Pout = IA = 4π 1.00 &times; 10 − 6 W/m 2 (35 m )
Substitute numerical values and
evaluate η:
η=
2
= 0.0154 W
0.0154 W
= 5.1%
0.30 W
67 ••• [SSM] The noise intensity level at some location in an empty
examination hall is 40 dB. When 100 students are writing an exam, the noise level
at that location increases to 60 dB. Assuming that the noise produced by each
student contributes an equal amount of acoustic power, find the noise intensity
level at that location when 50 students have left.
Picture the Problem Because the sound intensities are additive, we’ll find the
noise intensity level due to one student by subtracting the background noise
intensity from the intensity due to the students and dividing by 100. Then, we’ll
use this result to calculate the intensity level due to 50 students.
⎛ 50 I1 ⎞
⎟⎟
⎝ I0 ⎠
Express the intensity level due to 50
students:
β 50 = (10 dB) log⎜⎜
Find the sound intensity when 100
students are writing the exam:
⎛I ⎞
60 dB = (10 dB) log⎜⎜ 100 ⎟⎟
⎝ I0 ⎠
and
I100 = 10 6 I 0 = 10 −6 W/m 2
Find the sound intensity due to the
background noise:
⎛ I background ⎞
⎟⎟
40 dB = (10 dB) log⎜⎜
I
0
⎝
⎠
and
I background = 10 4 I 0 = 10 −8 W/m 2
Express the sound intensity due to
the 100 students:
I100 − I background = 10 −6 W/m 2 − 10 −8 W/m 2
Find the sound intensity due to 1
student:
I 100 − I background
Substitute numerical values and
evaluate the noise intensity level
due to 50 students:
β 50 = (10 dB)log
≈ 10 − 6 W/m 2
100
= 57 dB
= 10 −8 W/m 2
(
50 1.00 &times; 10 −8 W/m 2
10 −12 W/m 2
)
1496 Chapter 15
String Waves Experiencing Speed Changes
68 •
A 3.00-m-long piece of string with a mass of 25.0 g is tied to 4.00 m
of heavy twine with a mass of 75.0 g and the combination is put under a tension
of 100 N. If a transverse pulse is sent down the less dense string, determine the
reflection and transmission coefficients at the junction point.
Picture the Problem Let the subscript ″s″ refer to the string and the subscript ″t″
to the heavy twine. We can use the definitions of the reflection and transmission
coefficients and the expression for the speed of waves on a string (Equation 15-3)
to find the speeds of the pulse on the string and the heavy twine.
Use their definitions to express the
reflection and transmission
coefficients:
vs
v −v
vt
r= t s =
vt + vs 1 + vs
vt
1−
(1)
and
2v t
=
v t + vs
τ=
Use Equation 15-3 to express vt and
vs :
Dividing the expression for vs by the
expression for vt and simplifying
yields:
vt =
FT
2
v
1+ s
vt
and vs =
μt
(2)
FT
μs
FT
vs
=
vt
μs
FT
=
μt
μs
μt
Substitute for vs v t in equation (1) to
obtain:
Express the ratio of μt to μs:
1−
r=
μt
μs
μt
1+
μs
mt
μ t l t mt l s
=
=
μ s ms ms l t
ls
(3)
Traveling Waves 1497
Substituting in equation (3) yields:
1−
mt l s
ms l t
1+
mt l s
ms l t
r=
Substitute numerical values and
evaluate r:
(75.0 &times;10
(25.0 &times;10
(75.0 &times;10
(25.0 &times;10
1−
r=
1+
−3
−3
−3
−3
)
)
kg )(3.00 m )
kg )(4.00 m )
kg (3.00 m )
kg (4.00 m )
= − 0.20
Substitute for vs v t in equation (2) to
obtain:
Substitute numerical values and
evaluateτ:
2
τ=
1+
μt
μs
2
=
mt l s
ms l t
1+
2
τ=
1+
(75.0 &times;10
(25.0 &times;10
−3
−3
kg )(3.00 m )
kg )(4.00 m )
= 0.80
Remarks: Because r + τ = 1 , we could have used this relationship to find τ.
69 •
[SSM] Consider a taut string with a mass per unit length μ1, carrying
transverse wave pulses that are incident upon a point where the string connects to
a second string with a mass per unit length μ2. (a) Show that if μ2 = μ1, then the
reflection coefficient r equals zero and the transmission coefficient τ equals +1.
(b) Show that if μ2 &gt;&gt; μ1, then r ≈ –1 and τ ≈ 0; and (c) if μ2 &lt;&lt; μ1 then r ≈ +1
and τ ≈ +2.
Picture the Problem We can use the definitions of the reflection and
transmission coefficients and the expression for the speed of waves on a string
(Equation 15-3) to r and t in terms of the linear densities of the strings.
1498 Chapter 15
(a) Use their definitions to express
the reflection and transmission
coefficients:
v1
v −v
v2
r= 2 1 =
v
v2 + v1
1+ 1
v2
1−
(1)
and
τ=
Use Equation 15-3 to express v2 and
v1 :
Dividing the expression for v1 by the
expression for v2 and simplifying
yields:
2v2
=
v2 + v1
v2 =
FT
μ2
2
v
1+ 1
v2
and v1 =
(2)
FT
μ1
FT
v1
=
v2
μ1
μ2
μ1
=
FT
μ2
Substitute for v1 v2 in equation (1) to
obtain:
Substitute for v1 v2 in equation (2) to
obtain:
If μ2 = μ1:
μ2
μ1
1−
r=
τ=
r=
(3)
μ2
1+
μ1
2
(4)
μ2
1+
μ1
1− 1
1+ 1
= 0
and
τ=
2
μ2
1+
μ1
=
2
1+ 1
= 1
Traveling Waves 1499
(b) From equations (1) and (2), if
μ2 &gt;&gt; μ1 then v1 &gt;&gt; v2:
r=
v2 − v1 − v1
≈
= −1
v2 + v1
v1
and
τ=
(c) If μ2 &lt;&lt; μ1, then v2 &gt;&gt; v1:
2v2
=
v2 + v1
2
≈ 0
v
1+ 1
v2
v1
v −v
v2
r= 2 1 =
≈ 1
v1
v2 + v1
1+
v2
1−
and
τ=
2v 2
=
v2 + v1
2
≈ 2
v1
1+
v2
70 ••
Verify the validity of 1 = r 2 + (v1 v2 )τ 2 (Equation 15-36) by
substituting the expressions for r and τ into it.
Picture the Problem Making the indicated substitutions will lead us to the
identity 1 = 1.
2v2
v2 − v1
and τ =
v2 + v1
v2 + v1
The reflection and transmission
coefficients are given by:
r=
Substituting into the equation given
in the problem statement yields:
⎛ v − v ⎞ v ⎛ 2v2 ⎞
⎟⎟
1 = ⎜⎜ 2 1 ⎟⎟ + 1 ⎜⎜
⎝ v2 + v1 ⎠ v2 ⎝ v2 + v1 ⎠
2
2
Simplify this expression algebraically to obtain:
⎛
1=
⎞
(v2 − v1 )2 + 4v22 ⎜⎜ v1 ⎟⎟
2
2
2
2
⎝ v2 ⎠ = v2 − 2v2 v1 + v1 + 4v 2 v1 = v 2 + 2v2 v1 + v1 = (v2 + v1 )
(v2 + v1 )2
(v2 + v1 )2
(v2 + v1 )2
(v2 + v1 )2
2
= 1
71 ••• Consider a taut string that has a mass per unit length μ1 carrying
transverse wave pulses of the form y = f(x – v1t) that are incident upon a point P
where the string connects to a second string with mass per unit length μ2. Derive
1 = r 2 + (v1 v2 )τ 2 by equating the power incident on point P to the power
reflected at P plus the power transmitted at P.
1500 Chapter 15
Picture the Problem Choose the direction of propagation of the incident pulse as
the +x direction and let x = 0 at point P. Energy is conserved as the incident pulse
is partially reflected and partially transmitted at point P.
From the conservation of energy we
have:
Pin + Pr = Pt
From Equation 15-20, the power
transmitted in the direction of
increasing x is given by:
P = − FT
(1)
∂y ∂y
∂x ∂x
Substituting in equation (1) yields:
∂y ∂y ⎞
∂y ∂y ⎞ ⎛
∂y ∂y ⎞ ⎛
⎛
⎜ − FT in in ⎟ + ⎜ − FT r r ⎟ = ⎜ − FT t t ⎟
∂x ∂x ⎠
∂x ∂x ⎠ ⎝
∂x ∂x ⎠ ⎝
⎝
or, upon simplification,
∂yin ∂yin ∂y r ∂y r ∂y t ∂y t
=
+
(2)
∂x ∂x
∂x ∂x
∂x ∂x
Because the incident pulse is given
by yin = f ( x − v1t ) , the reflected and
transmitted pulses are given by:
y r = rf (− x − v1t )
and
⎛v
⎞
y t = τ f ⎜⎜ 1 [x − v2 t ]⎟⎟
⎝ v2
⎠
∂y
∂y
and using the
∂x
∂t
chain rule yields:
∂y df ∂η
∂y df ∂η
=
and =
∂x dη ∂x
∂t dη ∂t
where η is the argument of the wave
function.
For the transmitted pulse:
⎞
∂y t
df ∂ ⎛ v1
df v1
⎜⎜ [x − v2 t ]⎟⎟ = τ
=τ
∂x
dη ∂x ⎝ v2
dη v2
⎠
and
⎞
∂y t
df
df ∂ ⎛ v1
⎜⎜ [x − v2 t ]⎟⎟ = τ
(− v1 )
=τ
dη
dη ∂t ⎝ v2
∂t
⎠
Evaluating
= −τ v1
df
dη
Traveling Waves 1501
For the reflected pulse:
∂y r
df ∂
(− x − v1t ) = −r df
=r
dη
∂x
dη ∂x
and
∂yr
df ∂
(− x − v1t ) = r df (− v1 )
=r
dη ∂t
dη
∂t
df
= −rv1
dη
For the incident pulse:
∂yin df ∂
(x − v1t ) = df
=
dη
∂x
dη ∂x
and
∂yin df ∂
(x − v1t ) = df (− v1 )
=
dη ∂t
dη
∂t
df
= −v1
dη
Substitute in equation (2) to obtain:
df ⎛
df ⎞ ⎛
df ⎞ ⎛
df ⎞ ⎛ df v1 ⎞ ⎛
df ⎞
⎟⎟ ⎜⎜ − τ v1
⎜⎜ − v1
⎟⎟ + ⎜⎜ − r
⎟⎟ ⎜⎜ − rv1
⎟⎟ = ⎜⎜τ
⎟
dη ⎝
dη ⎠ ⎝ dη ⎠ ⎝
dη ⎠ ⎝ dη v2 ⎠ ⎝
dη ⎟⎠
Simplifying and rearranging terms
yields:
1= r2 +
v1 2
τ
v2
The Doppler Effect
72 •
A sound source is moving at 80 m/s toward a stationary listener that is
standing in still air. (a) Find the wavelength of the sound in the region between
the source and the listener. (b) Find the frequency heard by the listener.
Picture the Problem We can use Equation 15-38 ( λ =
v &plusmn; us
) to find the
fs
wavelength of the sound between the source and the listener and Equation15-41a
v &plusmn; ur
( fr =
f s ) to find the frequency heard by the listener.
v &plusmn; us
(a) Apply Equation 15-38 to find λ:
λ=
v &plusmn; us v − us 343 m/s − 80 m/s
=
=
fs
fs
200 s −1
= 1.32 m
1502 Chapter 15
(b) Apply Equation 15-41a to obtain
fr:
fr =
=
v &plusmn; ur
v&plusmn;0
fs
fs =
v − us
v &plusmn; us
(
343 m/s
200 s −1
343 m/s − 80 m/s
)
= 261 Hz
73 •
Consider the situation described in Problem 72 from the reference
frame of the source. In this frame, the listener and the air are moving toward the
source at 80 m/s and the source is at rest. (a) At what speed, relative to the source,
is the sound traveling in the region between the source and the listener? (b) Find
the wavelength of the sound in the region between the source and the listener.
(c) Find the frequency heard by the listener.
Picture the Problem (a) In the reference frame of the source, the speed of sound
from the source to the listener is reduced by the speed of the air. (b) We can find
the wavelength of the sound in the region between the source and the listener
from v = fλ. (c) Because the sound waves in the region between the source and the
listener will be compressed by the motion of the listener, the frequency of the
sound heard by the listener will be higher than the frequency emitted by the
v &plusmn; ur
source and can be calculated using f r =
f s (Equation 15-41a).
v &plusmn; us
(a) The speed of sound in the
reference frame of the source is:
v' = v − u wind = 343 m/s − 80 m/s
(b) Noting that the frequency is
unchanged, express the wavelength
of the sound:
λ=
v' 263 m/s
=
= 1.32 m
f
200 s −1
fr =
v' &plusmn; u r
⎛ v' + u r ⎞
fs = ⎜
⎟ fs
v' &plusmn; u s
⎝ v' &plusmn; 0 ⎠
(c) Apply Equation 15-41a to obtain:
= 263 m/s
⎛ 263 m/s + 80 m/s ⎞
⎟⎟ 200 s −1
= ⎜⎜
263 m/s
⎝
⎠
(
)
= 261 Hz
A sound source is moving away from the stationary listener at
74 •
80 m/s. (a) Find the wavelength of the sound waves in the region between the
source and the listener. (b) Find the frequency heard by the listener.
Traveling Waves 1503
Picture the Problem We can use λ = (v &plusmn; us ) f s ( Equation 15-38) to find the
wavelength of the sound in the region between the source and the listener
v &plusmn; ur
and f r =
f s (Equation 15-41a) to find the frequency heard by the listener.
v &plusmn; us
Because the sound waves in the region between the source and the listener will be
spread out by the motion of the listener, the frequency of the sound heard by the
listener will be lower than the frequency emitted by the source.
(a) Because the source is moving
away from the listener, use the
Equation 15-38 to find the
wavelength of the sound between the
source and the listener:
λ=
(b) Because the listener is at rest and
the source is receding, ur = 0 and the
denominator of Equation 15-41a is
the sum of the two speeds:
fr =
=
v &plusmn; us v + us
=
fs
fs
343 m/s + 80 m/s
200 s −1
= 2.12 m
=
v &plusmn; ur
v&plusmn;0
fs =
fs
v &plusmn; us
v + us
(
343 m/s
200 s −1
343 m/s + 80 m/s
)
= 162 Hz
75 •
The listener is moving 80 m/s from the stationary source that is in rest
relative to the air. Find the frequency heard by the listener.
Picture the Problem Because the listener is moving away from the source, we
know that the frequency he/she will hear will be less than the frequency emitted
v &plusmn; ur
by the source. We can use f r =
f s (Equation 15-41a), with us = 0 and the
v &plusmn; us
Relate the frequency heard by the
listener to that of the source:
fr =
v &plusmn; ur
⎛ v − ur
fs = ⎜
v &plusmn; us
⎝ v&plusmn;0
⎞
⎟ fs
⎠
⎛ 343 m/s − 80 m/s ⎞
⎟⎟ 200 s −1
= ⎜⎜
343 m/s
⎝
⎠
(
)
= 153 Hz
76 ••
You have made the trek to observe a Space Shuttle landing. Near the
end of its descent, the ship is traveling at Mach 2.50 at an altitude of 5000 m.
(a) What is the angle that the shock wave makes with the line of flight of the
shuttle? (b) How far are you from the shuttle by the time you hear its shock wave,
1504 Chapter 15
assuming the shuttle maintains both a constant heading and a constant 5000-m
Picture the Problem The diagram shows the position of the shuttle at time t after
it was directly over your head (located at point P) Let u represent the speed of the
shuttle and v the speed of sound. We can use trigonometry to determine the angle
of the shock wave as well as the location of the shuttle x when you hear the shock
wave.
(a) Referring to the diagram, express
θ in terms of v, u, and t:
Solving for θ yields:
(b) Using the diagram, relate θ to
the altitude h of the shuttle and the
distance x:
Substitute numerical values and
evaluate x:
sin θ =
vt
1
1
=
=
ut u v 2.5
⎛ 1 ⎞
⎟ = 23.58&deg; = 23.6&deg;
⎝ 2.50 ⎠
θ = sin −1 ⎜
tan θ =
x=
h
h
⇒ x=
tan θ
x
5000 m
= 11.5 km
tan23.58&deg;
77 ••
The SuperKamiokande neutrino detector in Japan is a water tank the
size of a 14-story building. When neutrinos collide with electrons in water, most
of their energy is transferred to the electrons. As a consequence, the electrons then
fly off at speeds that approach c. The neutrino is counted by detecting the shock
wave, called Cerenkov radiation, that is produced when the high-speed electrons
travel through the water at speeds greater than the speed of light in water. If the
maximum angle of the Cerenkov shock-wave cone is 48.75&deg;, what is the speed of
light in water?
Picture the Problem The angle θ of the Cerenkov shock wave is related to the
speed of light in water v and the speed of light in a vacuum c according to
sin θ = v c .
Traveling Waves 1505
Relate the speed of light in water v to
the angle of the Cerenkov cone:
Substitute numerical values and
evaluate v:
sin θ =
v
⇒ v = c sin θ
c
v = (2.998 &times; 108 m/s )sin 48.75&deg;
= 2.254 &times; 108 m/s
78 ••
You are in charge of calibrating the radar guns for a local police
department. One such device emits microwaves at a frequency of 2.00 GHz.
During the trials, you have it arranged so that these waves are reflected from a car
moving directly away from the stationary emitter. In this situation, you detect a
frequency difference (between the received microwaves and the ones sent out) of
293 Hz. Find the speed of the car.
Picture the Problem Because the car is moving away from the stationary emitter
at a speed ur, the frequency fr it receives will be less than the frequency emitted
by the emitter. The microwaves reflected from the car, moving away from a
stationary detector, will be of a still lower frequency fr′. We can use the Doppler
shift equations to derive an expression for the speed of the car in terms of
difference of these frequencies.
by the moving car in terms of fs, ur,
and c:
The waves reflected by the car are
like waves re-emitted by a source
moving away from the radar gun:
Substitute equation (1) in equation (2)
to eliminate fr:
−1
u
⎛ u ⎞
Because ur &lt;&lt; c , ⎜1 + r ⎟ ≈ 1 − r .
c⎠
c
⎝
−1
⎛ u ⎞
Substituting for ⎜1 + r ⎟ and
c⎠
⎝
simplifying yields:
fr =
c &plusmn; ur
⎛ c − ur ⎞
fs = ⎜
⎟ fs
c &plusmn; us
⎝ c&plusmn;0 ⎠
(1)
f r' =
⎛ c ⎞
c &plusmn; ur
⎟⎟ f r
f r = ⎜⎜
c &plusmn; us
⎝ c + us ⎠
(2)
⎛ c ⎞⎛ c − ur ⎞
⎛ c − ur ⎞
⎟⎟⎜
⎟⎟ f s
f r' = ⎜⎜
⎟ f s = ⎜⎜
⎝ c + us ⎠⎝ c ⎠
⎝ c + us ⎠
⎛ ur ⎞
−1
⎜1− ⎟
c ⎟ f = ⎛⎜1 − ur ⎞⎟ ⎛⎜1 + ur ⎞⎟ f
=⎜
s
s
c ⎠⎝
c⎠
⎜ 1 + ur ⎟
⎝
⎜
⎟
c ⎠
⎝
⎛ u ⎞⎛ u ⎞
f r' ≈ ⎜1 − r ⎟ ⎜1 − r ⎟ f s
c ⎠⎝
c⎠
⎝
2
⎛ u ⎞
= ⎜1 − r ⎟ f s
c⎠
⎝
1506 Chapter 15
2
2u
⎛ u ⎞
Because ur &lt;&lt; c , ⎜1 − r ⎟ ≈ 1 − r .
c⎠
c
⎝
⎛ 2u ⎞
f r' ≈ ⎜1 − r ⎟ f s
c ⎠
⎝
2
⎛ u ⎞
Substituting for ⎜1 − r ⎟ gives:
c⎠
⎝
The frequency difference detected at
the source is given by:
⎛ 2u
Δf = f s − f r' = f s − ⎜1 − r
c
⎝
2u
= r fs
c
⎞
⎟ fs
⎠
Solving for ur yields:
ur =
Substitute numerical values and
evaluate ur:
2.998 &times; 108 m/s
(293 Hz )
ur =
2(2.00 GHz )
c
Δf
2 fs
= 22.0
m 1 km 3600 s
&times;
&times;
s 10 3 m
h
= 79.1 km/h
79 ••
[SSM] The Doppler effect is routinely used to measure the speed of
winds in storm systems. As the manager of a weather monitoring station in the
Midwest, you are using a Doppler radar system that has a frequency of 625 MHz
to bounce a radar pulse off of the raindrops in a swirling thunderstorm system
50 km away. You measure the reflected radar pulse to be up-shifted in frequency
by 325 Hz. Assuming the wind is headed directly toward you, how fast are the
winds in the storm system moving? Hint: The radar system can only measure the
component of the wind velocity along its ″line of sight.″
Picture the Problem Because the wind is moving toward the weather station
(radar device), the frequency fr the raindrops receive will be greater than the
frequency emitted by the radar device. The radar waves reflected from the
raindrops, moving toward the stationary detector at the weather station, will be of
a still higher frequency fr′. We can use the Doppler shift equations to derive an
expression for the radial speed u of the wind in terms of difference of these
frequencies.
Use Equation 15-41a to express the
raindrops in terms of fs, ur, and c:
fr =
c &plusmn; ur
⎛ c + ur ⎞
fs = ⎜
⎟ fs
c &plusmn; us
⎝ c ⎠
(1)
Traveling Waves 1507
The waves reflected by the drops are
like waves re-emitted by a source
moving toward the source at the
weather station:
Substitute equation (1) in equation (2)
to eliminate fr:
−1
u
⎛ u ⎞
Because ur &lt;&lt; c , ⎜1 − r ⎟ ≈ 1 + r .
c⎠
c
⎝
−1
⎛ u ⎞
Substituting for ⎜1 − r ⎟ and
c⎠
⎝
simplifying yields:
2
2u
⎛ u ⎞
Because ur &lt;&lt; c , ⎜1 + r ⎟ ≈ 1 + r .
c⎠
c
⎝
f r' =
⎛ c ⎞
c &plusmn; ur
⎟⎟ f r
f r = ⎜⎜
c &plusmn; us
⎝ c − us ⎠
(2)
⎛ c ⎞ ⎛ c + ur ⎞
⎛ c + ur ⎞
⎟⎟ ⎜
⎟⎟ f s
f r' = ⎜⎜
⎟ f s = ⎜⎜
⎝ c − us ⎠ ⎝ c ⎠
⎝ c − us ⎠
⎛ ur ⎞
−1
⎜1+ ⎟
ur ⎞ ⎛ ur ⎞
⎛
c
⎟ f s = ⎜1 + ⎟ ⎜1 − ⎟ f s
=⎜
c ⎠⎝
c⎠
⎜ 1 − ur ⎟
⎝
⎜
⎟
c ⎠
⎝
⎛ u ⎞⎛ u ⎞
f r' ≈ ⎜1 + r ⎟ ⎜1 + r ⎟ f s
c ⎠⎝
c⎠
⎝
2
⎛ u ⎞
= ⎜1 + r ⎟ f s
c⎠
⎝
⎛ 2u ⎞
f r' ≈ ⎜1 + r ⎟ f s
c ⎠
⎝
2
⎛ u ⎞
Substituting for ⎜1 + r ⎟ gives:
c⎠
⎝
The frequency difference detected at
the source is:
⎛ 2u ⎞
Δf = f r' − f s = ⎜1 + r ⎟ f s − f s
c ⎠
⎝
2u
= r fs
c
Solving for ur yields:
ur =
Substitute numerical values and
evaluate ur:
c
Δf
2 fs
2.998 &times; 108 m/s
(325 Hz )
2(625 MHz )
1mi/h
= 77.95 m/s &times;
0.4470 m/s
ur =
= 174 mi/h
1508 Chapter 15
80 ••
A stationary destroyer is equipped with sonar that sends out 40-MHzpulses of sound. The destroyer receives pulses back from a submarine directly
below with a time delay of 80 ms at a frequency of 39.958 MHz. If the speed of
sound in seawater is 1.54 km/s, (a) what is the depth of the submarine? (b) What
is its vertical speed?
Picture the Problem Let the depth of the submarine be represented by D and its
vertical speed by u. The submarine acts as both a receiver and source. We can
apply the definition of average speed to determine the depth of the submarine and
use the Doppler shift equations to derive an expression for the vertical speed of
the submarine in terms of the frequency difference. The frequency received is
lower than the frequency of the source, so the sub is descending.
(a) Using the definition of average
speed, relate the depth of the
submarine to the time delay between
the transmitted and reflected pulses:
2 D = vΔt ⇒ D = 12 vΔt
Substitute numerical values and
evaluate D:
D=
⎛ v &plusmn; ur ⎞
⎟⎟ f s to express
(b) Use f r = ⎜⎜
⎝ v &plusmn; us ⎠
the frequency fsub received by the
submarine:
⎛v−u ⎞
f sub = ⎜
(1)
⎟ f0
⎝ v ⎠
where u is the vertical speed of the
submarine.
1
2
(1.54 km/s )(80 ms) =
62 m
⎛ v &plusmn; ur ⎞
⎟⎟ f s to express
Use f r = ⎜⎜
&plusmn;
v
u
s ⎠
⎝
the frequency fr′ received by the
destroyer:
⎛ v ⎞
f r' = ⎜
⎟ f sub
⎝v+u ⎠
Substitute equation (1) in equation
(2) to eliminate fsub:
f − f r'
⎛ v−u ⎞
fr '= ⎜
v
⎟ f0 ⇒ u = 0
f 0 + f r'
⎝v+u⎠
Substitute numerical values and evaluate u:
⎛ 40.000 MHz − 39.958 MHz ⎞
u=⎜
⎟(1.54 km/s ) = 0.81 m/s
⎝ 40.000 MHz + 39.958 MHz ⎠
and so, because u is positive, the sub is descending.
(2)
Traveling Waves 1509
81 ••
A police radar unit transmits microwaves of frequency 3.00 &times; 1010 Hz,
and their speed in air is 3.00 &times; 108 m/s. Suppose a car is receding from the
stationary police car at a speed of 140 km/h. (a) What is the frequency difference
between the transmitted signal and the signal received from the receding car?
(b) Suppose the police car is, instead, moving at a speed of 60 km/h in the same
direction as the other vehicle. What is the difference in frequency between the
emitted and the reflected signals?
Picture the Problem The radar wave strikes the speeding car at frequency fr .
This frequency is less than fs because the car is moving away from the source. The
frequency shift is given by Equation 15-42 (the low-speed, relative to light,
approximation). The car then acts as a moving source emitting waves of
frequency fr. The police car detects waves of frequency fr′ &lt; fr because the source
(the speeding car) is moving away from the police car. The total frequency shift
is the sum of the two frequency shifts.
(a) Express the frequency difference
Δf as the sum of the frequency
difference Δf1 = f r − f s and the
frequency difference Δf 2 = f r' − f r :
Δf = Δf1 + Δf 2
Using Equation 15-42, substitute for
the frequency differences in equation
(1):
u
u
u
f s − f r = − ( f s + f r ) (2)
c
c
c
where u = us &plusmn; u r is the speed of the
Δf = −
Apply Equation 15-42 to Δf1 to
obtain:
Δf1 f r − f s
u
=
=−
fs
fs
c
where we’ve used the minus sign
because we know the frequency
difference is a downshift.
Solving for fr yields:
⎛ u⎞
f r = ⎜1 − ⎟ f s
⎝ c⎠
Substitute for fr in equation (2) and
simplify to obtain:
u⎛
⎛ u⎞ ⎞
Δf = − ⎜⎜ f s + ⎜1 − ⎟ f s ⎟⎟
c⎝
⎝ c⎠ ⎠
u⎛
u⎞
= − ⎜ 2 − ⎟ fs
c⎝
c⎠
(1)
1510 Chapter 15
Because u/c is negligible compared
to 2:
Δf ≈ −2 f s
u
c
Substitute numerical values and evaluate Δf:
⎛
km
1h ⎞
⎜⎜140
⎟
&times;
h 3600 s ⎟⎠
⎝
10
Δf ≈ −2 3.00 &times; 10 Hz
= − 7.78 kHz
3.00 &times; 10 8 m/s
(
)
(b) Use the result derived in (a) with u equal to the difference in the speeds of the
car and the police cruiser to obtain:
km ⎞ ⎛ 1 h ⎞
km
⎛
⎟⎟
− 60
⎜140
⎟ ⎜⎜
h
h
3600
s
⎝
⎠
⎝
⎠ = − 4.4 kHz
Δf ≈ −2 3.00 &times; 1010 Hz
3.00 &times; 108 m/s
(
)
82 ••
In modern medicine, the Doppler effect is routinely used to measure
the rate and direction of blood flow in arteries and veins. High frequency
″ultrasound″ (sound at frequencies above the human hearing range) is typically
employed. Suppose you are in charge of measuring the blood flow in a vein
(located in the lower leg of an older patient) that returns blood upward to the
heart. Her varicose veins indicate that perhaps the one-way valves in the vein are
not working properly and that the blood is ″pooling″ in the veins and perhaps
even that the blood flow is backward toward her feet. Employing sound with a
frequency of 50.0 kHz, you point the sound source from above her thigh region
down towards her feet and measure the sound reflected from that vein area to be
lower than 50.0 kHz. (a) Was your diagnosis of the valve condition correct? If so,
explain. (b) Estimate the instrument’s frequency difference capability to enable
you to measure speeds down to 1.00 mm/s. Take the speed of sound in flesh to be
the same as that in water, 1500 m/s.
Picture the Problem (a) Whether your diagnosis was correct depends on whether
the signal reflected from the blood is upshifted or downshifted. (b) Applying the
Doppler shift equations with the source stationary and the receiver moving
initially and then a second time with the source moving and the receiver
stationary will allow us to estimate the instrument’s frequency difference
capability.
(a) Because the received sound frequency is less than the frequency that was sent
out, the blood it reflected from must have been moving away from the source, or
toward the feet of the patient. The blood is flowing the wrong way and your
diagnosis is correct.
Traveling Waves 1511
(b) The blood receives a frequency
that is downshifted according to:
⎛ v &plusmn; ur ⎞
⎛ v − ur ⎞
⎟⎟ f s = ⎜
f r = ⎜⎜
⎟ fs
⎝ v&plusmn;0 ⎠
⎝ v &plusmn; us ⎠
⎛ u ⎞
= ⎜1 − r ⎟ f s
v⎠
⎝
The frequency given by equation (1)
is the frequency emitted by the
moving blood back toward the
receiver. This results in a second
downshift:
⎛ v &plusmn; ur ⎞ ' ⎛ v &plusmn; 0
⎟⎟ f s = ⎜⎜
f r' = ⎜⎜
⎝ v &plusmn; us ⎠
⎝ v + us
The source frequency in equation (2)
is the received frequency given by
equation (1). Note that the receiver
and emitter speeds are the same …
the speed of the blood. Let this speed
be u to obtain:
u⎞
⎛ u ⎞⎛ u ⎞
⎛
f r' ≈ ⎜1 − ⎟ ⎜1 − ⎟ f s ≈ ⎜1 − 2 ⎟ f s
v⎠
⎝ v ⎠⎝ v ⎠
⎝
Express the magnitude of the
difference in frequencies:
u⎞
⎛
Δf = f s − f r' ≈ f s − ⎜1 − 2 ⎟ f s
v⎠
⎝
u
= 2 fs
v
Substitute numerical values and
evaluate Δf:
⎛ 1.00 m/s ⎞
Δf ≈ 2⎜
⎟ (50.0 kHz )
⎝ 1500 m/s ⎠
(1)
⎞ '
⎟⎟ f s
⎠
−1
⎛ u ⎞
⎛ u ⎞
= ⎜1 + s ⎟ f s' ≈ ⎜1 − s ⎟ f s'
v⎠
v⎠
⎝
⎝
(2)
= 0.033 Hz
83 ••
[SSM] A sound source of frequency fs moves with speed us relative
to still air toward a receiver who is moving away with speed ur relative to still air
away from the source. (a) Write an expression for the received frequency f′r. (b)
Use the result that (1 – x)–1 ≈ 1 + x to show that if both us and ur are small
compared to v, then the received frequency is approximately
⎛ u ⎞
f r ' = ⎜1 + rel ⎟ f s
v ⎠
⎝
where urel = us – ur is the velocity of the source relative to the receiver.
1512 Chapter 15
Picture the Problem The received and transmitted frequencies are related
v &plusmn; ur
f s (Equation 15-41a), where the variables have the meanings
through f r =
v &plusmn; us
given in the problem statement. Because the source and receiver are moving in the
same direction, we use the minus signs in both the numerator and denominator.
(a) Relate the received frequency fr
to the frequency fs of the source:
ur
v &plusmn; ur
v f
fs =
fr =
us s
v &plusmn; us
1−
v
1−
⎛ u
= ⎜1 − r
v
⎝
(b) Expand (1 − us v ) binomially
−1
Substitute to obtain:
−1
⎞⎛ u s ⎞
⎟⎜1 − ⎟ f s
v⎠
⎠⎝
−1
u
⎛ us ⎞
⎜1 − ⎟ ≈ 1 + s
v⎠
v
⎝
⎛ u ⎞⎛ u
f r = ⎜1 − r ⎟⎜1 + s
v ⎠⎝
v
⎝
⎞
⎟ fs
⎠
⎡ u u ⎛ u ⎞⎛ u ⎞⎤
= ⎢1 + s − r − ⎜ r ⎟⎜ s ⎟⎥ f s
⎣ v v ⎝ v ⎠⎝ v ⎠⎦
⎛ u − ur ⎞
≈ ⎜1 + s
⎟ fs
v ⎠
⎝
because both us and ur are small
compared to v.
Because urel = us − ur
⎛ u ⎞
f r' ≈ ⎜1 + rel ⎟ f s
v ⎠
⎝
84 ••• To study the Doppler shift on your own, you take an electronic tone
generating device that is set to a frequency of middle C (262Hz) to a campus
wishing well known as ″The Abyss. ″ When you hold the device at arm’s length
(1.00 m), you measure its intensity level to be 80.0 dB. You then drop the tuner
down the hole, listening to its sound as it falls. After the tuner has fallen for
5.50 s, what frequency do you hear?
Traveling Waves 1513
Picture the Problem As the tuner falls, its speed increases and so the frequency
you hear is Doppler shifted. Our concern, however, is with the frequency you hear
5.50 s after you’ve dropped the tuner. The sound that you hear at this time was
emitted sometime before the tuner had fallen for 5.50 s. Hence we must answer
the question ″At what time was the sound emitted that you heard 5.50 s after
dropping the tuner?″
v &plusmn; ur
v
fs =
fs
v &plusmn; us
v + vfall
Apply Equation 15-41a to express
the frequency you’ll hear after 5.50
s:
fr =
Relate the elapsed time of 5.50 s to
the time required for the tuner to fall
and the time for the sound to return
to you:
Δt tot = Δt fall + Δt return
Use a constant-acceleration equation
to relate the time-of-fall to the fall
distance:
Δy = 12 g (Δt fall ) ⇒ Δt fall =
The return time depends on how far
the tuner has fallen and on the speed
of sound in air:
Δt return =
Δy
⇒ Δy = vΔt return
v
Δt fall =
2vΔt return
g
Substituting for Δy in the expression
for Δtfall yields:
Square both sides of the equation,
use equation (2) to substitute for
Δtreturn, and write the result explicitly
as a quadratic equation to obtain:
(2)
2
g
g
(Δt fall )2 + 2(343 m/s2 ) (Δtfall ) − 2(343 m/s)(52.50 s ) = 0
or
2Δy
g
(Δt fall )2 + 2v (Δtfall ) − 2v Δt tot = 0
Substitute numerical values to obtain:
9.81 m/s
(1)
9.81 m/s
(Δt fall )2 + (69.93 s )(Δt fall ) − 384.6 s 2 = 0
1514 Chapter 15
Δt fall = 5.124 s
Use your graphing calculator or the
equation for its positive root:
Because vfall = gΔt fall , equation (1)
fr =
becomes:
v
fs
v + gΔt fall
Substitute numerical values and evaluate fr:
fr =
343 m/s
(262 Hz ) = 229 Hz
343 m/s + 9.81 m/s 2 (5.124 s )
(
)
85 ••
You are in a hot-air balloon carried along by a 36-km/h wind and have
a sound source with you that emits a sound of 800 Hz as it approaches a tall
building. (a) What is the frequency of the sound heard by an observer at the
window of this building? (b) What is the frequency of the reflected sound heard
by you?
Picture the Problem The simplest way to approach this problem is to transform
to a reference frame in which the balloon is at rest. In that reference frame, the
speed of sound is v = 343 m/s, and ur = 36 km/h = 10 m/s. Then, we can use the
equations for a moving receiver and a moving source to find the frequencies heard
at the window and on the balloon.
(a) Use Equation 15-41a to express
the received frequency in terms of
the frequency of the source:
v &plusmn; ur
fr =
fs =
v &plusmn; us
u
ur
1+ r
v f
v f =
0 s
us s
1&plusmn;
1&plusmn;
v
v
1&plusmn;
⎛ u ⎞
= ⎜1 + r ⎟ f s
v⎠
⎝
Substitute numerical values and
evaluate fr:
⎛
10 m/s ⎞
⎟⎟(800 Hz ) = 823 Hz
f r = ⎜⎜1 +
⎝ 343 m/s ⎠
= 0.82 kHz
Traveling Waves 1515
(b) Treating the tall building as a moving source, express the frequency of the
reflected sound heard by a person riding in the balloon:
v &plusmn; ur
f r' =
fr =
v &plusmn; us
ur
0
⎛
1&plusmn;
⎜
v f =
v f =⎜ 1
u r
u r ⎜ us
1&plusmn; s
1− s
⎜1−
v
v
v
⎝
1&plusmn;
Substitute numerical values and
evaluate fr′:
⎞
⎟
⎟ fr
⎟
⎟
⎠
⎞
⎛
⎟
⎜
1
⎟(823 Hz )
⎜
f r' =
⎜
10 m/s ⎟
⎜ 1 − 343 m/s ⎟
⎠
⎝
= 0.85 kHz
86 ••
A car is approaching a reflecting wall. A stationary observer behind
the car hears a sound of frequency 745 Hz from the car horn and a sound of
frequency 863 Hz from the wall. (a) How fast is the car traveling? (b) What is the
frequency of the car horn? (c) What frequency does the car driver hear reflected
from the wall?
Picture the Problem We can relate the frequencies fr and fr′ heard by the
stationary observer behind the car to the speed of the car ur and the frequency of
the car’s horn fs. Dividing these equations will eliminate the frequency of the car’s
horn and allow us to solve for the speed of the car. We can then substitute to find
the frequency of the car’s horn. We can find the frequency heard by the driver as
a moving receiver by relating this frequency to the frequency reflected from the
wall.
(a) Use Equation 15-41a to relate the
frequency heard by the observer
directly from the car’s horn to the
speed of the car:
v &plusmn; ur
fr =
fs =
v &plusmn; us
=
1
f
us s
1+
v
ur
0
1&plusmn;
v f
v f =
us s
us s
1&plusmn;
1+
v (1)
v
1&plusmn;
1516 Chapter 15
Relate the frequency reflected from
the wall to the speed of the car:
Divide equation (2) by equation (1)
to obtain:
Substitute numerical values and
evaluate us:
0
v &plusmn; ur
v f
f r' =
fs =
us s
v &plusmn; us
1−
v
1
f
=
u s
1− s
v
1&plusmn;
us
f r'
v ⇒ u = f r' − f r v
=
u
f r' + f r'
fr
1− s
v
1+
863 Hz − 745 Hz
(343 m/s)
863 Hz + 745 Hz
m 1km 3600 s
= 25.17 &times; 3 &times;
s 10 m
h
us =
= 90.6 km/h
(b) Solve equation (1) for fs to
obtain:
⎛ u
f s = ⎜1 + r
v
⎝
⎞
⎟ fr
⎠
Substitute numerical values and
evaluate fs:
⎛ 25.17 m/s ⎞
⎟(745 Hz )
f s = ⎜⎜1 +
343 m/s ⎟⎠
⎝
= 800 Hz
(c) The driver is a moving receiver
and so we can relate the frequency
heard by the driver to the frequency
reflected by the wall (the frequency
heard by the stationary observer):
⎛ u ⎞
f driver = ⎜1 + r ⎟ f r'
v⎠
⎝
Substitute numerical values and
evaluate fdriver:
⎛ 25.17 m/s ⎞
⎟⎟(863 Hz )
f driver = ⎜⎜1 +
343
m/s
⎝
⎠
= 926 Hz
(2)
Traveling Waves 1517
87 ••
The driver of a car traveling at 100 km/h toward a vertical wall briefly
sounds the horn. Exactly 1.00 s later she hears the echo and notes that its
frequency is 840 Hz. How far from the wall was the car when the driver sounded
the horn and what is the frequency of the horn?
Picture the Problem Let t = 0 when the driver sounds her horn and let the
distance to the wall at that instant be d. The received and transmitted frequencies
v &plusmn; ur
are related through f r =
f s (Equation 15-41a). Solving this equation for fs
v &plusmn; us
will allow us to determine the frequency of the car horn. We can use the total
distance the sound travels (car-to-wall plus wall-back to-car … now closer to the
wall) to determine the distance to the wall when the horn was briefly sounded.
Use Equation 15-41a to relate the
frequency heard by the driver to
her speed and to the frequency of
her horn:
u
ur
1+ r
v &plusmn; ur
v f
v f =
fr =
fs =
us s
us s
v &plusmn; us
1&plusmn;
1−
v
v
Solving for fs yields:
us
v f
fs =
ur r
1+
v
1&plusmn;
1−
Substitute numerical values and evaluate fs:
km
1h
&times;
27.78 m/s
h 3600 s
1−
1−
343 m/s
(840 Hz ) = 343 m/s (840 Hz ) = 714 Hz
fs =
27.78 m/s
km
1h
&times;
1+
100
343 m/s
h 3600 s
1+
343 m/s
100
Relate the distance d to the wall at
t = 0 to the distance she travels in
time Δt = 1 s, her speed us, and the
speed of sound v:
Substitute numerical values and
evaluate d:
d + (d − us Δt ) = vΔt ⇒ d =
d=
1
2
1
2
(us + v )Δt
(27.78 m/s + 343 m/s)(1.00 s )
= 185 m
1518 Chapter 15
88 ••
You are on a transatlantic flight traveling due west at 800 km/h. An
experimental plane flying at Mach 1.6 and 3.0 km to the north of your plane is
also on an east-to-west course. What is the distance between the two planes when
you hear the sonic boom from the experimental plane?
Picture the Problem You’ll hear the sonic boom when the surface of its cone
reaches your plane. In the following diagram, the experimental plane is at C and
your plane is at P. The distance h = 3.0 km. The distance between the planes
when you hear the sonic boom is d. We can use trigonometry to determine the
angle of the shock wave as well as the separation of the planes when you hear the
sonic boom.
Using the Pythagorean theorem,
relate the separation of the planes d,
to the distance h and the angle θ :
Express θ in terms of v, u, and t:
d 2 = h 2 + d 2 cos 2 θ ⇒ d = h
sin θ =
1
1 − cos 2 θ
vt
1
1
=
=
ut u v 1.6
so
⎛ 1 ⎞
⎟ = 38.7&deg;
⎝ 1.6 ⎠
θ = sin −1 ⎜
Substitute numerical values and
evaluate d:
d = (3.0 km )
1
= 4.8 km
1 − cos 2 38.7&deg;
89 ••• The Hubble space telescope has been used to determine the existence
of planets orbiting distant stars. The planet orbiting the star will cause the star to
″wobble″ with the same period as the planet’s orbit. Because of wobble, light
from the star will be Doppler-shifted up and down periodically. Estimate the
maximum and minimum wavelengths of light of nominal wavelength 500 nm
emitted by the Sun that is Doppler-shifted by the motion of the Sun due to the
planet Jupiter.
Traveling Waves 1519
Picture the Problem The Sun and Jupiter orbit about their effective mass located
at their common center of mass. We can apply Newton’s second law to the Sun to
obtain an expression for its orbital speed about the Sun-Jupiter center of mass and
then use this speed in the Doppler shift equation to estimate the maximum and
minimum wavelengths resulting from the Jupiter-induced motion of the Sun.
Letting v be the orbital speed of the
sun about the center of mass of the
Sun-Jupiter system, express the
Doppler shift of the light due to this
motion when the Sun is approaching
Earth:
v
v
1+
1+
c
c
c
c =
f' = = f
v
v λ
λ'
1−
1−
c
c
Solving for λ′ and simplifying
yields:
v
−1
c = λ ⎛⎜1 − v ⎞⎟⎛⎜1 + v ⎞⎟
λ' = λ
v
⎝ c ⎠⎝ c ⎠
1+
c
1−
12
⎛ v⎞ ⎛ v⎞
= λ ⎜1 − ⎟ ⎜1 + ⎟
⎝ c⎠ ⎝ c⎠
Because v &lt;&lt; c, we can expand
12
⎛ v⎞
⎛ v⎞
⎜1 − ⎟ and ⎜1 + ⎟
⎝ c⎠
⎝ c⎠
binomially to obtain:
−1 2
12
⎛ v⎞
Substitute for ⎜1 − ⎟ and
⎝ c⎠
⎛ v⎞
⎜1 + ⎟
⎝ c⎠
−1 2
to obtain:
When the Sun is receding from
Earth:
Hence the motion of the Sun will
give an observed Doppler shift of:
12
⎛ v⎞
⎜1 − ⎟
⎝ c⎠
and
⎛ v⎞
⎜1 + ⎟
⎝ c⎠
≈ 1−
−1 2
v
2c
−1 2
≈ 1−
v
2c
v
2
c = ⎛⎜1 − v ⎞⎟ ≈ 1 − v
v ⎝ 2c ⎠
c
1+
c
1−
v
2
c = ⎛⎜1 + v ⎞⎟ ≈ 1 + v
v ⎝ 2c ⎠
c
1−
c
1+
⎛
⎝
v⎞
c⎠
λ ' ≈ λ ⎜1 &plusmn; ⎟
(1)
1520 Chapter 15
Apply Newton’s second law to the
Sun:
GM S M eff
GM eff
v2
=
M
⇒v =
S
2
rcm
rcm
rcm
Measured from the center of the Sun,
the distance to the center of mass of
the Sun-Jupiter system is:
rcm =
The effective mass is related to the
masses of the Sun and Jupiter
according to:
M sM J
1
1
1
=
+
⇒ M eff =
M eff M s M J
Ms + MJ
Substitute for Meff and rcm to obtain:
(0)M S + rs-J M J
Ms + MJ
=
rs-J M J
Ms + MJ
MsM J
Ms + MJ
GM s
=
rs-J M J
rs-J
Ms + MJ
G
v=
Using rs-J = 7.78 &times; 1011 m as the mean orbital radius of Jupiter, substitute
numerical values and evaluate v:
v=
(6.673 &times;10
−11
N ⋅ m 2 / kg 2 )(1.99 &times; 1030 kg )
= 1.306 &times; 104 m/s
11
7.78 &times; 10 m
Substitute numerical values in
equation (1) to obtain:
⎛
λ ' ≈ (500 nm ) ⎜⎜1 &plusmn;
⎝
= (500 nm ) 1 &plusmn; 4.36 &times; 10 −5
(
The maximum and minimum
wavelengths are:
1.306 &times; 10 4 m/s ⎞
⎟
2.998 &times; 108 m/s ⎟⎠
)
λmax = 500.02 nm
and
λmin = 499.98 nm
General Problems
90 •
At time t = 0, the shape of a wave pulse on a string is given by the
2
function y ( x,0) = 0.120 m 3 / (2.00 m ) + x 2 , where x is in meters. (a) Sketch
y(x, 0) versus x. (b) Give the wave function y(x,t) at a general time t if the pulse is
moving in the +x direction with a speed of 10.0 m/s and if the pulse is moving in
the −x direction with a speed of 10.0 m/s.
(
)
Traveling Waves 1521
Picture the Problem The equation of a wave traveling in the +x direction is of
the form y (x,t ) = f ( x − vt ) and that of a wave traveling in the −x direction
is y (x,t ) = f ( x + vt ) .
(a) The pulse at t = 0 shown below was plotted using a spreadsheet program:
0.035
0.030
y (x ,0) (m)
0.025
0.020
0.015
0.010
0.005
0.000
-6
-4
-2
0
2
4
6
x (m)
(b) If the pulse is moving in the +x
direction, the wave function must be
of the form:
Replace x with x − (10.0 m/s )t to
obtain:
If the pulse is moving in the −x
direction, the wave function must
be of the form:
Replace x with x + (10.0 m/s )t to
obtain:
y ( x,t ) = f ( x − vt ) = f [x − (10.0 m/s )t ]
because v = 10.0 m/s
y ( x,t ) =
0.120 m 3
(2.00 m )2 + [x − (10.0 m/s )t ] 2
y ( x,t ) = f ( x + vt ) = f [x + (10.0 m/s )t ]
because v = 10.0 m/s
y ( x, t ) =
0.120 m 3
(2.00 m ) 2 + [x + (10.0 m/s )t ] 2
1522 Chapter 15
91 •
[SSM] A whistle that has a frequency of 500 Hz moves in a circle of
radius 1.00 m at 3.00 rev/s. What are the maximum and minimum frequencies
heard by a stationary listener in the plane of the circle and 5.00 m away from its
center?
Picture the Problem The pictorial
representation depicts the whistle
traveling in a circular path of radius
r = 1.00 m. The stationary listener will
hear the maximum frequency when the
whistle is at point 1 and the minimum
frequency when it is at point 2. These
maximum and minimum frequencies
are determined by f0 and the tangential
speed us = 2π r/T. We can relate the
frequencies heard at point P to the
speed of the approaching whistle at
point 1 and the speed of the receding
whistle at point 2.
Relate the frequency heard at point P
to the speed of the approaching
whistle at point 1:
Because us = rω :
r = 1.00 m
1
2
us
5.00 m
P
1
f
us s
1−
v
f max =
1
f
rω s
1−
v
f max =
Substitute numerical values and evaluate fmax:
f max =
1
(1.00 m )⎛⎜ 3.00 rev &times; 2π rad ⎞⎟
s
rev ⎠
⎝
1−
343 m/s
Relate the frequency heard at point P
to the speed of the receding whistle
at point 2:
f min =
(500 Hz ) =
1
f
us s
1+
v
529 Hz
us
Traveling Waves 1523
Substitute numerical values and evaluate fmin:
f min =
1
(1.00 m )⎛⎜ 3.00 rev &times; 2π rad ⎞⎟
s
rev ⎠
⎝
1+
343 m/s
(500 Hz ) =
474 Hz
92 •
Ocean waves move toward the beach with a speed of 8.90 m/s and a
crest-to-crest separation of 15.0 m. You are in a small boat anchored off shore.
(a) At what frequency do the wave crests reach you boat? (b) You now lift anchor
and head out to sea at a speed of 15.0 m/s. At what frequency do the wave crests
Picture the Problem (a) The crest-to-crest separation of the waves is their
wavelength. We can find the frequency of the waves from v = fλ. (b) When you
lift anchor and head out to sea you’ll become a moving receiver and we can
v &plusmn; ur
apply f r =
f s to calculate the frequency you’ll observe.
v &plusmn; us
v
8.90 m/s
= 0.593 Hz
15.0 m
(a) The frequency of the ocean
waves is the ratio of their speed to
their wavelength:
f0 =
(b) Express the frequency of the
waves in terms of their speed and the
fr =
Substitute numerical values and
evaluate fr:
⎛ 15.0 m/s ⎞
⎟⎟ (0.593 Hz )
f r = ⎜⎜1 +
⎝ 8.90 m/s ⎠
λ
=
v &plusmn; ur
v + ur
⎛ u ⎞
fs =
f s = ⎜1 + r ⎟ f s
v &plusmn; us
v
v⎠
⎝
= 1.59 Hz
93 ••
A 12.0-m-long wire that has an 85.0-g mass is stretched under a
tension of 180 N. A pulse is generated at the left end of the wire, and 25.0 ms
later a second pulse is generated at the right end of the wire. Where do the pulses
first meet?
1524 Chapter 15
Picture the Problem Let t be the time of travel of the left-hand pulse and the
subscripts L and R refer to the pulse coming from the left and right, respectively.
Because the pulse traveling from the right starts later than the pulse from the left,
its travel time is t − Δt, where Δt = 25.0 ms. Both pulses travel at the same speed
and the sum of the distances they travel is 12.0 m.
Express the total distance the two
pulses travel:
Solve for vt to obtain:
The speed of the pulse is given by:
d = d L + d R = vt + v(t − Δt )
vt =
v=
1
2
(d + vΔt )
FT
μ
=
FT
m L
⎛
⎞
FT
Δt ⎟⎟
vt = 12 ⎜⎜ d +
m L ⎠
⎝
Substitute numerical values and evaluate vt:
Substituting for v yields:
⎡
(180 N )(12.0 m ) (25.0 &times;10 −3 s )⎤ = 7.99 m
vt = 12 ⎢12.0 m +
⎥
0.085 kg
⎣
⎦
from the left end of the wire.
94 ••
You are parked on the shoulder of a highway. Find the speed of a car
in which the tone of the car’s horn drops by 10 percent as it passes you. (In other
words, the total drop in frequency between the ″approach″ value and the
″recession″ value is 10%.)
Picture the Problem Let the frequency of the car’s horn be fs, the frequency you
hear as the car approaches be fr, and the frequency you hear as the car recedes be
v &plusmn; ur
fr′. We can use f r =
f s to express the frequencies heard as the car
v &plusmn; us
approaches and recedes and then use these frequencies to express the fractional
change in frequency as the car passes you.
Express the fractional change in
frequency as the car passes you:
Δf
= 0.10
fr
Relate the frequency heard as the car
approaches to the speed of the car:
fr =
v &plusmn; ur
v&plusmn;0
fs =
fs =
v − us
v &plusmn; us
1
f
us s
1−
v
Traveling Waves 1525
Express the frequency heard as the
car recedes in terms of the speed of
the car:
Divide the second of these frequency
equations by the first to obtain:
f r' =
1
f
us s
1+
v
us
f r'
v
=
us
fr
1+
v
and
1−
us
f r f r' Δf
v = 0.10
−
=
= 1−
u
fr fr
fr
1+ s
v
1−
Solving for us yields:
us =
Substitute numerical values and
evaluate us:
us =
0.10
v
1.9
0.10
(343 m/s)
1.9
m 1 km 3600 s
= 18.05 &times; 3 &times;
s 10
h
= 65 km/h
95 ••
[SSM] A loudspeaker driver 20.0 cm in diameter is vibrating at
800 Hz with an amplitude of 0.0250 mm. Assuming that the air molecules in the
vicinity have the same amplitude of vibration, find (a) the pressure amplitude
immediately in front of the driver, (b) the sound intensity, and (c) the acoustic
power being radiated by the front surface of the driver.
Picture the Problem (a) and (b) The pressure amplitude can be calculated
directly from p0 = ρωvs0 , and the intensity from I = 12 ρω 2 s02 v. (c) The power
radiated is the intensity times the area of the driver.
(a) Relate the pressure amplitude to
the displacement amplitude, angular
frequency, wave velocity, and air
density:
p0 = ρωvs0
Substitute numerical values and
evaluate p0:
p0 = 1.29 kg/m 3 2π 800 s −1
)[ (
)]
&times; (343 m/s) (0.0250 &times;10 m )
(
−3
= 55.6 N/m 2
1526 Chapter 15
(b) Relate the intensity to these same
quantities:
I = 12 ρω 2 s02 v
Substitute numerical values and
evaluate I:
I = 12 1.29 kg/m 3 2π 800 s −1
(
)[ (
(
&times; 0.0250 &times; 10 −3 m
)]
2
) (343 m/s)
2
= 3.494 W/m 2 = 3.49 W/m 2
(c) Express the power in terms of the
intensity and the area of the driver:
P = IA = π r 2 I
Substitute numerical values and
evaluate P:
P = π (0.100 m ) 3.494 W/m 2
2
(
)
= 0.110 W
96 ••
A plane, harmonic, sound wave in air has an amplitude of 1.00 μm and
an intensity of 10.0 mW/m2. What is the frequency of the wave?
Picture the Problem The frequency of the wave is related to the density of the
air, displacement amplitude, and velocity by I = 12 ρω 2 s02 v.
Relate the intensity of the wave to
the density of the air, displacement
amplitude, velocity, and angular
frequency:
I = 12 ρω 2 s02 v ⇒ ω =
Because ω = 2πf :
f =
1
2π s0
1
s0
2I
ρv
2I
ρv
Substitute numerical values and evaluate f:
f =
1
2π 1.00 &times;10 − 6 m
(
(
)
2 1.00 &times;10 − 2 W/m 2
= 1.07 kHz
1.29 kg/m 3 (343 m/s )
) (
)
97 ••
Water flows at 7.0 m/s in a pipe of radius 5.0 cm. A plate with area
equal to the cross-sectional area of the pipe is suddenly inserted to stop the flow.
Find the force exerted on the plate. Take the speed of sound in water to be
1.4 km/s. Hint: When the plate is inserted, a pressure wave propagates through
the water at the speed of sound vs. The mass of water brought to a stop in time Δt
is the water in a length of pipe equal to vsΔt.
Traveling Waves 1527
Picture the Problem The force exerted on the plate is due to the change in
momentum of the water. We can use Newton’s second law in the form F = Δp/Δt
to relate F to the mass of water in a length of pipe equal to vsΔt and to the speed
of the water. This mass of water, in turn, is given by the product of its density and
the volume of water in a length of pipe equal to vsΔt.
Δp Δmv w
=
Δt
Δt
Relate the force exerted on the plate
to the change in momentum of the
water:
F=
Express Δm in terms of the mass of
water in a length of pipe equal to
vsΔt:
Δm = ρΔV = ρvs AΔt
Substitute for Δm to obtain:
F = ρvs Avw
Substitute numerical values and evaluate F:
(
)
[
F = 1.00 &times; 10 3 kg/m 3 (1.4 km/s ) π (0.050 m )
2
](7.0 m/s) =
77 kN
98 ••
A high-speed flash photography setup to capture a picture of a bullet
exploding a soap bubble is shown in Figure 15-33. The shock wave from the
bullet is to be detected by a microphone that will trigger the flash. The
microphone is placed on a track that is parallel to, and 0.350 m below, the path of
the bullet. The track is used to adjust the position of the microphone. If the bullet
is traveling at 1.25 times the speed of sound, how far back from the soap bubble
must the microphone be set to trigger the flash? (Assume that the flash itself is
instantaneous once the microphone is triggered.)
Picture the Problem Let d be the horizontal distance from the soap bubble to the
position of the microphone. The angle θ of the shock wave is related to the speed
of sound in air u and the speed of the bullet v according to sin θ = u v . We can
determine θ from the given information and then use this angle to find d.
0.350 m
tan θ
Express d in terms of the angle of the
shock wave and the distance from
the soap bubble to the laboratory
bench:
d=
Relate the speed of the bullet to the
angle of the shock-wave cone:
sin θ =
u
⎛u⎞
⇒ θ = sin −1 ⎜ ⎟
v
⎝v⎠
(1)
1528 Chapter 15
Substitute for θ in equation (1) to
obtain:
d=
Substitute numerical values and
evaluate d:
d=
0.350 m
⎡
⎛ u ⎞⎤
tan ⎢sin −1 ⎜ ⎟⎥
⎝ v ⎠⎦
⎣
0.350 m
= 26.3 cm
⎡ −1 ⎛ u ⎞⎤
tan ⎢sin ⎜
⎟⎥
⎝ 1.25u ⎠⎦
⎣
99 ••
A column of precision marchers keeps in step by listening to the band
positioned at the head of the column. The beat of the music is for 100 paces/min.
A television camera shows that only the marchers at the front and the rear of the
column are actually in step. The marchers in the middle section are striding
forward with the left foot when those at the front and rear are striding forward
with the right foot. The marchers are so well trained, however, that they are all
certain that they are in proper step with the music. How long is the column?
Picture the Problem The source of the problem is that it takes a finite time for
the sound to travel from the front of the line of marchers to the back. We can use
the given data to determine the time required for the beat to reach the marchers in
the back of the column and then use this time and the speed of sound to find the
length of the column.
Express the length of the column in
terms of the speed of sound and the
time required for the beat to travel
the length of the column:
L = vΔt
Calculate the time for the sound to
travel the length of the column:
Δt =
Substitute numerical values and
evaluate L:
L = (343 m/s )(0.600 s ) = 206 m
1
min = 0.600 s
100
100 ••
A bat flying toward a stationary obstacle at 12.0 m/s emits brief, highfrequency sound pulses at a repetition frequency of 80.0 Hz. What is the interval
between the arrival times of the reflected pulses heard by the bat?
Picture the Problem The interval between the arrival times of the reflected
pulses heard by the bat is the reciprocal of the frequency of the reflected pulses.
v &plusmn; ur
We can use f r =
f s to relate the frequency of the reflected pulses to the
v &plusmn; us
speed of the bat and the frequency it emits.
Traveling Waves 1529
1
fr
Relate the interval between the
arrival times of the echo pulses heard
by the bat to frequency of the
reflected pulses:
Δt =
Relate the frequency of the pulses
received by the bat to its speed and
the frequency it emits:
u
1+ r
v &plusmn; ur
v + ur
v f
fr =
fs =
fs =
us s
v &plusmn; us
v − us
1−
v
u
1− s
v
Δt =
⎛ ur ⎞
⎜1 + ⎟ f s
v⎠
⎝
Substitute for fr to obtain:
Substitute numerical values and
evaluate Δt:
12.0 m/s
343 m/s
Δt =
= 11.7 ms
⎛ 12.0 m/s ⎞
−1
⎜⎜1 +
⎟⎟ 80.0 s
⎝ 343 m/s ⎠
1−
(
)
101 ••
Laser ranging to the moon is done routinely to accurately determine
the Earth–moon distance. However, to determine the distance accurately,
corrections must be made for the average speed of light in Earth’s atmosphere,
which is 99.997 percent of the speed of light in vacuum. Assuming that Earth’s
atmosphere is 8.00 km high, estimate the length of the correction.
Picture the Problem Let d be the distance to the moon, h be the height of Earth's
atmosphere, and v be the average speed of light in Earth’s atmosphere. We can
express d ′, the distance measured when Earth’s atmosphere is ignored, in terms
of the time for a pulse of light to make a round-trip from Earth to the moon and
solve this equation for the length of correction d ′ − d.
Express the roundtrip time for a
pulse of light to reach the moon and
return:
Express the &quot;measured&quot; distance d ′
when we do not account for the
atmosphere:
t = t Earth's atmosphere + t out of Earth's atmosphere
h
d −h
=2 +2
v
c
1
1 ⎛ h
d −h⎞
ct = c⎜ 2 + 2
⎟
2
2 ⎝ v
c ⎠
c
= h+d −h
v
d' =
1530 Chapter 15
Solve for the length of correction
d ′ − d:
⎛c ⎞
d' − d = h⎜ − 1⎟
⎝v ⎠
Substitute numerical values and
evaluate d ′ − d:
c
⎛
⎞
d' − d = (8.00 km )⎜
− 1⎟
⎝ 0.99997c ⎠
≈ 0.2 m
Remarks: This is larger than the accuracy of the measurements, which is
102 ••
A tuning fork attached to a taut string generates transverse waves. The
vibration of the fork is perpendicular to the string. Its frequency is 400 Hz and the
amplitude of its oscillation is 0.50 mm. The string has a linear mass density of
0.010 kg/m and is under a tension of 1.0 kN. Assume that there are no waves
reflected at the far end of the string. (a) What are the period and frequency of
waves on the string? (b) What is the speed of the waves? (c) What are the
wavelength and wave number? (d) What is a suitable wave function for the waves
on the string? (e) What is the maximum speed and acceleration of a point on the
string? (f) At what minimum average rate must energy be supplied to the fork to
keep it oscillating at a steady amplitude?
Picture the Problem (a) The frequency of the waves on the string is the same as
the frequency of the tuning fork and their period is the reciprocal of the
frequency. (b) We can find the speed of the waves from the tension in the string
and its linear density. (c) The wavelength can be determined from the frequency
and the speed of the waves and the wave number from its definition. (d) The
general form of the wave function for waves on a string is y (x,t ) = A sin (kx &plusmn; ωt ) ,
so, once we know k and ω, because A is given, we can write a suitable wave
function for the waves on this string. (e) The maximum speed and acceleration of
a point on the string can be found from the angular frequency and amplitude of
the waves. (f) Finally, we can use Pav = 12 μω 2 A2 v to find the minimum average
rate at which energy must be supplied to the tuning fork to keep it oscillating with
(a) The frequency of the waves on
the string is the same as the
frequency of the tuning fork:
f = 400 Hz
The period of the waves on the wire
is the reciprocal of their frequency:
T=
1
1
=
= 2.50 ms
f 400 s −1
Traveling Waves 1531
(b) Relate the speed of the waves to
the tension in the string and its linear
density:
v=
FT
μ
=
1.0 kN
= 316.2 m/s
0.010 kg/m
= 0.32 km/s
(c) Use the relationship between the
wavelength, speed and frequency of
a wave to find λ:
λ=
Using its definition, express and
evaluate the wave number:
k=
v 316.2 m/s
=
= 79.05 cm
f
400 s −1
= 79 cm
2π
λ
=
2π
= 7.95 m −1
79.05 &times; 10 − 2 m
= 7.9 m −1
(d) Determine the angular frequency
of the waves:
ω = 2πf = 2π (400 s −1 ) = 2.51&times; 10 3 s −1
Substitute for A, k, and ω in the general form of the wave function to obtain:
y ( x,t ) =
(0.50 mm )sin[(7.9 m −1 ) x − (2.51&times;10 3 s −1 ) t ]
(e) Relate the maximum speed of a
point on the string to the amplitude
of the waves and the angular
frequency of the tuning fork:
vmax = Aω
Express the maximum acceleration
of a point on the string in terms of
the amplitude of the waves and the
angular frequency of the tuning fork:
amax = Aω 2
(f) Express the minimum average
power required to keep the tuning
fork oscillating at a steady amplitude
in terms of the linear density of the
string, the amplitude of its
vibrations, and the speed of the
waves on the string:
Pav = 12 μω 2 A2 v
(
)(
)
)(
)
= 0.50 &times; 10−3 m 2.51&times; 103 s −1
= 1.3 m/s
(
= 0.50 &times; 10− 3 m 2.51&times; 103 s −1
= 3.2 km/s2
2
1532 Chapter 15
Substitute numerical values and evaluate Pav:
Pav =
1
2
(0.010 kg/m )(2.51&times;103 s −1 )2 (0.50 &times;10−3 m )2 (316 m/s) =
2.5 W
103 ••• A long rope with a mass per unit length of 0.100 kg/m is under a
constant tension of 10.0 N. A motor drives one end of the rope with transverse
simple harmonic motion at 5.00 cycles per second and an amplitude of
40.0 mm. (a) What is the wave speed? (b) What is the wavelength? (c) What is
the maximum transverse linear momentum of a 1.00-mm segment of the rope?
(d) What is the maximum net force on a 1.00-mm segment of the rope?
Picture the Problem Let Δm represent the mass of the segment of length
Δx = 1.00 mm. We can find the wave speed from the given data for the tension in
the rope and its linear density. The wavelength can be found from v = fλ. We’ll
use the definition of linear momentum to find the maximum transverse linear
momentum of the 1-mm segment and apply Newton’s second law to the segment
to find the maximum net force on it.
(a) Find the wave speed from the
tension and linear density:
v=
FT
μ
=
10.0 N
= 10.0 m/s
0.100 kg/m
v 10.0 m/s
=
= 2.00 m
5.00 s −1
f
(b) Express the wavelength in terms
of the speed and frequency of the
wave:
λ=
(c) Relate the maximum transverse
linear momentum of the 1.00-mm
segment to the maximum transverse
speed of the wave:
pmax = Δmvmax = μΔxAω = 2πfμΔxA
Substitute numerical values and
evaluate pmax:
pmax = 2π 5.00 s −1 (0.100 kg/m )
(
)
&times; (1.00 &times; 10
−3
)
m (0.0400 m )
−4
= 1.257 &times; 10 kg ⋅ m/s
= 1.26 &times; 10− 4 kg ⋅ m/s
(d) The maximum net force acting on
the segment is the product of the mass
of the segment and its maximum
acceleration:
Fmax = Δmamax = μΔxAω 2 = ωpmax
= 2πfpmax
Traveling Waves 1533
Substitute numerical values and
evaluate Fmax:
(
)(
Fmax = 2π 5.00 s −1 1.257 &times; 10 −4 kg ⋅ m/s
)
= 3.95 mN
104 ••• In this problem, you will derive an expression for the potential energy
of a segment of a string carrying a traveling wave (Figure 15-34). The potential
energy of a segment equals the work done by the tension in stretching the string,
which is ΔU = FT (Δl − Δx ), where FT is the tension, Δl is the length of the
stretched segment, and Δx is its original length. (a) Use the binomial expansion to
2
⎛ Δy ⎞
show that Δl − Δx ≈ (Δy Δx ) Δx , and therefore ΔU ≈ FT ⎜ ⎟ Δx .
⎝ Δx ⎠
(b) Compute ∂y/∂x from the wave function y ( x, t ) = A sin (kx − ωt ) (Equation 1515) and show that ΔU ≈ 12 FT A 2 k 2 cos 2 (kx − ωt )Δx .
1
2
2
1
2
Picture the Problem We can follow the step-by-step instructions outlined above
to obtain the given expressions for ΔU.
(a) Express the potential energy of a
segment of the string:
For Δy/Δx &lt;&lt; 1:
ΔU = FT (Δl − Δx )
⎡ 1 ⎛ Δy ⎞ 2 ⎤
Δl = Δx ⎢1 + 2 ⎜ ⎟ ⎥
⎝ Δx ⎠ ⎦⎥
⎣⎢
and
⎡ 1 ⎛ Δy ⎞ 2 ⎤
Δl − Δx = Δx ⎢1 + 2 ⎜ ⎟ ⎥ − Δx
⎝ Δx ⎠ ⎥⎦
⎢⎣
2
=
1
2
⎛ Δy ⎞
⎜ ⎟ Δx
⎝ Δx ⎠
Substitute to obtain:
⎡ ⎛ Δy ⎞ 2 ⎤
ΔU = FT ⎢ 12 ⎜ ⎟ ⎥ =
⎢⎣ ⎝ Δx ⎠ ⎥⎦
(b) Differentiate
y ( x,t ) = A sin (kx − ωt ) to obtain:
∂y
= kA cos(kx − ωt )
∂x
Approximate Δy/Δx by ∂y/∂x and
substitute in our result from Part (a):
ΔU = 12 FT (kA cos(kx − ωt )) Δx
2
1
2
⎛ Δy ⎞
FT ⎜ ⎟ Δx
⎝ Δx ⎠
2
=
1
2
FT A 2 k 2 Δx cos 2 (kx − ωt )
1534 Chapter 15
105 ••• One end of a heavy, 3.00-m-long rope is attached to a high ceiling and
the rest of the rope is allowed to hang freely. Show that transverse waves on the
∂ ⎛ ∂y ⎞ 1 ⎛ ∂ 2 y ⎞
∂2 y 1 ∂2 y
=
.
⎜ x ⎟ = ⎜⎜ 2 ⎟⎟ instead of
∂x ⎝ ∂x ⎠ g ⎝ ∂t ⎠
∂x 2 v 2 ∂t 2
Picture the Problem Let the +x
direction be straight upward and choose
the origin 3.00 m below the ceiling. Let
y be the transverse direction in which
the rope is displaced. Applying
Newton’s second law to an element of
length of the rope will lead to the given
equation, which relates the spatial
derivatives of y(x,t) to its time
derivatives. We’ll consider only small
values of the angles θ1 and θ2 (see the
diagram to the right) between the
tangent to the rope and the x axis.
Apply
∑F
x
= ma x to the segment
whose mass is Δm and whose length
is Δx to obtain:
The mass of the segment is the
product of its linear density and
length:
θ2
x
FT
x2
Δm
Δx
Fg = (Δm) g
x1
θ1
0
FT ( x2 ) cos θ 2 − FT ( x1 ) cos θ1 − (Δm )g = 0
Note that the segment is not accelerated
vertically.
Δm = μΔx
Substituting for Δm yields:
FT ( x2 ) cosθ 2 − FT ( x1 ) cos θ1 − gμΔx = 0
Apply the small-angle
approximation ( cosθ = 1 for θ &lt;&lt; 1 )
and let x1 = x and x2 = x1 + Δx to
obtain:
FT
(1)
FT ( x2 ) − FT ( x1 ) − gμΔx = 0
or
FT ( x + Δx ) − FT ( x ) = gμΔx
Traveling Waves 1535
Noting that the tension in the rope is
zero at x = 0, solve for the tension at
x2 :
FT (Δx ) = gμΔx ⇒ FT ( x2 ) = gμx2
Using the differential approximation we have:
⎞
⎛ ∂F
⎟Δx = FT ( x1 ) + ∂FT Δx
FT ( x2 ) = FT ( x1 ) + ⎜ T
(2)
⎜ ∂x x = x ⎟
∂
x
1
1 ⎠
⎝
∂F
∂F
in order to be more concise.
where we’ve written T in place of T
∂x1
∂x x = x1
∂FT
Δx − FT ( x1 ) − gμΔx = 0
∂x1
Substituting equation (2) in equation
(1) yields:
FT ( x1 ) +
Simplify this equation to obtain:
∂FT
− gμ = 0
∂x1
The segment moves horizontally, and
the net force acting in this direction is:
∑F
y
(3)
= FT ( x2 )sin θ 2 − FT ( x1 )sin θ1
Using the small-angle approximation sin θ ≈ tan θ yields:
∑F
y
∂y
∂y
− FT ( x1 )
∂x2
∂x1
∂y
where we’ve also used tan θ = slope =
.
∂x
≈ FT ( x2 ) tan θ 2 − FT ( x1 ) tan θ1 = FT ( x2 )
Applying the differential
approximation gives:
Substituting for FT ( x2 ) and
(4)
∂y
∂y ⎛ ∂ ∂y ⎞
∂y ∂ 2 y
⎟⎟Δx =
=
+ ⎜⎜
+
Δx
∂x2 ∂x1 ⎝ ∂x ∂x1 ⎠
∂x1 ∂x12
∂y
in equation (4) yields:
∂x2
⎛
⎞ ⎛ ∂y ∂ 2 y ⎞
∂FT
∂y
⎜
(
)
=
+
Δ
F
F
x
x
∑ y ⎜ T 1 ∂x ⎟⎟ ⎜⎜ ∂x + ∂x 2 Δx ⎟⎟ − FT (x1 ) ∂x
1
1
1
⎝
⎠⎝ 1
⎠
1536 Chapter 15
Expand and simplify this expression to obtain:
∑ Fy = FT (x1 )
≈ FT ( x1 )
∂F ∂y
∂F ∂ 2 y
∂y
∂2 y
∂y
2
+ FT ( x1 ) 2 Δx + T
Δx + T 2 (Δx ) − FT (x1 )
∂x1
∂x1
∂x1 ∂x1
∂x1 ∂x1
∂x1
∂F ∂y
∂2 y
Δx + T
Δx
2
∂x1 ∂x1
∂x1
where we have neglected the term with (Δx ) .
2
∑F
= ma y to our length
FT ( x1 )
∂FT ∂y
∂2 y
∂2 y
Δ
+
Δ
=
Δ
x
x
x
μ
∂x1 ∂x1
∂x12
∂t 2
Divide both sides of this equation by
Δx to obtain:
FT ( x1 )
∂ 2 y ∂FT ∂y
∂2 y
+
=
μ
(5)
∂x12 ∂x1 ∂x1
∂t 2
Substituting μgx for FT yields:
∂2 y
∂ 2 y ∂ (μgx ) ∂y
=μ 2
μgx 2 +
∂x ∂x
∂t
∂x
where we have dropped the subscript 1
by replacing x1 with x. This can be done
without loss of generality because all
the x′s in equation (5) are x1′s.
Applying
y
element yields:
Divide both sides of the equation by
μg to obtain:
Note that the expression on the lefthand side of the equation can be
written as:
Substituting in equation (6) yields:
x
∂ 2 y ∂y 1 ∂ 2 y
+
=
∂x 2 ∂x g ∂t 2
(6)
∂ 2 y ∂y ∂ ⎛ ∂y ⎞
+
= ⎜ x ⎟ , a result you
∂x 2 ∂x ∂x ⎝ ∂x ⎠
can verify by using the chain rule to
differentiate the right-hand side of this
equation.
x
∂ ⎛ ∂y ⎞ 1 ∂ 2 y
⎜x ⎟ =
∂x ⎝ ∂x ⎠ g ∂t 2
```