Chapter 15 Wave Motion
FTB ⎞2
⎠⎟
fB = vB λ = vB = mL FTB → FTB = ⎛ fB = ⎛ 392 ⎞2 = 0.630
fA vA λ vA = FTA FTA ⎝⎜ fA ⎝⎜ 494 ⎟⎠
FTA
mL
82. Relative to the fixed needle position, the ripples are moving with a linear velocity given by
v = ⎛ 33 rev ⎞ ⎛ 1min ⎞⎛ 2π (0.108 m) ⎞ = 0.3732 m s
⎝⎜ min ⎟⎠ ⎜⎝ 60 s ⎟⎠ ⎜⎝
1 rev ⎠⎟
This speed is the speed of the ripple waves moving past the needle. The frequency of the waves is
f = v 0.3732 m s = 240.77 Hz ≈ 240 Hz
λ = 1.55 ×10−3 m
83. The speed of the pulses is found from the tension and mass per unit length of the wire.
v= FT = 255 N = 129.52 m s
μ 0.152 kg 10.0 m
The total distance traveled by the two pulses will be the length of the wire. The second pulse has a
shorter time of travel than the first pulse, by 20.0 ms.
( )l = d1 + d2 = vt1 + vt2 = vt1 + v t1 − 2.00 ×10−2
t1 = l + 2.00 ×10−2 v = (10.0 m) + 2.00 ×10−2 (129.52 m s) = 4.8604 ×10−2s
2v 2(129.52 m s)
( )d1 = vt1 = (129.52 m s) 4.8604 ×10−2s = 6.30 m
The two pulses meet 6.30 m from the end where the first pulse originated.
84. We take the wave function to be D ( x,t) = Asin (kx − ωt) . The wave speed is given by v = ω = λ ,
k f
while the speed of particles on the cord is given by ∂D .
∂t
∂D = −ω Acos (kx − ωt ) → ⎛ ∂D ⎞ = ωA
∂t ⎝⎜ ∂t ⎟⎠max
ω A = v = ω → A= 1 = λ 10.0 cm = 1.59 cm
k k 2π = 2π
85. For a resonant
condition, the free 1
end of the string
n=1
will be an 0 n=5 n=3
antinode, and the 0
fixed end of the 1
string will be a -1
node. The
minimum distance
from a node to an antinode is λ 4 . Other wave patterns that fit the boundary conditions of a node at
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Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual
one end and an antinode at the other end include 3λ 4 , 5λ 4 , … . See the diagrams. The general
relationship is l = (2n −1) λ 4 , n = 1,2,3, . Solving for the wavelength gives
λ = 4l , n = 1, 2,3, .
2n −1
86. The addition of the support will force the bridge to have its lowest mode of oscillation to have a node
at the center of the span, which would be the first overtone of the fundamental frequency. If the
wave speed in the bridge material remains constant, then the resonant frequency will double, to
6.0 Hz. Since earthquakes don’t do significant shaking at that frequency, the modifications would be
effective at keeping the bridge from having large oscillations during an earthquake.
87. From the figure, we can see that the amplitude is 3.5 cm, and the wavelength is 20 cm. The
maximum of the wave at x = 0 has moved to x = 12 cm at t = 0.80 s, which is used to find the
velocity. The wave is moving to the right. Finally, since the displacement is a maximum at x = 0
and t = 0, we can use a cosine function without a phase angle.
A = 3.5cm; λ = 20cm → k = 2π = 0.10π cm−1; v = 12 cm = 15cm s; ω = vk = 1.5π rad s
λ 0.80 s
D ( x,t) = Acos(kx − ωt) = (3.5cm) cos(0.10π x −1.5π t) , x in cm, t in s
88. From the given data, A = 0.50 m and v = 2.5 m 4.0s = 0.625 m s. We use Eq. 15-6 for the average
power, with the density of sea water from Table 13-1. We estimate the area of the chest as
(0.30 m)2 . Answers may vary according to the approximation used for the area of the chest.
( )P = 2π 2ρSvf 2 A2 = 2π 2 1025kg m3 (0.30 m)2 (0.625m s) (0.25 Hz)2 (0.50 m)2
= 18 W
89. The unusual decrease of water corresponds to a trough in Figure 15-4. The crest or peak of the wave
is then one-half wavelength from the shore. The peak is 107.5 km away, traveling at 550 km/hr.
Δx = vt → t = Δx = 1 ( 215 km ) ⎛ 60 min ⎞ = 11.7 min ≈ 12 min
v 2 ⎝⎜ 1hr ⎟⎠
550 km hr
90. At t = 1.0 s, the leading edge of t = 1.0 s
each wave is 1.0 cm from the
other wave. They have not yet
interfered. The leading edge of
the wider wave is at 22 cm, and
the leading edge of the narrower
wave is at 23 cm.
0 5 10 15 20 25 30 35 40
x (cm)
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500
Chapter 15 Wave Motion
At t = 2.0 s, the waves are t = 2.0 s
overlapping. The diagram uses
dashed lines to show the parts of
the original waves that are
undergoing interference.
At t = 3.0 s, the waves have 0 5 10 15 20 25 30 35 40
“passed through” each other,
and are no longer interfering. x (cm)
t = 3.0 s
0 5 10 15 20 25 30 35 40
x (cm)
91. Because the radiation is uniform, the same energy must pass through every spherical surface, which
has the surface area 4π r2. Thus the intensity must decrease as 1 r2 . Since the intensity is
proportional to the square of the amplitude, the amplitude will decrease as 1 r. The radial motion
will be sinusoidal, and so we have D = ⎝⎜⎛ A ⎟⎠⎞ sin (kr − ωt).
r
92. The wavelength is to be 1.0 m. Use Eq. 15-1.
v= fλ → f = v = 344 m s = 340 Hz
λ 1.0 m
There will be significant diffraction only for wavelengths larger than the width of the window, and
so waves with frequencies lower than 340 Hz would diffract when passing through this window.
93. The value of k was taken to be 1.0 m−1 for this problem. The peak of the wave moves to the right by
0.50 m during each second that elapses. This can be seen qualitatively from the graph, and
quantitatively from the spreadsheet data. Thus the wave speed is given by the constant c, 0.50 m s .
The direction of motion is in the positive x direction. The wavelength is seen to be λ = π m . Note
that this doesn’t agree with the relationship λ = 2π . The period of the function sin2 θ is π , not 2π
k
as is the case for sinθ . In a similar fashion the period of this function is T = 2π s . Note that this
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Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual
doesn’t agree with the relationship 5
kv = ω = 2π , again because of the t=0
T
4
behavior of the sin2 θ function. But
t = 1.0 s
the relationship λ = v is still true for D (m)
T 3
this wave function. The spreadsheet 2
used for this problem can be found
on the Media Manager, with t = 2.0 s
filename “PSE4_ISM_CH15.XLS,”
on tab “Problem 15.93.” 1
0
-6 -5 -4 -3 -2 -1 0 1 2 3 4 5
x (m)
Further insight is gained by re-writing the function using the trigonometric identity
sin2 θ = 1 − 1 cos 2θ , because function cos 2θ has a period of π.
2 2
94. (a) The graph shows the wave moving D (x, t ) (m) 10
3.0 m to the right each second, which t=0
is the expected amount since the
speed of the wave is 3.0 m/s and the 8 t = 1.0
form of the wave function says the t = 2.0
wave is moving to the right. The
spreadsheet used for this problem 6
can be found on the Media Manager,
with filename 4
“PSE4_ISM_CH15.XLS,” on tab
“Problem 15.94a.” 2
0
-8 -6 -4 -2 0 2 4 6 8 10
x (m)
(b) The graph shows the wave moving D (x, t ) (m) 10
3.0 m to the left each second, which t=0
is the expected amount since the
speed of the wave is 3.0 m/s and the 8 t = 1.0
form of the wave function says the 6 t = 2.0
wave is moving to the left. The
spreadsheet used for this problem 4
can be found on the Media Manager,
with filename 2
“PSE4_ISM_CH15.XLS,” on tab
“Problem 15.94b.” 0
-10 -8 -6 -4 -2 0 2 4 6 8
x (m)
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502
CHAPTER 16: Sound
Responses to Questions
1. Sound exhibits diffraction, refraction, and interference effects that are characteristic of waves. Sound
also requires a medium, a characteristic of mechanical waves.
2. Sound can cause objects to vibrate, which is evidence that sound is a form of energy. In extreme
cases, sound waves can even break objects. (See Figure 14-24 showing a goblet shattering from the
sound of a trumpet.)
3. Sound waves generated in the first cup cause the bottom of the cup to vibrate. These vibrations
excite vibrations in the stretched string which are transmitted down the string to the second cup,
where they cause the bottom of the second cup to vibrate, generating sound waves which are heard
by the second child.
4. The wavelength will change. The frequency cannot change at the boundary since the media on both
sides of the boundary are oscillating together. If the frequency were to somehow change, there
would be a “pile-up” of wave crests on one side of the boundary.
5. If the speed of sound in air depended significantly on frequency, then the sounds that we hear would
be separated in time according to frequency. For example, if a chord were played by an orchestra, we
would hear the high notes at one time, the middle notes at another, and the lower notes at still
another. This effect is not heard for a large range of distances, indicating that the speed of sound in
air does not depend significantly on frequency.
6. Helium is much less dense than air, so the speed of sound in the helium is higher than in air. The
wavelength of the sound produced does not change, because it is determined by the length of the
vocal cords and other properties of the resonating cavity. The frequency therefore increases,
increasing the pitch of the voice.
7. The speed of sound in a medium is equal to v = B ρ , where B is the bulk modulus and ρ is the
density of the medium. The bulk moduli of air and hydrogen are very nearly the same. The density
of hydrogen is less than the density of air. The reduced density is the main reason why sound travels
faster in hydrogen than in air.
8. The intensity of a sound wave is proportional to the square of the frequency, so the higher-frequency
tuning fork will produce more intense sound.
9. Variations in temperature will cause changes in the speed of sound and in the length of the pipes. As
the temperature rises, the speed of sound in air increases, increasing the resonance frequency of the
pipes, and raising the pitch of the sound. But the pipes get slightly longer, increasing the resonance
wavelength and decreasing the resonance frequency of the pipes and lowering the pitch. As the
temperature decreases, the speed of sound decreases, decreasing the resonance frequency of the
pipes, and lowering the pitch of the sound. But the pipes contract, decreasing the resonance
wavelength and increasing the resonance frequency of the pipes and raising the pitch. These effects
compete, but the effect of temperature change on the speed of sound dominates.
10. A tube will have certain resonance frequencies associated with it, depending on the length of the
tube and the temperature of the air in the tube. Sounds at frequencies far from the resonance
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503
Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual
frequencies will not undergo resonance and will not persist. By choosing a length for the tube that
isn’t resonant for specific frequencies you can reduce the amplitude of those frequencies.
11. As you press on frets closer to the bridge, you are generating higher frequency (and shorter
wavelength) sounds. The difference in the wavelength of the resonant standing waves decreases as
the wavelengths decrease, so the frets must be closer together as you move toward the bridge.
12. Sound waves can diffract around obstacles such as buildings if the wavelength of the wave is large
enough in comparison to the size of the obstacle. Higher frequency corresponds to shorter
wavelength. When the truck is behind the building, the lower frequency (longer wavelength) waves
bend around the building and reach you, but the higher frequency (shorter wavelength) waves do
not. Once the truck has emerged from behind the building, all the different frequencies can reach
you.
13. Standing waves are generated by a wave and its reflection. The two waves have a constant phase
relationship with each other. The interference depends only on where you are along the string, on
your position in space. Beats are generated by two waves whose frequencies are close but not equal.
The two waves have a varying phase relationship, and the interference varies with time rather than
position.
14. The points would move farther apart. A lower frequency corresponds to a longer wavelength, so the
distance between points where destructive and constructive interference occur would increase.
15. According to the principle of superposition, adding a wave and its inverse produces zero
displacement of the medium. Adding a sound wave and its inverse effectively cancels out the sound
wave and substantially reduces the sound level heard by the worker.
16. (a) The closer the two component frequencies are to each other, the longer the wavelength of the
beat. If the two frequencies are very close together, then the waves very nearly overlap, and the
distance between a point where the waves interfere constructively and a point where they interfere
destructively will be very large.
17. No. The Doppler shift is caused by relative motion between the source and observer.
18. No. The Doppler shift is caused by relative motion between the source and observer. If the wind is
blowing, both the wavelength and the velocity of the sound will change, but the frequency of the
sound will not.
19. The child will hear the highest frequency at position C, where her speed toward the whistle is the
greatest.
20. The human ear can detect frequencies from about 20 Hz to about 20,000 Hz. One octave corresponds
to a doubling of frequency. Beginning with 20 Hz, it takes about 10 doublings to reach 20,000 Hz.
So, there are approximately 10 octaves in the human audible range.
21. If the frequency of the sound is halved, then the ratio of the frequency of the sound as the car recedes
to the frequency of the sound as the car approaches is equal to ½. Substituting the appropriate
Doppler shift equations in for the frequencies yields a speed for the car of 1/3 the speed of sound.
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504
Chapter 16 Sound
Solutions to Problems
In these solutions, we usually treat frequencies as if they are significant to the whole number of units. For
example, 20 Hz is taken as to the nearest Hz, and 20 kHz is taken as to the nearest kHz. We also treat all
decibel values as good to whole number of decibels. So 120 dB is good to the nearest decibel.
1. The round trip time for sound is 2.0 seconds, so the time for sound to travel the length of the lake is
1.0 seconds. Use the time and the speed of sound to determine the length of the lake.
d = vt = (343m s) (1.0 s) = 343 m ≈ 340m
2. The round trip time for sound is 2.5 seconds, so the time for sound to travel the length of the lake is
1.25 seconds. Use the time and the speed of sound in water to determine the depth of the lake.
d = vt = (1560 m s)(1.25 s) = 1950 m = 2.0×103m
3. (a) λ20 Hz = v = 343 m s = 17 m λ20 kHz = v = 343 m s = 1.7 ×10−2 m
f 20 Hz f 2.0 ×104 Hz
So the range is from 1.7 cm to 17 m.
(b) λ = v 343m s = 2.3 ×10−5 m
f = 15 ×106 Hz
4. The distance that the sounds travels is the same on both days. That distance is equal to the speed of
sound times the elapsed time. Use the temperature-dependent relationships for the speed of sound in
air.
d = v1t1 = v2t2 → [(331 + 0.6(27)) m s](4.70s) = [(331 + 0.6(T2 )) m s](5.20s) →
T2 = −29°C
5. (a) The ultrasonic pulse travels at the speed of sound, and the round trip distance is twice the
distance d to the object.
( )( )2dmin = vtmin
→ d min = 1 vtmin = 1 343m s 1.0 ×10−3s = 0.17 m
2 2
(b) The measurement must take no longer than 1/15 s. Again, the round trip distance is twice the
distance to the object.
( ) ( )2dmax = vtmax
→ d max = 1 vtmax = 1 343m s s1 = 11m
2 2
15
(c) The distance is proportional to the speed of sound. So the percentage error in distance is the
same as the percentage error in the speed of sound. We assume the device is calibrated to work
at 20°C.
[ ]Δd
d
= Δv = v − v23°C 20°C = 331 + 0.60(23) m s − 343m s = 0.005248 ≈ 0.5 %
v
v20°C 343m s
6. (a) For the fish, the speed of sound in seawater must be used.
d = vt → t = d 1350 m = 0.865s
v = 1560 m s
(b) For the fishermen, the speed of sound in air must be used.
d = vt → t= d = 1350 m = 3.94 s
v 343m s
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Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual
7. The total time T is the time for the stone to fall ( )tdown plus the time for the sound to come back to
( )the top of the cliff tup : T = tup + tdown . Use constant acceleration relationships for an object
dropped from rest that falls a distance h in order to find tdown , with down as the positive direction.
Use the constant speed of sound to find tup for the sound to travel a distance h.
down: y= y0 + v0tdown + 1 atd2own → h = 1 gtd2own up: h = vsndtup → tup = h
2 2 vsnd
( )h 2 ⎛ h ⎞2 ⎛ vsnd ⎞
= 1 gtd2own = 1 g T − tup = 1 g ⎜ T − vsnd ⎟ → h2 − 2vsnd ⎜⎝ g +T ⎟⎠ h + T 2 v2 = 0
2 2 2 ⎠ snd
⎝
This is a quadratic equation for the height. This can be solved with the quadratic formula, but be
sure to keep several significant digits in the calculations.
h2 − 2 (343m s) ⎛ 343 m s + 3.0 s ⎞ h + (3.0s)2 (343m s)2 = 0 →
⎝⎜ 9.80 m s2 ⎠⎟
h2 − (26068 m) h + 1.0588 ×106 m2 = 0 → h = 26028 m , 41m
The larger root is impossible since it takes more than 3.0 sec for the rock to fall that distance, so the
correct result is h = 41m .
8. The two sound waves travel the same distance. The sound will travel faster in the concrete, and thus
take a shorter time.
( )d = v t = v t = vair air vconcrete
tair − 0.75s → tair = 0.75s
concrete concrete concrete v − vconcrete
air
d = v tair air = vair ⎛ vconcrete 0.75s ⎞
⎜ ⎟
⎝ v − vconcrete ⎠
air
The speed of sound in concrete is obtained from Table 16-1 as 3000 m/s.
d = (343m s) ⎛ 3000 m s m s (0.75s) ⎞ = 290 m
⎝⎜ 3000 m s − 343 ⎠⎟
9. The “5 second rule” says that for every 5 seconds between seeing a lightning strike and hearing the
associated sound, the lightning is 1 mile distant. We assume that there are 5 seconds between seeing
the lightning and hearing the sound.
(a) At 30oC, the speed of sound is [331+ 0.60(30)]m s = 349 m s . The actual distance to the
lightning is therefore d = vt = (349 m s) (5s) = 1745 m . A mile is 1610 m.
% error = 1745 −1610 (100) ≈ 8%
1745
(b) At 10oC, the speed of sound is [331+ 0.60(10)] m s = 337 m s . The actual distance to the
lightning is therefore d = vt = (337 m s) (5s) = 1685 m . A mile is 1610 m.
% error = 1685 −1610 (100) ≈ 4%
1685
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506
Chapter 16 Sound
10. The relationship between the pressure and displacement amplitudes is given by Eq. 16-5.
ΔPM = 2πρvAf A ΔPM 3.0 ×10−3 Pa 7.5 ×10−9 m
2πρvf
( )(a) → = = 2π m3 (331m s) (150 Hz) =
1.29 kg
A ΔPM 3.0 ×10−3 Pa = 7.5 ×10−11 m
2πρvf
1.29 kg m3 (331m s) 15 ×103 Hz
( ) ( )(b) = = 2π
11. The pressure amplitude is found from Eq. 16-5. The density of air is 1.29 kg m3 .
( ) ( )(a) ΔPM = 2πρvAf = 2π 1.29 kg m3 (331m s) 3.0 ×10−10 m (55 Hz) = 4.4 ×10−5 Pa
( ) ( )(b) ΔPM = 2πρvAf = 2π 1.29 kg m3 (331m s) 3.0 ×10−10 m (5500 Hz) = 4.4 ×10−3 Pa
12. The pressure wave can be written as Eq. 16-4.
(a) ΔP = −ΔPM cos (kx − ωt)
ΔPM = 4.4 × 10−5 Pa ; ω = 2π f = 2π (55 Hz) = 110π rad s ;k = ω = 110π rad s = 0.33π m−1
v 331m s
( ) ( )ΔP = − 4.4 ×10−5 Pa cos ⎡⎣ 0.33π m−1 x − (110π rad s) t⎦⎤
(b) All is the same except for the amplitude and ω = 2π f = 2π (5500 Hz) = 1.1×104π rad s.
( ) ( ) ( )ΔP = − 4.4 ×10−3 Pa cos ⎣⎡ 0.33π m−1 x − 1.1×104π rad s t⎤⎦
( ) ( )13. The pressure wave is ΔP = (0.0035 Pa) sin ⎣⎡ 0.38π m−1 x − 1350π s−1 t⎦⎤.
(a) λ = 2π = 2π = 5.3 m
k 0.38π m−1
(b) f = ω = 1350π s−1 = 675 Hz
2π 2π
(c) v = ω = 1350π s−1 = 3553m s≈ 3600 m s
k 0.38π m−1
(d) Use Eq. 16-5 to find the displacement amplitude.
ΔPM = 2πρvAf →
ΔPM (0.0035 Pa) 1.0 ×10−13 m
2πρ vf m3 (3553m
( )A= = 2π s) (675 Hz) =
2300 kg
( )14. 120 dB = 10 log I120
I0
→ I120 = 1012 I0 = 1012 1.0 ×10−12 W m2 = 1.0 W m2
( )20 I 20
dB = 10 log I0 → I20 = 102 I0 = 102 1.0 ×10−12 W m2 = 1.0 ×10−10 W m2
The pain level is 1010 times more intense than the whisper.
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507
Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual
15. β = 10log I = 10 log 2.0 ×10−6 W m2 = 63 dB
I0 1.0 ×10−12 W m2
16. From Figure 16-6, at 40 dB the low frequency threshold of hearing is about 70 − 80 Hz . There is
no intersection of the threshold of hearing with the 40 dB level on the high frequency side of the
chart, and so a 40 dB signal can be heard all the way up to the highest frequency that a human can
hear, 20, 000 Hz .
17. (a) From Figure 16-6, at 100 Hz, the threshold of hearing (the lowest detectable intensity by the
ear) is approximately 5 ×10−9 W m2 . The threshold of pain is about 5 W m2 . The ratio of
highest to lowest intensity is thus 5W m2 = 109 .
5 ×10−9 W m2
(b) At 5000 Hz, the threshold of hearing is about 10−13 W m2 , and the threshold of pain is about
10−1 W m2 . The ratio of highest to lowest intensity is 10−1 W m2 = 1012 .
10−13 W m2
Answers may vary due to estimation in the reading of the graph.
18. Compare the two power output ratings using the definition of decibels.
β = 10 log P150 = 10log 150 W = 1.8 dB
P100 100 W
This would barely be perceptible.
19. The intensity can be found from the decibel value.
( )β = 10log I
I0
→ I = 10β /10I0 = 1012 10−12 W m2 = 1.0 W m2
Consider a square perpendicular to the direction of travel of the sound wave. The intensity is the
energy transported by the wave across a unit area perpendicular to the direction of travel, per unit
time. So I = ΔE , where S is the area of the square. Since the energy is “moving” with the wave,
SΔt
the “speed” of the energy is v, the wave speed. In a time Δt , a volume equal to ΔV = SvΔt would
contain all of the energy that had been transported across the area S. Combine these relationships to
find the energy in the volume.
( )I m2 (0.010 m)3
= ΔE → ΔE = ISΔt = I ΔV = 1.0 W = 2.9 ×10−9 J
S Δt v 343m s
20. From Example 12-4, we see that a sound level decrease of 3 dB corresponds to a halving of intensity.
Thus the sound level for one firecracker will be 95 dB − 3 dB = 92 dB .
21. From Example 16-4, we see that a sound level decrease of 3 dB corresponds to a halving of intensity.
Thus, if two engines are shut down, the intensity will be cut in half, and the sound level will be 127
dB. Then, if one more engine is shut down, the intensity will be cut in half again, and the sound
level will drop by 3 more dB, to a final value of 124 dB .
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508
Chapter 16 Sound
( ) ( )22. 62 dB = 10 log ISignal INoise tape → = 106.2 = 1.6 ×106
ISignal I Noise
tape
( ) ( )98dB = 10 log ISignal INoise tape → = 109.8 = 6.3 ×109
I Signal I Noise
tape
23. (a) According to Table 16-2, the intensity in normal conversation, when about 50 cm from the
speaker, is about 3×10−6 W m2 . The intensity is the power output per unit area, and so the
power output can be found. The area is that of a sphere.
P → P = IA = I 4π r2 = 3×10−6 W m2 4π (0.50 m)2 = 9.425×10−6 W ≈ 9.4 ×10−6 W
( ) ( )I=A
(b) 75 W ⎛ 1 person W ⎞ = 7.96 ×106 ≈ 8.0 ×106 people
⎝⎜ 9.425 ×10−6 ⎠⎟
24. (a) The energy absorbed per second is the power of the wave, which is the intensity times the area.
( )50 I
dB = 10 log I0 → I = 105 I0 = 105 1.0×10−12 W m2 = 1.0×10−7 W m2
( ) ( )P = IA = 1.0×10−7 W m2 5.0×10−5 m2 = 5.0×10−12 W
(b) 1 J ⎛ 5.0 1s J ⎞ ⎛ 1 yr s ⎞ = 6.3×103 yr
⎝⎜ ×10−12 ⎟⎠ ⎜⎝ 3.16 ×107 ⎠⎟
25. The intensity of the sound is defined to be the power per unit area. We assume that the sound
spreads out spherically from the loudspeaker.
(a) I 250 = 250 W = 1.624 W m2 β250 = 10log I 250 = 10 log 1.624 W m2 = 122 dB
I0 1.0 ×10−12 W m2
4π (3.5m)2
I 45 = 45 W = 0.2923W m2 β45 = 10log I 45 = 10 log 0.2923W m2 = 115dB
I0 1.0 ×10−12 W m2
4π (3.5m)2
(b) According to the textbook, for a sound to be perceived as twice as loud as another means that
the intensities need to differ by a factor of 10. That is not the case here – they differ only by a
factor of 1.624 ≈ 6. The expensive amp will not sound twice as loud as the cheaper one.
0.2598
26. (a) Find the intensity from the 130 dB value, and then find the power output corresponding to that
intensity at that distance from the speaker.
( )β I 2.8 m
= 130 dB = 10log I0 → I2.8m = 1013 I0 = 1013 1.0 ×10−12 W m2 = 10 W m2
( )P = IA = 4π r2I = 4π (2.2 m)2 10 W m2 = 608 W ≈ 610 W
(b) Find the intensity from the 85 dB value, and then from the power output, find the distance
corresponding to that intensity.
( )β I
= 85 dB = 10log I0 → I = 108.5 I0 = 108.5 1.0 ×10−12 W m2 = 3.16 ×10−4 W m2
P 608 W
4π I = 4π 3.16 ×10−4 W m2
( )P = 4π r2I
→ r= = 390 m
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27. The first person is a distance of r1 = 100 m from the explosion, while the second person is a distance
r2 = 5 (100 m) from the explosion. The intensity detected away from the explosion is inversely
proportional to the square of the distance from the explosion.
I1 = r22 = ⎡ 5 (100 m) ⎤2 =5 ; β = 10 log I1 = 10log 5 = 7.0 dB
I2 r12 ⎢ ⎥ I2
⎣ 100 m ⎦
28. (a) The intensity is proportional to the square of the amplitude, so if the amplitude is 2.5 times
greater, the intensity will increase by a factor of 6.25 ≈ 6.3 .
(b) β = 10log I I0 = 10log 6.25 = 8dB
29. (a) The pressure amplitude is seen in Eq. 16-5 to be proportional to the displacement amplitude and
to the frequency. Thus the higher frequency wave has the larger pressure amplitude, by a factor
of 2.6.
(b) The intensity is proportional to the square of the frequency. Thus the ratio of the intensities is
the square of the frequency ratio.
I 2.6 f = (2.6 f )2 = 6.76 ≈ 6.8
If
f2
30. The intensity is given by Eq. 15-7, I = 2π 2vρ f 2 A2, using the density of air and the speed of sound
in air.
( ) ( )I = 2ρvπ 2 f 2 A2 = 2 1.29 kg m3 (343m s)π 2 (380 Hz)2 1.3 ×10−4 m 2 = 21.31W m2
β = 10 log I = 10 log 21.31W m 2 2 = 133dB ≈ 130 dB
I0 1.0 ×10−12 W
m
Note that this is above the threshold of pain.
31. (a) We find the intensity of the sound from the decibel value, and then calculate the displacement
amplitude from Eq. 15-7.
( )β I
= 10 log I0 → I = 10β /10I0 = 1012 10−12 W m2 = 1.0 W m2
I = 2π 2vρ f 2 A2 →
1 I1 1.0 W m2 3.2 ×10−5 m
πf 1.29 kg m3
2ρv = π (330 Hz) ( 343 m
( )A= 2 s) =
(b) The pressure amplitude can be found from Eq. 16-7.
I = (ΔPM )2 →
2vρ
( ) ( )ΔPM = 2vρ I = 2 (343m s) 1.29 kg m3 1.0 W m2 = 30 Pa (2 sig. fig.)
32. (a) We assume that there has been no appreciable absorption in this 25 meter distance. The
intensity is the power divide by the area of a sphere of radius 25 meters. We express the sound
level in dB.
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510
Chapter 16 Sound
( ( ) )I
= P ; β = 10 log I = 10 log 4π P I0 = 10 log 4π ( 25 5.0 ×105 W m2 = 138dB
4π r2 I0 r2
)m 2 10−12 W
(b) We find the intensity level at the new distance, and subtract due to absorption.
( ( ) )β
= 10 log P I0 = 10 log 4π 5.0 ×105 W W m2 = 106 dB
4π r2
(1000 m)2 10−12
βwith = 106 dB − (1.00 km) (7.0 dB km) = 99 dB
absorption
(c) We find the intensity level at the new distance, and subtract due to absorption.
( ( ) )β
= 10 log 4π P I0 = 10 log 4π 5.0 ×105 W W m2 = 88.5dB
r2
(7500 m)2 10−12
βwith = 88.5dB − (7.50 km) (7.0 dB km) = 36 dB
absorption
33. For a closed tube, Figure 16-12 indicates that f1 = v . We assume the bass clarinet is at room
4l
temperature.
f1 = v → l = v = 343m s = 1.24 m
4l 4 f1
4 (69.3Hz)
34. For a vibrating string, the frequency of the fundamental mode is given by f v1 FT .
= 2L = 2L mL
( )f 1 FT
= 2L mL → FT =4Lf 2m = 4(0.32 m) (440 Hz)2 3.5×10−4 kg = 87 N
35. (a) If the pipe is closed at one end, only the odd harmonic frequencies are present, and are given by
fn = nv = nf1, n = 1, 3, 5 .
4L
f1 = v = 343m s = 69.2 Hz
4L
4 (1.24 m)
f3 = 3 f1 = 207 Hz f5 = 5 f1 = 346 Hz f7 = 7 f1 = 484 Hz
(b) If the pipe is open at both ends, all the harmonic frequencies are present, and are given by
fn = nv = nf1 .
2l
f1 = v = 343 m s = 138.3 Hz ≈ 138 Hz
2l
2 (1.24 m)
f2 = 2 f1 = v = 277 Hz f3 = 3 f1 = 3v = 415 Hz f4 = 4 f1 = 2v = 553 Hz
l 2l l
36. (a) The length of the tube is one-fourth of a wavelength for this (one end closed) tube, and so the
wavelength is four times the length of the tube.
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f = v = 343m s = 410 Hz
λ
4 (0.21m)
(b) If the bottle is one-third full, then the effective length of the air column is reduced to 14 cm.
f = v = 343m s = 610 Hz
λ
4 (0.14 m)
37. For a pipe open at both ends, the fundamental frequency is given by f1 = v , and so the length for a
2l
given fundamental frequency is l = v.
2 f1
l 20 Hz = 343m s = 8.6 m l 20 kHz = 343m s = 8.6 ×10−3 m
2 (20 Hz) 2 (20, 000 Hz)
38. We approximate the shell as a closed tube of length 20 cm, and calculate the fundamental frequency.
f v = 343m s = 429 Hz ≈ 430 Hz
= 4l
4 (0.20 m)
39. (a) We assume that the speed of waves on the guitar string does not change when the string is
fretted. The fundamental frequency is given by f v , and so the frequency is inversely
= 2l
proportional to the length.
f ∝ 1 → f l = constant
l
fEl E = fAl A → lA = lE fE = (0.73 m) ⎛ 330 Hz ⎞ = 0.5475 m
fA ⎜⎝ 440 Hz ⎠⎟
The string should be fretted a distance 0.73 m − 0.5475 m = 0.1825 m ≈ 0.18 m from the nut
of the guitar.
(b) The string is fixed at both ends and is vibrating in its fundamental mode. Thus the wavelength
is twice the length of the string (see Fig. 16-7).
λ = 2l = 2 (0.5475 m) = 1.095 m ≈ 1.1 m
(c) The frequency of the sound will be the same as that of the string, 440 Hz . The wavelength is
given by the following.
λ = v = 343 m s = 0.78 m
f 440 Hz
40. (a) At T = 15o C , the speed of sound is given by v = (331 + 0.60(15)) m s = 340 m s (with 3
significant figures). For an open pipe, the fundamental frequency is given by f v .
= 2l
f = v → l = v = 340 m s = 0.649 m
2l 2f
2 (262 Hz)
(b) The frequency of the standing wave in the tube is 262 Hz . The wavelength is twice the
length of the pipe, 1.30 m .
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512
Chapter 16 Sound
(c) The wavelength and frequency are the same in the air, because it is air that is resonating in the
organ pipe. The frequency is 262 Hz and the wavelength is 1.30 m .
41. The speed of sound will change as the temperature changes, and that will change the frequency of
the organ. Assume that the length of the pipe (and thus the resonant wavelength) does not change.
f22 = v22 f5.0 = v5.0 Δf = f5.0 − f22 = v5.0 − v22
λ λ λ
Δf v5.0 − v22 v5.0 331 + 0.60(5.0)
f v22 331 + 0.60 (22)
= λ = −1= − 1 = −2.96 × 10−2 = −3.0%
v22
λ
42. A flute is a tube that is open at both ends, and so the fundamental frequency is given by f = v ,
2l
where l is the distance from the mouthpiece (antinode) to the first open side hole in the flute tube
(antinode).
f = v → l v = 343m s = 0.491 m
2l = 2f
2 (349 Hz)
43. For a tube open at both ends, all harmonics are allowed, with fn = nf1 . Thus consecutive harmonics
differ by the fundamental frequency. The four consecutive harmonics give the following values for
the fundamental frequency.
f1 = 523 Hz − 392 Hz = 131Hz, 659 Hz − 523 Hz = 136 Hz, 784 Hz − 659 Hz = 125 Hz
The average of these is f1 = 1 (131 Hz + 136 Hz + 125 Hz) ≈ 131 Hz. We use that for the fundamental
3
frequency.
(a) f1 = v → l = v = 343m s = 1.31 m
2l 2 f1
2 (131Hz)
Note that the bugle is coiled like a trumpet so that the full length fits in a smaller distance.
(b) fn = nf1 → nG4 = fG4 = 392 Hz = 2.99 ; nC5 = fC5 = 523 Hz = 3.99 ;
f1 131 Hz f1 131 Hz
nE5 = f E5 = 659 Hz = 5.03 ; nG5 = fG5 = 784 Hz = 5.98
f1 131 Hz f1 131 Hz
The harmonics are 3, 4, 5, and 6 .
44. (a) The difference between successive overtones for this pipe is 176 Hz. The difference between
successive overtones for an open pipe is the fundamental frequency, and each overtone is an
integer multiple of the fundamental. Since 264 Hz is not a multiple of 176 Hz, 176 Hz cannot
be the fundamental, and so the pipe cannot be open. Thus it must be a closed pipe.
(b) For a closed pipe, the successive overtones differ by twice the fundamental frequency. Thus
176 Hz must be twice the fundamental, so the fundamental is 88 Hz . This is verified since
264 Hz is 3 times the fundamental, 440 Hz is 5 times the fundamental, and 616 Hz is 7 times the
fundamental.
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Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual
45. The tension and mass density of the string do not change, so the wave speed is constant. The
frequency ratio for two adjacent notes is to be 21/12. The frequency is given by f = v .
2l
v
f = v f1st 2l 1st = 21/12 → l 1st = l unfingered = 65.0 cm = 61.35cm
2l 21 / 12 21 / 12
fret fret fret
→ f =unfingered v
2l unfingered
l 1st unfingered l unfingered − n /12
lfret 2 /12
l = =2nd 1/12 → l = ; x = l − l = l (1 − 2 )nth n /12nth
2 2fret 2fret
unfingered nth unfingered
fret fret
( ) ( )x1 = (65.0 cm) 1 − 2−1/12 = 3.6 cm ; x2 = (65.0 cm) 1 − 2−2/12 = 7.1cm
( ) ( )x3 = (65.0 cm) 1 − 2−3/12 = 10.3cm ; x4 = (65.0 cm) 1 − 2−4/12 = 13.4 cm
( ) ( )x5 = (65.0 cm) 1 − 2−5/12 = 16.3cm ; x6 = (65.0 cm) 1 − 2−6/12 = 19.0 cm
46. (a) The difference between successive overtones for an open pipe is the fundamental frequency.
f1 = 330 Hz − 275 Hz = 55 Hz
(b) The fundamental frequency is given by f1 = v . Solve this for the speed of sound.
2l
v = 2l f1 = 2 (1.80 m) (55 Hz) = 198 m s ≈ 2.0 ×102 m s
47. The difference in frequency for two successive harmonics is 40 Hz. For an open pipe, two
successive harmonics differ by the fundamental, so the fundamental could be 40 Hz, with 240 Hz
being the 6th harmonic and 280 Hz being the 7th harmonic. For a closed pipe, two successive
harmonics differ by twice the fundamental, so the fundamental could be 20 Hz. But the overtones of
a closed pipe are odd multiples of the fundamental, and both overtones are even multiples of 30 Hz.
So the pipe must be an open pipe .
f = v → l v = [331 + 0.60(23.0)]m s = 4.3 m
2l = 2f
2 (40 Hz)
48. (a) The harmonics for the open pipe are fn = nv . To be audible, they must be below 20 kHz.
2l
( )nv
2l
< 2 ×104 Hz → n < 2 (2.48 m) 2 ×104 Hz = 289.2
343 m s
Since there are 289 harmonics, there are 288 overtones .
(b) The harmonics for the closed pipe are fn = nv , n odd. Again, they must be below 20 kHz.
4l
( )nv
4l
< 2 ×104 Hz → n < 4 (2.48 m) 2 ×104 Hz = 578.4
343 m s
The values of n must be odd, so n = 1, 3, 5, …, 577. There are 289 harmonics, and so there are
288 overtones .
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514
Chapter 16 Sound
49. A tube closed at both ends will have standing waves with displacement nodes at each end, and so has
the same harmonic structure as a string that is fastened at both ends. Thus the wavelength of the
fundamental frequency is twice the length of the hallway, λ1 = 2l = 16.0 m.
f1 = v = 343 m s = 21.4 Hz ; f2 = 2 f1 = 42.8 Hz
λ1
16.0 m
50. To operate with the first harmonic, we see from the figure that the thickness must be half of a
wavelength, so the wavelength is twice the thickness. The speed of sound in the quartz is given by
v = G ρ , analogous to Eqs. 15-3 and 15-4.
( ) ( )t
= 1 λ = 1 v = 1 G ρ = 1 2.95 ×1010 N m2 2650 kg m2 = 1.39 ×10−4 m
2 2 f 2 f 2
12.0 ×106 Hz
51. The ear canal can be modeled as a closed pipe of length 2.5 cm. The resonant frequencies are given
by fn = nv , n odd . The first several frequencies are calculated here.
4l
nv n (343m s) = n (3430 Hz) , n odd
4l
4 2.5 ×10−2 m
( )fn= =
f1 = 3430 Hz f3 = 10, 300 Hz f5 = 17, 200 Hz
In the graph, the most sensitive frequency is between 3000 and 4000 Hz. This corresponds to the
fundamental resonant frequency of the ear canal. The sensitivity decrease above 4000 Hz, but is
seen to “flatten out” around 10,000 Hz again, indicating higher sensitivity near 10,000 Hz than at
surrounding frequencies. This 10,000 Hz relatively sensitive region corresponds to the first overtone
resonant frequency of the ear canal.
52. From Eq. 15-7, the intensity is proportional to the square of the amplitude and the square of the
frequency. From Fig. 16-14, the relative amplitudes are A2 ≈ 0.4 and A3 ≈ 0.15.
A1 A1
I2 2π 2vρ f 2 A22 f 2 A22 ⎛ f2 ⎞2 ⎛ A2 ⎞2 0.4 2 = 0.64
( )I = 2π 2vρ f 2 A2 I1 2 2 = ⎝⎜ f1 ⎟⎠ ⎝⎜ A1 ⎟⎠
→ = = = 22
2π 2vρ f12 A12 f12 A12
I3 ⎛ f3 ⎞2 ⎛ A3 ⎞2 = 32 (0.15)2 = 0.20
I1 =⎜ f1 ⎟ ⎜ A1 ⎟
⎠ ⎝ ⎠
⎝
β 2 −1 = 10 log I2 = 10 log 0.64 = −2 dB ; β 3−1 = 10 log I3 = 10 log 0.24 = −7 dB
I1 I1
53. The beat period is 2.0 seconds, so the beat frequency is the reciprocal of that, 0.50 Hz. Thus the
other string is off in frequency by ±0.50 Hz . The beating does not tell the tuner whether the
second string is too high or too low.
54. The beat frequency is the difference in the two frequencies, or 277 Hz − 262 Hz = 15 Hz . If the
frequencies are both reduced by a factor of 4, then the difference between the two frequencies will
also be reduced by a factor of 4, and so the beat frequency will be 1 (15 Hz ) = 3.75 Hz ≈ 3.8 Hz .
4
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Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual
55. Since there are 4 beats/s when sounded with the 350 Hz tuning fork, the guitar string must have a
frequency of either 346 Hz or 354 Hz. Since there are 9 beats/s when sounded with the 355 Hz
tuning fork, the guitar string must have a frequency of either 346 Hz or 364 Hz. The common value
is 346 Hz .
56. (a) Since the sounds are initially 180° out of phase, x d−x B
another 180° of phase must be added by a path A
length difference. Thus the difference of the distances from the speakers to the point of
constructive interference must be half of a wavelength. See the diagram.
(d − x) − x = 1 λ → d = 2x + 1 λ → d min = 1 λ = v 343 m s 0.583 m
2 2 2 2f
= 2 (294 Hz) =
This minimum distance occurs when the observer is right at one of the speakers. If the speakers
are separated by more than 0.583 m, the location of constructive interference will be moved
away from the speakers, along the line between the speakers.
(b) Since the sounds are already 180° out of phase, as long as the listener is equidistant from the
speakers, there will be completely destructive interference. So even if the speakers have a tiny
separation, the point midway between them will be a point of completely destructive
interference. The minimum separation between the speakers is 0.
57. Beats will be heard because the difference in the speed of sound for the two flutes will result in two
different frequencies.
f1 = v1 = [331 + 0.60(28)]m s = 263.4 Hz
2l
2 (0.66 m)
f2 = v2 = [331 + 0.60(5.0)]m s = 253.0 Hz Δf = 263.4 Hz − 253.0 Hz = 10 beats sec
2l
2 (0.66 m)
58. (a) The microphone must be moved to the right until the difference D
in distances from the two sources is half a wavelength. See the
diagram. We square the expression, collect terms, isolate the
remaining square root, and square again.
S2 − S1 = 1 λ → S2 l S1
2 x
(1 D + x )2 +l 2 − ( )1D − x 2 +l2 = 1 λ →
2 2 2
( )1 D + x 2 +l 2 = 1 λ + ( 1 D − x )2 + l 2 →
2 2 2
( 1 D + x )2 + l 2 = 1 λ2 + 2( 1 λ) (1 D − x )2 +l 2 + (1 D − x )2 + l 2 →
2 4 2 2 2
2Dx − 1 λ2 = λ ( 1 D − x )2 + l 2 → 4D2x2 − 2 (2Dx) 1 λ2 + 1 λ4 = λ2 ⎣⎡( 1 D − x )2 + l 2 ⎦⎤
4 2 4 16 2
( ( ) )4D2x2 1 D2 + l 2 − 1 λ2
4 16
− Dxλ 2 + 1 λ4 = 1 D2λ 2 − Dxλ 2 + x2λ 2 + λ 2l 2 → x=λ
16 4
4D2 − λ2
The values are D = 3.00 m, l = 3.20 m, and λ = v f = (343m s) (494 Hz) = 0.694 m.
x = (0.694 m) 1 (3.00 m)2 + (3.20 m)2 − 1 (0.694 m)2 = 0.411 m
4 16
4 (3.00 m)2 − (0.694 m)2
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516
Chapter 16 Sound
(b) When the speakers are exactly out of phase, the maxima and minima will be interchanged. The
intensity maxima are 0.411 m to the left or right of the midpoint, and the intensity minimum is
at the midpoint.
59. The beat frequency is 3 beats per 2 seconds, or 1.5 Hz. We assume the strings are the same length
and the same mass density.
(a) The other string must be either 220.0 Hz − 1.5 Hz = 218.5 Hz or 220.0 Hz + 1.5 Hz
= 221.5 Hz .
(b) Since f = v 1 FT , we have f ∝ FT → f f′ → FT′ = FT ⎛ f ′ ⎞2 .
2l = 2l μ = FT′ ⎝⎜ f ⎠⎟
FT
To change 218.5 Hz to 220.0 Hz: F′ = FT ⎛⎜⎝ 220.0 2 = 1.014 , 1.4% increase .
218.5
⎞
⎠⎟
To change 221.5 Hz to 220.0 Hz: F′ = FT ⎛⎜⎝ 220.0 ⎞2 = 0.9865, 1.3% decrease .
221.5 ⎟⎠
60. (a) To find the beat frequency, calculate the frequency of each sound, and then subtract the two
frequencies.
fbeat = f1 − f2 = vv = ( 343 m s) 1 − 1 = 3.821Hz ≈ 4 Hz
− 2.64 m 2.72 m
λ1 λ2
(b) The speed of sound is 343 m/s, and the beat frequency is 3.821 Hz. The regions of maximum
intensity are one “beat wavelength” apart.
λ = v = 343m s = 89.79 m ≈ 90 m (2 sig. fig.)
f 3.821 Hz
61. (a) Observer moving towards stationary source.
f ′ = ⎛⎜1 + vobs ⎞ f = ⎛⎝⎜1 + 30.0 m s ⎞ (1350 Hz) = 1470 Hz
⎝ vsnd ⎟ 343 m s ⎠⎟
⎠
(b) Observer moving away from stationary source.
f ′ = ⎝⎛⎜1 − vobs ⎞ f = ⎝⎜⎛1 − 30.0 m s ⎞ (1350 Hz ) = 1230 Hz
vsnd ⎟ 343 m s ⎟⎠
⎠
62. The moving object can be treated as a moving “observer” for calculating the frequency it receives
and reflects. The bat (the source) is stationary.
f ′object = fbat ⎛⎜1 − vobject ⎞
⎝ vsnd ⎟
⎠
Then the object can be treated as a moving source emitting the frequency f ′object , and the bat as a
stationary observer.
⎛⎝⎜1 − vobject ⎞
vsnd ⎠⎟
(( ))fb′a′t
= f ′object = f bat = f bat v − vsnd object
v + vsnd object
⎛⎜⎝1 + vobject ⎞ ⎝⎜⎛1 + vobject ⎞
vsnd ⎟⎠ vsnd ⎟⎠
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Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual
( )= 343 m s − 30.0 m s
5.00 ×104 Hz 343 m s + 30.0 m s = 4.20 ×104 Hz
63. (a) For the 18 m/s relative velocity:
f ′source = f 1 = ( 2300 Hz ) 1 = 2427 Hz ≈ 2430 Hz
18 m
moving ⎜⎛⎝1 − vsrc ⎞ ⎛⎜⎝1 − 343 m s ⎞
vsnd ⎟⎠ s ⎟⎠
f ′observer = f ⎛⎜1 + vsrc ⎞ = ( 2300 Hz ) ⎛ 1 + 18 m s ⎞ = 2421 Hz ≈ 2420 Hz
⎝ vsnd ⎟ ⎜⎝ 343 m s ⎠⎟
moving ⎠
The frequency shifts are slightly different, with f ′source > f ′ .observer The two frequencies are
moving moving
close, but they are not identical. As a means of comparison, calculate the spread in frequencies
divided by the original frequency.
f ′ − f ′source observer 2427 Hz − 2421Hz
2300 Hz
moving moving = = 0.0026 = 0.26%
fsource
(b) For the 160 m/s relative velocity:
f ′source = f 1 = (2300 Hz ) 1 = 4311Hz ≈ 4310 Hz
160
moving ⎜⎛⎝1 − vsrc ⎞ ⎜⎛⎝1 − 343 m s⎞
vsnd ⎟⎠ m s ⎟⎠
f ′observer = f ⎜⎛⎝1 + vsrc ⎞ = (2300 Hz ) ⎛⎜⎝1 + 160 m s ⎞ = 3372 Hz ≈ 3370 Hz
vsnd ⎟ 343 m s ⎟⎠
moving ⎠
The difference in the frequency shifts is much larger this time, still with f ′source > f ′ .observer
moving moving
f ′ − f ′source observer 4311Hz − 3372 Hz
2300 Hz
moving moving = = 0.4083 = 41%
fsource
(c) For the 320 m/s relative velocity:
f ′source = f 1 = ( 2300 Hz ) 1 = 34, 300 Hz
320
moving ⎝⎛⎜1 − vsrc ⎞ ⎛⎜⎝1 − 343 m s⎞
vsnd ⎠⎟ m s ⎟⎠
f ′observer = f ⎛⎜1 + vsrc ⎞ = ( 2300 Hz ) ⎝⎜⎛1 + 320 m s ⎞ = 4446 Hz ≈ 4450 Hz
⎝ vsnd ⎠⎟ 343 m s ⎠⎟
moving
The difference in the frequency shifts is quite large, still with f ′source > f ′ .observer
moving moving
f ′ − f ′source observer 34, 300 Hz − 4446 Hz
2300 Hz
moving moving = = 12.98 = 1300%
fsource
(d) The Doppler formulas are asymmetric, with a larger shift for the moving source than for the
moving observer, when the two are getting closer to each other. In the following derivation,
assume vsrc vsnd , and use the binomial expansion.
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Chapter 16 Sound
f ′source = f 1 = f ⎛ 1 − vsrc ⎞−1 ≈ f ⎝⎜⎛1 + vsrc ⎞ = f ′observer
⎝⎜ vsnd ⎠⎟ vsnd ⎠⎟
moving ⎛⎝⎜1 − vsrc ⎞ moving
vsnd ⎠⎟
64. The frequency received by the stationary car is higher than the frequency emitted by the stationary
car, by Δf = 4.5 Hz .
f obs = fsource + Δf = fsource →
⎛⎜⎝1 − vsource ⎞
vsnd ⎠⎟
fsource = Δf ⎛ vsnd − 1⎟⎞ = ( 4.5 Hz ) ⎛ 343 m s − 1⎟⎞⎠ = 98 Hz
⎜⎝ vsource ⎠ ⎝⎜ 15 m s
65. (a) The observer is stationary, and the source is moving. First the source is approaching, then the
source is receding.
120.0 km h ⎛ 1m s h ⎞ = 33.33 m s
⎜⎝ 3.6 km ⎟⎠
1 1
( )f ′source f
moving 33.33m s ⎞
towards 343m s ⎟⎠
= ⎝⎜⎛1 vsrc ⎞ = 1280 Hz ⎜⎛⎝1 − = 1420 Hz
vsnd ⎟⎠
−
f ′source = f 1 = (1280 Hz ) 1 = 1170 Hz
33.33m s
moving ⎛⎝⎜1 + vsrc ⎞ ⎛ 1 + 343m s ⎞
away vsnd ⎟⎠ ⎜⎝ ⎟⎠
(b) Both the observer and the source are moving, and so use Eq. 16-11.
90.0 km h ⎛ 1m s h ⎞ = 25 m s
⎜⎝ 3.6 km ⎠⎟
f ′approaching = f ( )vsnd + vobs = (1280 Hz ) ( 343 m s + 25 m s ) = 1520 Hz
( )vsnd − vsrc (343m s − 33.33m s)
f ′receding = f ( )vsnd − vobs = (1280 Hz ) ( 343 m s − 25 m s) = 1080 Hz
( )vsnd + vsrc (343m s + 33.33m s)
(c) Both the observer and the source are moving, and so again use Eq. 16-11.
80.0 km h ⎛ 1m s h ⎞ = 22.22 m s
⎜⎝ 3.6 km ⎠⎟
f ′police = f ( )vsnd − vobs = (1280 Hz ) ( 343 m s − 22.22 m s) = 1330 Hz
( )vsnd − vsrc ( 343 m s − 33.33m s)
car
approaching
f ′police = f ( )vsnd + vobs = (1280 Hz ) ( 343 m s + 22.22 m s) = 1240 Hz
( )vsnd + vsrc ( 343 m s + 33.33m s)
car
receding
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Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual
66. The wall can be treated as a stationary “observer” for calculating the frequency it receives. The bat
is flying toward the wall.
f ′wall = fbat 1
⎜⎛1 − vbat ⎞
⎝ vsnd ⎟⎠
Then the wall can be treated as a stationary source emitting the frequency fw′all , and the bat as a
moving observer, flying toward the wall.
(( ))fb′a′t
= f ′wall ⎛⎜1 + vbat ⎞ = f bat 1 ⎞ ⎝⎛⎜1 + vbat ⎞ = f bat vsnd + vbat
⎝ vsnd ⎟ ⎠⎟ vsnd ⎟ vsnd − vbat
⎠ ⎛⎜⎝1 − vbat ⎠
vsnd
( )= 343 m s + 7.0 m s
3.00 ×104 Hz 343 m s − 7.0 m s = 3.13 ×104 Hz
67. We assume that the comparison is to be made from the frame of reference of the stationary tuba.
The stationary observers would observe a frequency from the moving tuba of
f obs = fsource = 75 Hz = 78 Hz fbeat = 78 Hz − 75 Hz = 3 Hz .
⎝⎜⎛1 − vsource ⎞ ⎛⎜⎝1 − 12.0 m s ⎞
vsnd ⎠⎟ 343 m s ⎟⎠
( )68. For the sound to be shifted up by one note, we must have f ′source = f 21/12 .
moving
( )f ′source= f 1 = f 21 / 12 →
moving
⎜⎛1 − vsrc ⎞
⎝ vsnd ⎠⎟
vsrc = ⎛⎝⎜1 − 1 ⎠⎟⎞ vsnd = ⎜⎛⎝1 − 1 ⎞⎠⎟ (343 m s) = 19.25 m s ⎛ 3.6 km h ⎞ = 69.3 km h
21 / 12 21 / 12 ⎜⎝ ms ⎟⎠
69. The ocean wave has λ = 44 m and v = 18 m s relative to the ocean floor. The frequency of the
ocean wave is then f v = 18 m s = 0.409 Hz.
= 44 m
λ
(a) For the boat traveling west, the boat will encounter a Doppler shifted frequency, for an observer
moving towards a stationary source. The speed v = 18 m s represents the speed of the waves in
the stationary medium, and so corresponds to the speed of sound in the Doppler formula. The
time between encountering waves is the period of the Doppler shifted frequency.
f ′observer = ⎛ 1 + vobs ⎞ f = ⎜⎛⎝1 + 15 m s ⎞ (0.409 Hz ) = 0.750 Hz →
⎜ vsnd ⎟ 18 m s ⎟⎠
moving ⎝ ⎠
T = 1 = 1 = 1.3s
f 0.750 Hz
(b) For the boat traveling east, the boat will encounter a Doppler shifted frequency, for an observer
moving away from a stationary source.
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520
Chapter 16 Sound
f ′observer = ⎛ 1 − vobs ⎞ f = ⎛⎝⎜1 − 15 m s ⎞ ( 0.409 Hz ) = 0.0682 Hz →
⎜ vsnd ⎠⎟ 18 m s ⎠⎟
moving ⎝
T = 1 = 1 = 15 s
f 0.0682 Hz
70. The Doppler effect occurs only when there is relative motion of the source and the observer along
the line connecting them. In the first four parts of this problem, the whistle and the observer are not
moving relative to each other and so there is no Doppler shift. The wind speed increases (or
decreases) the velocity of the waves in the direction of the wind, as if the speed of sound were
different, but the frequency of the waves doesn’t change. We do a detailed analysis of this claim in
part (a).
(a) The wind velocity is a movement of the medium, and so adds or subtracts from the speed of
sound in the medium. Because the wind is blowing away from the observer, the effective speed
of sound is vsnd − vwind. The wavelength of the waves traveling towards the observer is
( )λa = vsnd − vwind f0 , where f0 is the frequency emitted by the factory whistle. This
wavelength approaches the observer at a relative speed of vsnd − vwind. Thus the observer hears
the frequency calculated here.
fa = vsnd − vwind = vsnd − vwind = f0 = 720 Hz
λa vsnd − vwind
f0
(b) Because the wind is blowing towards the observer, the effective speed of sound is vsnd + vwind.
The same kind of analysis as applied in part (a) gives that fb = 720 Hz .
(c) Because the wind is blowing perpendicular to the line towards the observer, the effective speed
of sound along that line is vsnd. Since there is no relative motion of the whistle and observer,
there will be no change in frequency, and so fc = 720 Hz .
(d) This is just like part (c), and so fd = 720 Hz .
(e) Because the wind is blowing toward the cyclist, the effective speed of sound is vsnd + .vwind The
wavelength traveling toward the cyclist is λe = (vsnd + )vwind f0 . This wavelength approaches
the cyclist at a relative speed of vsnd + vwind + vcycle. The cyclist will hear the following
frequency.
( ) ( )( ) ( ( )) ( )fe =
vsnd + vwind + vcycle = vsnd + vwind + vcycle f0 = 343 + 15.0 + 12.0 ms 720 Hz
λe vsnd + vwind 343 + 15.0
= 744 Hz
(f) Since the wind is not changing the speed of the sound waves moving towards the cyclist, the
speed of sound is 343 m/s. The observer is moving towards a stationary source with a speed of
12.0 m/s.
f′= f ⎜⎛1 + vobs ⎞ = (720 Hz ) ⎛⎝⎜1 + 12.0 m s ⎞ = 745 Hz
⎝ vsns ⎟ 343 m s ⎟⎠
⎠
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521
Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual
71. The maximum Doppler shift occurs when the heart has its maximum velocity. Assume that the heart
is moving away from the original source of sound. The beats arise from the combining of the
original 2.25 MHz frequency with the reflected signal which has been Doppler shifted. There are
two Doppler shifts – one for the heart receiving the original signal (observer moving away from
stationary source) and one for the detector receiving the reflected signal (source moving away from
stationary observer).
(( ))f ′′detector ⎜⎛1 − vheart ⎞
f ′heart = ⎜⎝⎛foriginal 1− vheart ⎞ = f ′heart = foriginal ⎝ vsnd ⎟ = f original v − vsnd heart
vsnd ⎟ ⎛⎜1 + ⎠ v + vsnd heart
⎠ ⎜⎛1 + vheart ⎞ ⎝ vheart ⎞
⎝ vsnd ⎠⎟ vsnd ⎟⎠
(( )) ( )Δf = f − f ′′ = f − foriginal
detector original original v − vsnd blood = foriginal 2vblood →
v + vsnd blood v + vsnd blood
Δf 260 Hz
2 f −original 2.25 ×106 Hz − 260 Hz
( ) ( )vblood 8.9 ×10−2 m
= vsnd Δf = 1.54 ×103 m s 2 = s
If instead we had assumed that the heart was moving towards the original source of sound, we would
get vblood = vsnd Δf . Since the beat frequency is much smaller than the original frequency,
2 foriginal + Δf
the Δf term in the denominator does not significantly affect the answer.
72. (a) The angle of the shock wave front relative to the direction of motion is given by Eq. 16-12.
sin θ = vsnd = vsnd = 1 → θ = sin −1 1 = 30o (2 sig. fig.)
vobj 2.0vsnd 2.0 2.0
( )(b) The displacement of the plane vobjt from the time it vobjt
θ
passes overhead to the time the shock wave reaches the
observer is shown, along with the shock wave front. From the h
displacement and height of the plane, the time is found.
tan θ = h → t = vobj h
vobjt tan θ
= 6500 m = 18 s
(2.0) (310 m s) tan 30o
73. (a) The Mach number is the ratio of the object’s speed to the speed of sound.
( )M ⎛ 1m s ⎞
= vobs = 1.5 ×104 km hr ⎝⎜ 3.6 km hr ⎠⎟ = 92.59 ≈ 93
vsound 45 m s
(b) Use Eq. 16-125 to find the angle.
θ = sin −1 vsnd = sin −1 1 = sin −1 1 = 0.62°
vobj M 92.59
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522
Chapter 16 Sound
74. From Eq. 16-12, sin θ = vsnd .
vobj
(a) θ = sin−1 vsnd = sin−1 343m s = 2.2o
vobj 8800 m s
(b) θ = sin −1 vsnd = sin−1 1560 m s = 10o (2 sig. fig.)
vobj 8800 m s
75. Consider one particular wave as shown in the diagram, created at the vobjt
location of the black dot. After a time t has elapsed from the creation
θ
of that wave, the supersonic source has moved a distance vobjt , and the vsndt
wave front has moved a distance vsndt . The line from the position of
the source at time t is tangent to all of the wave fronts, showing the
location of the shock wave. A tangent to a circle at a point is perpendicular to the radius connecting
that point to the center, and so a right angle is formed. From the right triangle, the angle θ can be
defined.
sin θ = vsndt = vsnd
vobjt vobj
76. (a) The displacement of the plane from the time it passes 2.0 km θ
1.25 km
overhead to the time the shock wave reaches the listener
is shown, along with the shock wave front. From the
displacement and height of the plane, the angle of the
shock wave front relative to the direction of motion can
be found. Then use Eq. 16-12.
tan θ = 1.25 km → θ = tan −1 1.25 = 32o
2.0 km 2.0
(b) M = vobj = 1 = 1 = 1.9
vsnd sin θ sin 32o
77. Find the angle of the shock wave, and then find the distance the D θ
plane has traveled when the shock wave reaches the observer.
Use Eq. 16-12. 9500 m
θ = sin −1 vsnd = sin −1 vsnd = sin −1 1 = 27°
vobj 2.2vsnd 2.2
tan θ = 9500 m → D = 9500 m = 18616 m = 19 km
D tan 27°
78. The minimum time between pulses would be the time for a pulse to travel from the boat to the
maximum distance and back again. The total distance traveled by the pulse will be 150 m, at the
speed of sound in fresh water, 1440 m/s.
d = vt → t = d 150 m = 0.10 s
v = 1440 m s
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523
Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual
79. Assume that only the fundamental frequency is heard. The fundamental frequency of an open pipe is
given by f = v .
2L
(a) f3.0 = v = 343 m s = 57 Hz f2.5 = v = 343 m s = 69 Hz
2L 2L
2(3.0 m) 2(2.5 m)
f2.0 = v = 343 m s = 86 Hz f1.5 = v = 343 m s = 114.3 Hz ≈ 110 Hz
2L 2L
2(2.0 m) 2(1.5 m)
f1.0 = v = 343 m s = 171.5 Hz ≈ 170 Hz
2L
2(1.0 m)
(b) On a noisy day, there are a large number of component frequencies to the sounds that are being
made – more people walking, more people talking, etc. Thus it is more likely that the
frequencies listed above will be a component of the overall sound, and then the resonance will
be more prominent to the hearer. If the day is quiet, there might be very little sound at the
desired frequencies, and then the tubes will not have any standing waves in them to detect.
80. The single mosquito creates a sound intensity of I0 = 1×10−12 W m2 . Thus 100 mosquitoes will
create a sound intensity of 100 times that of a single mosquito.
I = 100I0 β = 10 log 100I 0 = 10 log100 = 20 dB .
I0
81. The two sound level values must be converted to intensities, then the intensities added, and then
converted back to sound level.
I82 : 82 dB = 10 log I 82 → I82 = 108.2 I0 = 1.585 × 108 I0
I0
I89 : 89 dB = 10 log I 87 → I89 = 108.9 I0 = 7.943 × 108 I0
I0
( )Itotal = I82 + I89 = 9.528 ×108 I0 →
β total = 10 log 9.528 ×108 I0 = 10 log 6.597 ×108 = 89.8dB ≈ 90 dB (2 sig. fig.)
I0
82. The power output is found from the intensity, which is the power radiated per unit area.
( )115 I
dB = 10 log I0 → I = 1011.5I0 = 1011.5 1.0 ×10−12 W m2 = 3.162 × 10−1 W m2
P P → P = 4π r2I = 4π (9.00 m)2 3.162 ×10−1 W m2 = 322 W
=( )I A = 4π r2
83. Relative to the 1000 Hz output, the 15 kHz output is –12 dB.
−12 dB = 10 log P15 kHz → − 1.2 = log P15 kHz → 10−1.2 = P15 kHz → P15 kHz = 11W
175 W 175 W 175 W
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524
Chapter 16 Sound
84. The 130 dB level is used to find the intensity, and the intensity is used to find the power. It is
assumed that the jet airplane engine radiates equally in all directions.
( )β I
= 130 dB = 10 log I0 → I = 1013 I0 = 1013 1.0 × 10−12 W m2 = 1.0 ×101 W m2
( ) ( )P = IA = Iπ r2 = 1.0 ×101 W m2 π 2.0 ×10−2 2 = 0.013W
85. The gain is given by β = 10 log Pout = 10 log 125 W W = 51dB .
Pin 1.0 ×10−3
86. It is desired that the sound from the speaker arrives at a listener 30 ms after the sound from the singer
arrives. The fact that the speakers are 3.0 m behind the singer adds in a delay of 3.0 m =
343m s
8.7 ×10−3s, or about 9 ms. Thus there must be 21 ms of delay added into the electronic circuitry.
87. The strings are both tuned to the same frequency, and both have the same length. The mass per unit
length is the density times the cross sectional area. The frequency is related to the tension by Eqs.
15-1 and 15-2.
f = v ;v= FT → f = 1 FT = 1 FT → FT = 4l 2ρ f 2π r2 →
2l μ 2l μ 2l ρπ r2
FT high 4l 2ρ f π r2 2 ⎛ rhigh 2 ⎛ d1 2 ⎛ 0.724 mm 2 1.07
FT low high ⎜ rlow
⎝ ⎞ 2 high ⎞ ⎞
= = ⎟ = ⎜ ⎟ = ⎝⎜ 0.699 mm ⎟⎠ =
4l 2ρ f π r2 2 ⎠ ⎝ d1 ⎠
low
2 low
88. The strings are both tuned to the same frequency, and both have the same length. The mass per unit
length is the density times the cross sectional area. The frequency is related to the tension by Eqs.
15-1 and 15-2.
f = v ;v= FT → f = 1 FT = 1 FT → FT = 4l 2ρ f 2π r2 →
2l μ 2l μ 2l ρπ r2
2 2 2 2 2
acoustic 2 acoustic acoustic acoustic acoustic
2
2 electric 2 electric
electric electric electric
F 4l ρ f π r ρ r ⎛ ρ ⎞ ⎛ d ⎞T acoustic acoustic
F = 4l ρ f π r = ρ r = ⎜⎝ ρ ⎟⎠ ⎝⎜ d ⎟⎠T electric electric
= ⎛ 7760 kg m3 ⎞ ⎛ 0.33 m ⎞2 = 1.7
⎜⎝ 7990 kg m3 ⎟⎠ ⎜⎝ 0.25 m ⎠⎟
89. (a) The wave speed on the string can be found from the length and the fundamental frequency.
f = v → v = 2l f = 2 (0.32 m) (440 Hz) = 281.6 m s ≈ 280 m s
2l
The tension is found from the wave speed and the mass per unit length.
( )v = FT
μ → FT = μ v2 = 7.21×10−4 kg m (281.6 m s)2 = 57 N
(b) The length of the pipe can be found from the fundamental frequency and the speed of sound.
f = v → l = v = 343 m s = 0.1949 m ≈ 0.19 m
4l 4f
4 (440 Hz)
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525
Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual
(c) The first overtone for the string is twice the fundamental. 880 Hz
The first overtone for the open pipe is 3 times the fundamental. 1320 Hz
90. The apparatus is a closed tube. The water level is the closed end, and so is a node of air
displacement. As the water level lowers, the distance from one resonance level to the next
corresponds to the distance between adjacent nodes, which is one-half wavelength.
Δl = 1 λ → λ = 2Δl = 2 (0.395 m − 0.125 m) = 0.540 m
2
f = v = 343m s = 635 Hz
λ 0.540 m
91. The fundamental frequency of a tube closed at one end is given by f1 = v . The change in air
4l
temperature will change the speed of sound, resulting in two different frequencies.
v30.0°C ⎝⎜⎛ ⎠⎟⎞→ f = f vv30.0°C
f 4l v30.0°C
f = v = v25.0°C 30.0°C 25.0°C 30.0°C
25.0°C 25.0°C
25.0°C
4l
Δf = f −30.0°C f 25.0° C = f ⎛⎝⎜ vv25.0°C 30.0°C − 1⎟⎠⎞ = (349 Hz ) ⎛ 331 + 0.60 ( 30.0) − 1⎟⎞ = 3 Hz
25.0°C ⎜ 331 + 0.60 ( 25.0) ⎠
⎝
92. Call the frequencies of four strings of the violin fA , fB , fC , fD with fA the lowest pitch. The mass
per unit length will be named μ . All strings are the same length and have the same tension. For a
string with both ends fixed, the fundamental frequency is given by f1 = v 1 FT .
2l = 2l μ
fB = 1.5 fA → 1 FT = 1.5 1 FT → μB = μA = 0.44μA
2l μB 2l μA
(1.5)2
fC = 1.5 fB = (1.5)2 fA → 1 FT = (1.5)2 1 FT → μC = μA = 0.20μA
2l μC 2l μA
(1.5)4
fD = 1.5 fC = (1.5)3 fA → 1 FT = (1.5)3 1 FT → μD = μA = 0.088μA
2l μD 2l μA
(1.5)6
93. The effective length of the tube is l eff. = l + 1 D = 0.60 m + 1 ( 0.030 m ) = 0.61 m.
3 3
Uncorrected frequencies: fn = (2n − 1) v ,n = 1, 2, 3… →
4l
f1−4 = (2n − 1) 343m s = 143 Hz, 429 Hz, 715 Hz, 1000 Hz
4 (0.60 m)
Corrected frequencies: fn = (2n − 1) v ,n = 1, 2, 3… →
4l eff
f1−4 = (2n − 1) 343m s = 141 Hz, 422 Hz, 703 Hz, 984 Hz
4 (0.61m)
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526
Chapter 16 Sound
94. Since the sound is loudest at points equidistant from the two sources, the two sources must be in
phase. The difference in distance from the two sources must be an odd number of half-wavelengths
for destructive interference.
0.28 m = λ 2 → λ = 0.56 m f = v λ = 343m s 0.56 m = 610 Hz
0.28 m = 3λ 2 → λ = 0.187 m f = v λ = 343 m s 0.187 m = 1840 Hz (out of range)
95. As the train approaches, the observed frequency is given by ff ′ =approach ⎜⎛1 − vtrain ⎞ . As the train
⎝ vsnd ⎟
⎠
recedes, the observed frequency is given by f ′recede = f ⎛⎜1 + vtrain ⎞ . Solve each expression for f,
equate them, and then solve for vtrain . ⎝ vsnd ⎟
⎠
⎝⎛⎜f ′approach 1− vtrain ⎞ = f ′recede ⎛⎜⎝1 + vtrain ⎞ →
vsnd ⎟ vsnd ⎟
⎠ ⎠
(( )) ( ) (( ))v = vtrain snd
f ′ − f ′approach recede = 343m s 552 Hz − 486 Hz = 22 m s
552 Hz + 486 Hz
f ′ + f ′approach recede
96. The Doppler shift is 3.5 Hz, and the emitted frequency from both trains is 516 Hz. Thus the
frequency received by the conductor on the stationary train is 519.5 Hz. Use this to find the moving
train’s speed.
vsnd vsource = ⎜⎛⎝1 − f ⎞ vsnd ⎝⎜⎛1 516 Hz ⎞ (343 m s) = s
v − vsnd source f ⎠⎟ 519.5 Hz ⎟⎠
( )f ′ = f → ′ = − 2.31m
97. (a) Since both speakers are moving towards the observer at the same speed, both frequencies have
the same Doppler shift, and the observer hears no beats .
(b) The observer will detect an increased frequency from the speaker moving towards him and a
decreased frequency from the speaker moving away. The difference in those two frequencies
will be the beat frequency that is heard.
f ′towards = f 1 f ′away = f 1
⎝⎛⎜1 − vtrain ⎞ ⎝⎜⎛1 + vtrain ⎞
vsnd ⎠⎟ vsnd ⎟⎠
f 1 f 1 f ⎡ vsnd vsnd ⎤
⎢ v − vsnd train v + vsnd train ⎥
( ) ( )f ′towards − ⎣ ⎦
fa′way = ⎜⎝⎛1 − vtrain ⎞ − ⎝⎛⎜1 + vtrain ⎞ = −
vsnd ⎟⎠ vsnd ⎟⎠
(348 Hz ) ⎡ 343m s − 343m s ⎤ = 20 Hz (2 sig. fig.)
⎢ m s − 10.0
⎣ (343 m s) (343m s + 10.0 m s) ⎥
⎦
(c) Since both speakers are moving away from the observer at the same speed, both frequencies
have the same Doppler shift, and the observer hears no beats .
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527
Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual
98. For each pipe, the fundamental frequency is given by f = v . Find the frequency of the shortest
2l
pipe.
f v = 343m s = 71.46 Hz
= 2l
2 (2.40 m)
The longer pipe has a lower frequency. Since the beat frequency is 8.0 Hz, the frequency of the
longer pipe must be 63.46 Hz. Use that frequency to find the length of the longer pipe.
f = v → l = v = 343m s = 2.70 m
2l 2f
2 (63.46 Hz)
99. Use Eq. 16-11, which applies when both source and observer are in motion. There will be two
Doppler shifts in this problem – first for the emitted sound with the bat as the source and the moth as
the observer, and then the reflected sound with the moth as the source and the bat as the observer.
(( ))f ′moth = fbat (( )) (( )) (( ))fb′a′t = f ′moth
vsnd + vmoth vsnd + vbat = fbat vsnd + vmoth vsnd + vbat
vsnd − vbat vsnd − vmoth vsnd − vbat vsnd − vmoth
= (51.35 kHz ) (343 + 5.0) (343 + 7.5) = 55.23 kHz
(343 − 7.5) (343 − 5.0)
100. The beats arise from the combining of the original 3.80 MHz frequency with the reflected signal
which has been Doppler shifted. There are two Doppler shifts – one for the blood cells receiving the
original frequency (observer moving away from stationary source) and one for the detector receiving
the reflected frequency (source moving away from stationary observer).
(( ))f ′′detector ⎛⎜1 − vblood ⎞
fb′lood = ⎝⎜⎛foriginal 1− vblood ⎞ = f ′blood = foriginal ⎝ vsnd ⎟ = foriginal v − vsnd blood
vsnd ⎟ ⎛⎜1 + vblood ⎞ ⎜⎛1 + ⎠ v + vsnd blood
⎠ ⎝ vsnd ⎟ ⎝ vblood ⎞
⎠ vsnd ⎟
⎠
(( )) ( )Δf = f − f ′′ = f − foriginal
detector original original v − vsnd blood = foriginal 2vblood
v + vsnd blood v + vsnd blood
2 (0.32 m s)
1.54 ×103 m s + 0.32 m s
( ) ( )= 3.80 ×106 Hz = 1600 Hz
101. It is 70.0 ms from the start of one chirp to the start of the next. Since the chirp itself is 3.0 ms long, it
is 67.0 ms from the end of a chirp to the start of the next. Thus the time for the pulse to travel to the
moth and back again is 67.0 ms. The distance to the moth is half the distance that the sound can
travel in 67.0 ms, since the sound must reach the moth and return during the 67.0 ms.
( )d
= vsndt = (343m )s 1 67.0 ×10−3 s = 11.5 m
2
102. (a) We assume that vsrc vsnd , and use the binomial expansion.
f ′source = f 1 = f ⎛⎜1 − vsrc ⎞−1 ≈ f ⎜⎛1 + vsrc ⎞ = f ′observer
⎝ vsnd ⎟ ⎝ vsnd ⎟
moving ⎛⎝⎜1 − vsrc ⎞ ⎠ ⎠ moving
vsnd ⎠⎟
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528
Chapter 16 Sound
(b) We calculate the percent error in general, and then substitute in the given relative velocity.
⎛ f ⎛⎜1 + vsrc ⎞ − f 1 ⎞
⎜ ⎝ vsnd ⎟ ⎟
⎜ ⎠ ⎛ vsrc ⎞ ⎟
⎜⎝ 1 − vsnd ⎟⎠ ⎟
⎜⎝⎛ approx. − exact ⎟⎠⎞100 ⎜ ⎟
% error = exact = 100 ⎜ f1 ⎟
⎟
⎜ ⎛⎜1 − vsrc ⎞ ⎟⎠
⎝ vsnd ⎟⎠
⎜
⎜⎝
= 100 ⎣⎢⎡⎝⎛⎜1 + vsrc ⎞ ⎝⎜⎛1 − vsrc ⎞ − 1⎤⎥ = −100 ⎛ vsrc ⎞2 = −100 ⎛ 18.0 m s ⎞2 = −0.28%
vsnd ⎠⎟ vsnd ⎠⎟ ⎦ ⎝⎜ vsnd ⎠⎟ ⎜⎝ 343 m s ⎠⎟
The negative sign indicates that the approximate value is less than the exact value.
103. The person will hear a frequency f ′towards = f ⎝⎛⎜1 + vwalk ⎞ from the speaker that they walk towards.
vsnd ⎟
⎠
The person will hear a frequency f ′away = f ⎛⎜1 − vwalk ⎞ from the speaker that they walk away from.
⎝ vsnd ⎟
⎠
The beat frequency is the difference in those two frequencies.
( )f ′towards − f ⎛⎜1 + vwalk ⎞ f ⎛⎜1 − vwalk ⎞ vwalk 1.4 m s
fa′way = ⎝ vsnd ⎟ − ⎝ vsnd ⎟ = 2 f vsnd =2 282 Hz 343m s = 2.3 Hz
⎠ ⎠
104. There will be two Doppler shifts in this problem – first for a stationary source with a moving
“observer” (the blood cells), and then for a moving source (the blood cells) and a stationary
“observer” (the receiver). Note that the velocity component of the blood parallel to the sound
transmission is vblood cos 45° = v .1 It is that component that causes the Doppler shift.
2 blood
⎛⎜f ′ = fblood 1− ⎞v1
original ⎝
2 blood
⎟⎠vsnd
(( ))f ′′detector ⎛⎜⎝1 − ⎞v1
= ⎜⎝⎛1 + fb′lood ⎞ = foriginal ⎝⎛⎜1 + = foriginal v − v1 →
v1 ⎠⎟ 2 blood snd 2 blood
2 blood ⎟⎠vsnd v + v1
⎞v1 snd 2 blood
vsnd
2 blood
⎟⎠vsnd
vblood = ( )f − f ′′original detector snd
( )2 f ′′ + f vdetectororiginal
Since the cells are moving away from the transmitter / receiver combination, the final frequency
received is less than the original frequency, by 780 Hz. Thus f ′′detector = foriginal − 780 Hz.
( )f − f ′′original ( )( )2
vblood = ( )2 f ′′ + f v =detectordetector snd 780 Hz vsnd
original 2foriginal − 780 Hz
2 (780 Hz) (1540 m s) = 0.17 m s
106 Hz −
( )= ⎣⎡2 5.0 × 780 Hz ⎤⎦
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
529
Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual
105. The apex angle is 15o, so the shock wave angle is 7.5o. The angle of the shock wave is also given by
sin θ = vwave v .object
sin θ = vwave vobject → v = vobject wave sinθ = 2.2 km h sin 7.5o = 17 km h
106. First, find the path difference in the original configuration. Then move the obstacle to the right by
Δd so that the path difference increases by 1 λ . Note that the path difference change must be on the
2
same order as the wavelength, and so Δd d , l since λ l , d.
( )ΔD initial = 2 d2 (+ 1 l )2 −l ; ( )ΔD final = 2 (d + Δd )2 + (1 l )2 − l
2 2
( ) ( )( ) ( )( ) ( ) ( )ΔD −final 2 −l 2 −l
ΔD initial = 1 λ = 2 d + Δd 2 + 1 l −2 d2 + 1 l →
2 2 2
2 (d + Δd )2 + ( 1 l )2 = 1 λ +2 ( )d 2 + 1 l 2
2 2 2
Square the last equation above.
4 ⎣⎡d 2 + 2d Δd + (Δd )2 + (1 l )2 ⎤⎦ = 1 λ2 + (2 1 λ)2 d2 (+ 1 l )2 + 4 ⎣⎡d 2 + ( 1 l )2 ⎦⎤
2 4 2 2 2
We delete terms that are second order in the small quantities Δd and λ.
8d Δd = 2λ ( )d 2 + 1 l 2 → Δd = λ ( )d 2 + 1 l 2
2 4d 2
107. (a) The “singing” rod is manifesting standing waves. By holding the rod at its midpoint, it has a
node at its midpoint, and antinodes at its ends. Thus the length of the rod is a half wavelength.
The speed of sound in aluminum is found in Table 16-1.
f = v = v = 5100 m s = 3400 Hz
λ 2L 1.50 m
(b) The wavelength of sound in the rod is twice the length of the rod, 1.50 m .
(c) The wavelength of the sound in air is determined by the frequency and the speed of sound in air.
λ = v = 343 m s = 0.10 m
f 3400 Hz
108. The displacement amplitude is related to the intensity by Eq. 15-7. The intensity can be calculated
from the decibel value. The medium is air.
( ) ( )β I
= 10 log I0 → I = 10β 10 I0 = 1010.5 10−12 W m2 = 0.0316 W m2
(a) I = 2π 2vρ f 2 A2 →
1 I1 0.0316 W m2 = 2.4 ×10−7 m
πf
2 (343m s) 1.29 kg m3
( ) ( )A= =
2vρ π 8.0 ×103 Hz
1 I1 0.0316 W m2 = 5.4 ×10−5 m
πf
2vρ = π (35 Hz) 2 (343m s) 1.29 kg m3
A( )(b)=
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
530
Chapter 16 Sound
109. (a) The spreadsheet used for this problem can be found on the Media Manager, with filename
“PSE4_ISM_CH16.XLS,” on tab “Problem 16.109a.”
D (x) 1.2
1.0
0.8 0.1 0.2 0.3 0.4 0.5
0.6
0.4 x (m)
0.2
0.0
0
(b) The spreadsheet used for this problem can be found on the Media Manager, with filename \
“PSE4_ISM_CH16.XLS,” on tab “Problem 16.109b.”
D (x) 1.2
1.0
0.8 0.1 0.2 0.3 0.4 0.5
0.6
0.4 x (m)
0.2
0.0
0
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
531
CHAPTER 17: Temperature, Thermal Expansion, and the Ideal Gas Law
Responses to Questions
1. 1 kg of aluminum will have more atoms. Aluminum has an atomic mass less than iron. Since each
Al atom is less massive than each Fe atom, there will be more Al atoms than Fe atoms in 1 kg.
2. Properties of materials that could be exploited in making a thermometer include:
a. thermal expansion, both linear and volume
b. the proportionality between temperature and pressure (for an ideal gas when volume is held
constant)
c. temperature dependence of resistivity
d. frequency of emitted radiation from a heated object (blackbody radiation curve).
(Note: resistivity and blackbody radiation are defined in Volume II.)
3. 1Cº is larger. Between the freezing and boiling points of water there are 100 Celsius degrees and
180 Fahrenheit degrees, so the Celsius degrees must be larger.
4. A and B have the same temperature; the temperature of C is different.
5. No. We can only infer that the temperature of C is different from that of A and B. We cannot infer
anything about the relationship of the temperatures of A and B.
6. The initial length should be l 0. However, since α is very small, the absolute value of Δl will be
about the same whether the initial or final length is used.
7. Aluminum. Al has a larger coefficient of linear expansion than Fe, and so will expand more than Fe
when heated and will be on the outside of the curve.
8. As the pipe changes temperature due to the presence or absence of steam it will expand and contract.
The bend allows the pipe to increase or decrease slightly in length without applying too much stress
to the fixed ends.
9. Lower. Mercury has a larger coefficient of volume expansion than lead. When the temperature rises,
mercury will expand more than lead. The density of mercury will decrease more than the density of
lead will decrease, and so the lead will need to displace more mercury in order to balance its weight.
10. The bimetallic strip is made of two types of metal joined together. The metal of the outside strip has
a higher coefficient of linear expansion than that of the inside strip, so it will expand and contract
more dramatically. If the temperature goes above the thermostat setting, the outer strip will expand
more than the inner, causing the spiral to wind more tightly and tilt the glass vessel so that the liquid
mercury flows away from the contact wires and the heater turns off. If the temperature goes below
the thermostat setting, the vessel tilts back as the outer strip contracts more than the inner and the
spiral opens, and the heater turns on. Moving the temperature setting lever changes the initial
position of the glass vessel. For instance, if the lever is set at 50, the vessel is tilted with the mercury
far from the contact wires. The outer strip has to shrink significantly to uncurl the spiral enough to
tilt the vessel back.
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532
Chapter 17 Temperature, Thermal Expansion, and the Ideal Gas Law
11. If water is added quickly to an overheated engine, it comes into contact with the very hot metal parts
of the engine. Some areas of the metal parts will cool off very rapidly; others will not. Some of the
water will quickly turn to steam and will expand rapidly. The net result can be a cracked engine
block or radiator, due to the thermal stress, and/or the emission of high temperature steam from the
radiator. Water should always be added slowly, with the engine running. The water will mix with the
hotter water already in the system, and will circulate through the engine, gradually cooling all parts
at about the same rate.
12. No. Whatever units are used for the initial length (l 0 ) will be the units of the change in length
(Δl ). The ratio Δl l 0 does not depend on the units used.
13. When the cold thermometer is placed in the hot water, the glass part of the thermometer will expand
first, as heat is transferred to it first. This will cause the mercury level in the thermometer to
decrease. As heat is transferred to the mercury inside the thermometer, the mercury will expand at a
rate greater than the glass, and the level of mercury in the thermometer will rise.
14. Since Pyrex glass has a smaller coefficient of linear expansion than ordinary glass, it will expand
less than ordinary glass when heated, making it less likely to crack from internal stresses. Pyrex
glass is therefore more suitable for applications involving heating and cooling. An ordinary glass
mug may expand to the point of cracking if boiling water is poured in it, whereas a Pyrex mug will
not.
15. Slow. On a hot day, the brass rod holding the pendulum will expand and lengthen, increasing the
length and therefore the period of the pendulum, causing the clock to run slow.
16. Soda is essentially water, and water (unlike most other substances) expands when it freezes. If the
can is full while the soda is liquid, then as the soda freezes and expands, it will push on the inside
surfaces of the can and the ends will bulge out.
17. The coefficient of volume expansion is much greater for alcohol than for mercury. A given
temperature change will therefore result in a greater change in volume for alcohol than for mercury.
This means that smaller temperature changes can be measured with an alcohol thermometer.
18. Decrease. As the temperature changes, both the aluminum sphere and the water will expand,
decreasing in density. The coefficient of volume expansion of water is greater than that of
aluminum, so the density of the water will decrease more than the density of the aluminum will
decrease. Even though the sphere displaces a greater volume of water at a higher temperature, the
weight of the water displaced (the buoyant force) will decrease because of the greater decrease in the
density of the water.
19. Helium. If we take the atomic mass of 6.7 ×10−27 kg and divide by the conversion factor from kg to u
(atomic mass units), which is 1.66 ×10−27 kg u , we get 4.03 u. This corresponds to the atomic mass
of helium.
20. Not really, as long as the pressure is very low. At low pressure, most gases will behave like an ideal
gas. Some practical considerations would be the volatility of the gas and its corrosive properties.
Light monatomic or diatomic gases are best.
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533
Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual
21. Fresh water is less dense than sea water, requiring the ship to displace more water for the same
buoyant force. (The buoyant force has to equal the weight of the ship for the ship to float.) The ship
sat lower in the fresh water than in sea water, and was therefore more likely to be swamped by
waves in a storm and sink.
Solutions to Problems
In solving these problems, the authors did not always follow the rules of significant figures rigidly. We
tended to take quoted temperatures as correct to the number of digits shown, especially where other
values might indicate that.
1. The number of atoms in a pure substance can be found by dividing the mass of the substance by the
mass of a single atom. Take the atomic masses of gold and silver from the periodic table.
2.15 ×10−2 kg
( )NAu
N Ag
= (196.96655u atom) 1.66 ×10−27 kg u = 107.8682 = 0.548 → NAu = 0.548NAg
196.96655
2.15 ×10−2 kg
( )(107.8682 u atom) 1.66 ×10−27 kg u
Because a gold atom is heavier than a silver atom, there are fewer gold atoms in the given mass.
2. The number of atoms is found by dividing the mass of the substance by the mass of a single atom.
Take the atomic mass of copper from the periodic table.
3.4 ×10−3 kg = 3.2 ×1022 atoms of Cu
atom) 1.66 ×10−27 kg
( )NCu =
(63.546 u u
3. (a) T (°C) = 5 [T ( °F ) − 32] = 5 [68 − 32] = 20°C
9 9
(b) T (°F) = 9 T ( °C) + 32 = 9 (1900) + 32 = 3452°F ≈ 3500°F
5 5
4. High: T (°C) = 5 [T ( °F) − 32] = 5 [136 − 32] = 57.8°C
Low: 9 9
T (°C) = 5 [T ( °F) − 32] = 5 [−129 − 32] = −89.4°C
9 9
5. T (°F) = 9 T (°C) + 32 = 9 (39.4°C) + 32 = 102.9°F
5 5
6. Assume that the temperature and the length are linearly related. The change in temperature per unit
length change is as follows.
ΔT = 100.0°C − 0.0°C = 9.970C° cm
Δl 21.85 cm − 11.82 cm
Then the temperature corresponding to length L is T (l ) = 0.0°C + (l − 11.82 cm) (9.970 C° cm) .
(a) T (18.70 cm) = 0.0°C + (18.70 cm − 11.82 cm) (9.970C° cm) = 68.6°C
(b) T (14.60 cm) = 0.0°C + (14.60 cm − 11.82 cm) (9.970C° cm) = 27.7°C
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534
Chapter 17 Temperature, Thermal Expansion, and the Ideal Gas Law
7. Take the 300 m height to be the height in January. Then the increase in the height of the tower is
given by Eq. 17-1a.
( )Δl = αl 0ΔT = 12 ×10−6 C° (300 m) (25°C − 2°C) = 0.08 m
8. When the concrete cools in the winter, it will contract, and there will be no danger of buckling. Thus
the low temperature in the winter is not a factor in the design of the highway. But when the concrete
warms in the summer, it will expand. A crack must be left between the slabs equal to the increase in
length of the concrete as it heats from 15oC to 50oC.
( )Δl = αl 0ΔT = 12 ×10−6 C° (12 m) (50°C − 15°C) = 5.0 ×10−3m
9. The increase in length of the table is given by Eq. 17-1a.
( )Δl = αl 0ΔT = 0.2 ×10−6 C° (1.6 m) (5.0C°) = 1.6 ×10−6 m
( )For steel, Δl = αl 0ΔT = 12 ×10−6 C° (1.6 m) (5.0C°) = 9.6 ×10−5 m .
The change for Super Invar is only 1 of the change for steel.
60
10. The increase in length of the rod is given by Eq. 17-1a.
Δl = αl 0ΔT → ΔT = Δl → Tf = Ti + Δl = 25°C+ 19 0.010 C° = 551.3°C ≈ 550°C
αl 0 αl 0 × 10−6
11. The density at 4oC is ρ = M = 1.00 ×103 kg . When the water is warmed, the mass will stay the
V 1.00 m3
same, but the volume will increase according to Eq. 17-2.
( ) ( )ΔV = βV0ΔT = 210 ×10−6 C° 1.00 m3 (94°C − 4°C) = 1.89 ×10−2 m3
The density at the higher temperature is ρ = M = 1.00 1.00 ×103 kg m3 = 981kg m3
V m3 +1.89 ×10−2
12. We assume that all of the expansion of the water is in the thickness of the mixed layer. We also
assume that the volume of the water can be modeled as a shell of constant radius, equal to the radius
of the earth, and so the volume of the shell is the surface area of the shell times its thickness.
V = 4π RE2d . Use Eq. 17-2.
V = 4π RE2d → ΔV = 4π RE2Δd ; ΔV = βV ΔT = β 4π RE2dΔT
4π RE2Δd = β 4π RE2dΔT →
( )Δd = β dΔT = 210 ×10−6 C° (50 m) (0.5°C) = 0.00525m ≈ 5mm
13. The rivet must be cooled so that its diameter becomes the same as the diameter of the hole.
Δl = αl 0ΔT → l − l 0 = α L0 (T − T0 )
= T0 l −l0 20°C + 1.870 cm − 1.872 cm −69°C
αl 0
12 ×10−6 C° (1.872 cm)
( )T + = =
The temperature of “dry ice” is about −80°C, so this process will be successful.
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535
Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual
14. Assume that each dimension of the plate changes according to Eq. 17-1a.
ΔA = A − A0 = (l + Δl ) (w + Δw) − l w = l w + l Δw + wΔl + Δl Δw − l w
= l Δw + wΔl + Δl Δw
Neglect the very small quantity Δl Δw .
ΔA = l Δw + wΔl = l (αwΔT ) + w(α l ΔT ) = 2αl wΔT
15. The change in volume of the aluminum is given by the volume expansion formula, Eq. 17-2.
( )ΔV = βV0ΔT = ⎛ ⎛⎝⎜ 8.75 cm ⎞3 ⎞ 3.9 cm3
75 ×10−6 C° ⎜ 4 π 2 ⎟⎠ ⎟ (180°C − 30°C) =
⎝ 3 ⎠
16. Since the coefficient of volume expansion is much larger for the coolant than for the aluminum and
the steel, the coolant will expand more than the aluminum and steel, and so coolant will overflow the
cooling system. Use Eq. 17-2.
β β βΔV = ΔVcoolant − ΔValuminum − ΔVsteel = V ΔT −coolant coolant V ΔT −aluminum aluminum V ΔTsteel steel
( β β β )= V − V − V ΔTcoolant coolant
aluminum aluminum steel steel
( ) ( ) ( )= ⎡⎣ 410 ×10−6 C° (17 L) − 75 ×10−6 C° (3.5L) − 35×10−6 C° (13.5L)⎤⎦ (12C°)
= 0.0748 L ≈ 75mL
17. (a) The amount of water lost is the final volume of the water minus the final volume of the
container. Also note that the original volumes of the water and the container are the same.
( ) ( )Vlost =
V0 + ΔV −H2O V0 + ΔV β β= ΔV − ΔV = V ΔT − V ΔTcontainer
H2O container H2O 0 container 0
Vlost (0.35g) ⎛ 1mL g ⎞
V0ΔT ⎜⎝ 0.98324 ⎟⎠
βcontainer = β H2O − = 210 ×10−6 C° − = 5.0 ×10−5 C°
( 55.50 mL) (60°C − 20°C)
(b) From Table 17-1, the most likely material is copper .
18. (a) The sum of the original diameter plus the expansion must be the same for both the plug and the
ring.
( ) ( ) l l l ll l l l α α+ Δ = + Δ → + ΔT = + ΔT0 iron 0 brass
iron iron iron brass brass brass
l − lbrass 8.753 cm − 8.743 cm
8.743 cm − 19 ×10−6 C°
( ) ( )ΔT = iron =
12 ×10−6 C°
( ) ( )α l − α liron iron
brass brass 8.753 cm
= −163C° = Tfinal − Tinitial = Tfinal − 15°C → Tfinal = −148°C ≈ −150°C
(b) Simply switch the initial values in the above calculation.
l − lbrass 8.743 cm − 8.753 cm
8.753 cm − 19 ×10−6 C°
( ) ( )ΔT = iron = 8.743 cm =
12 ×10−6 C°
( ) ( )α l − α liron iron
brass brass
= 164C° = Tfinal − Tinitial = Tfinal − 15°C → Tfinal = 179°C ≈ 180°C
19. We model the vessel as having a constant cross-sectional area A. Then a volume V0 of fluid will
occupy a length l 0 of the tube, given that V0 = Al 0 . Likewise V = Al .
ΔV = V −V0 = Al − Al 0 = AΔl and ΔV = βV0ΔT = β Al 0ΔT .
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536
Chapter 17 Temperature, Thermal Expansion, and the Ideal Gas Law
Equate the two expressions for ΔV , and get AΔl = β Al 0ΔT → Δl = βl 0ΔT . But Δl = αl 0ΔT ,
so we see that under the conditions of the problem, α = β .
20. (a) When a substance changes temperature, its volume will change by an amount given by
Eq. 17-2. This causes the density to change.
Δρ = ρf −ρ = M M = V0 M M M M = M ⎛1 − 1⎞⎠⎟
V − V0 + ΔV − V0 = V0 + βV0ΔT − V0 V0 ⎜⎝ 1 + βΔT
= ρ ⎛ 1 − 1+ βΔT ⎞ = ρ ⎛ −βΔT ⎞
⎜⎝ 1 + βΔT 1+ βΔT ⎠⎟ ⎝⎜ 1 + βΔT ⎠⎟
If we assume that βΔT 1 , then the denominator is approximately 1, so Δρ = −ρβΔT .
(b) The fractional change in density is
( )Δρ = −ρβΔT = −βΔT = − 87 ×10−6 °C (−55°C − 25°C) = 6.96 ×10−3
ρρ
This is a 0.70% increase .
21. As the wine contracts or expands, its volume changes. We assume that the volume change can only
occur by a corresponding change in the headspace. Note that if the volume increases, the headspace
decreases, so their changes are of opposite signs. Use Eq. 17-2.
(a) The temperature decreases, so the headspace should increase.
ΔV = βV0ΔT = −π r2ΔH →
( )ΔH ⎛ 10−3 m3 ⎞
βV0ΔT 420 ×10−6 C° ( 0.750 L) ⎜⎝ 1L ⎟⎠ (10 C°)
π r2
= − = π (0.00925m)2 = 0.0117 m
H = 1.5cm + 1.17 cm = 2.67 cm ≈ 2.7 cm
(b) The temperature increases, so the headspace should decrease.
( )ΔH ⎛ 10−3 m3 ⎞
βV0ΔT 420 ×10−6 C° ( 0.750 L) ⎜⎝ 1L ⎟⎠ ( −10 C° )
π r2
= − = π (0.00925m)2 = −0.0117 m
H = 1.5cm − 1.17 cm = 0.33cm ≈ 0.3cm
22. (a) The original surface area of the sphere is given by A = 4π r2 . The radius will expand with
temperature according to Eq. 17-1b, rnew = r (1 + αΔT ) . The final surface area is Anew = 4π rn2ew ,
and so the change in area is ΔA = Anew − A .
ΔA = Anew − A = 4π r2 (1 + αΔT )2 − 4π r2 = 4π r2 ⎡⎣(1 + αΔT )2 − 1⎦⎤
= 4π r2 ⎡⎣1 + 2αΔT + α 2 ( ΔT )2 − 1⎦⎤ = 4π r2 (2αΔT ) [1 + 1 αΔT ]
2
If the temperature change is not large, 1 αΔT 1, and so ΔA = 8π r2αΔT
2
(b) Evaluate the above expression for the solid iron sphere.
( ) ( )ΔA = 8π r2αΔT = 8π 60.0 ×10−2 m 2 12 ×10−6 °C (275°C − 15°C) = 2.8 ×10−2 m2
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537
Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual
23. The pendulum has a period of τ0 = 2π l 0 g at 17oC, and a period of τ = 2π l g at 28oC. Notice
that τ > τ 0 since l > l 0 . With every swing of the clock, the clock face will indicate that a time τ 0
has passed, but the actual amount of time that has passed is τ . Thus the clock face is “losing time”
by an amount of Δτ =τ −τ0 every swing. The fractional loss is given by Δτ , and the length at the
τ0
higher temperature is given by
Δτ = τ −τ0 = 2π l g − 2π l 0 g = l − l 0 = l 0 + Δl − l 0 = l 0 + αl 0ΔT − l 0
τ0 τ0 2π l 0 g l0 l0 l0
( )= 1 + αΔT − 1 = 1 + 19 ×10−6 C° (11C°) − 1 = 1.04 ×10−4
( )Thus the amount of time lost in any time period τ 0 is Δτ = 1.04 ×10−4 τ0 . For one year, we have
the following.
( ) ( )Δτ = 1.04 ×10−4 3.16 ×107s = 3286s ≈ 55 min
24. The change in radius with heating does not cause a torque on the rotating wheel, and so the wheel’s
angular momentum does not change. Also recall that for a cylindrical wheel rotating about its axis,
the moment of inertia is I = 1 mr2 .
2
L0 = Lfinal ω ω→ I = I0 0 → ωfinal = I0ω0 = 1 mr02ω0 = r02ω0
I final 2 r2
final final 1 mr 2
2
ωfinal − ω0 r02ω0 − ω0 r02 r02 r02 r02
ω0 r2 r2 + Δr r0 + αr0ΔT r0 + αr0ΔT
=Δω = = −1= −1= −1= 2 −1
ω0 r0 2 2
( ) ( ) ( )ω
( )1
= (1 + αΔT )2
1− 1 + 2αΔT + (αΔT )2 = −2αΔT − (αΔT )2 = −αΔT 2 + αΔT
−1= (1 + αΔT )2 (1 + αΔT )2
(1 + αΔT )2
Now assume that αΔT 1, and so Δω = −αΔT 2 +αΔT ≈ −2αΔT . Evaluate at the given values.
ω
(1+αΔT )2
( )−2αΔT = −2 25 ×10−6 C° (75.0C°) = −3.8 ×10−3
25. The thermal stress must compensate for the thermal expansion. E is Young’s modulus for the
aluminum.
( ) ( )Stress = F A = α EΔT = 25 ×10−6 C° 70 ×109 N m2 (35°C − 18°C) = 3.0 ×107 N m2
26. (a) Since the beam cannot shrink while cooling, the tensile stress must compensate in order to keep
the length constant.
( ) ( )Stress = F A = αEΔT = 12 ×10−6 C° 200 ×109 N m2 (50C°) = 1.2 ×108 N m2
(b) The ultimate tensile strength of steel (from Table 12-2) is 5×108 N m2 , and so
the ultimate strength is not exceeded . There would only be a safety factor of about 4.2.
(c) For concrete, repeat the calculation with the expansion coefficient and elastic modulus for
concrete.
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538
Chapter 17 Temperature, Thermal Expansion, and the Ideal Gas Law
( ) ( )Stress = F A = αEΔT = 12 ×10−6 C° 20 ×109 N m2 (50C°) = 1.2 ×107 N m2
The ultimate tensile strength of concrete is 2×106 N m2 , and so the concrete will fracture .
27. (a) Calculate the change in temperature needed to increase the diameter of the iron band so that it
fits over the barrel. Assume that the barrel does not change in dimensions.
Δl = αl 0ΔT → l − l 0 = αl 0 (T − T0 )
T0 l −l0 134.122 cm − 134.110 cm 27.457°C
αl 0 12 ×10−6
( )T C° (134.110 cm) 27°C
= + = 20°C + = ≈
(b) Since the band cannot shrink while cooling, the thermal stress must compensate in order to keep
the length at a constant 132.122 cm. E is Young’s modulus for the material.
Stress = F A = αEΔT → F = AE Δl = AEαΔT
l0
( ) ( ) ( ) ( )= 9.4 ×10−2 m 6.5 ×10−3 m 100 ×109 N m2 12 ×10−6 C° (7.457 C°) = 5500 N
28. Use the relationships T (K) = T (°C) + 273.15 and T (K) = [5 T (°F) − 32] + 273.15 .
9
(a) T (K) = T (°C) + 273.15 = 66 + 273.15 = 339 K
(b) T (K) = 5 [T ( °F ) − 32] + 273.15 = 5 [92 − 32] + 273.15 = 306 K
9 9
(c) T (K) = T (°C) + 273.15 = −55 + 273.15 = 218 K
(d) T (K) = T (°C) + 273.15 = 5500 + 273.15 = 5773.15 K ≈ 5800 K
29. Use the relationship that T (K) = 5 [T (°F) − 32] + 273.15 .
9
T (K) = 5 [T (°F) − 32] + 273.15 →
9
T (°F) = 9 [T ( K ) − 273.15] + 32 = 9 [0 − 273.15] + 32 = −459.67°F
5 5
30. Use the relationship that T (K) = T (°C) + 273.15 .
(a) T (K) = T (°C) + 273.15 = 4270 K ≈ 4300 K ; T (K) = T (°C) + 273.15 = 15 ×106 K
(b) % error = ΔT ) ×100 = 273 ×100
T (K T (K)
4000°C: 273 ×100 ≈ 7% 15 ×106 °C: 273 × 100 ≈ 2 ×10−3%
4000 15 ×106
31. Assume the gas is ideal. Since the amount of gas is constant, the value of PV is constant.
T
( )P1V1 P2V2 P1 T2 ⎜⎝⎛ 1.00 atm ⎟⎞⎠ (273 + 38.0) K
T2 P2 T1 3.20 atm
T1 273 K
= → V2 = V1 = 3.80 m3 = 1.35 m3
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539
Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual
32. Assume the air is an ideal gas. Since the amount of air is constant, the value of PV is constant.
T
P1V1 = P2V2 → T2 = T1 P2 V2 = ( 293 K) ⎛ 40 atm ⎞ ⎛ 1 ⎞ = 1465 K = 1192°C ≈ 1200°C
T1 T2 P1 V1 ⎝⎜ 1 atm ⎠⎟ ⎜⎝ 8 ⎟⎠
33. Assume the nitrogen is an ideal gas. From Example 17-10, the volume of one mole of nitrogen gas
at STP is 22.4 ×10−3 m3 . The mass of one mole of nitrogen, with a molecular mass of 28.0 u, is 28.0
grams. Use these values to calculate the density of the oxygen gas.
ρ = M = 28.0 ×10−3 kg = 1.25 kg m3
V 22.4 ×10−3 m3
34. (a) Assume that the helium is an ideal gas, and then use the ideal gas law to calculate the volume.
Absolute pressure must be used, even though gauge pressure is given.
nRT (14.00 mol) (8.314 J moliK) (283.15 K) 0.2410 m3
P (1.350atm) 1.013 ×105 Pa atm
( )PV = nRT → V = = =
(b) Since the amount of gas is not changed, the value of PV T is constant.
P1V1 = P2V2 → T2 = T1 P2 V2 = ( 283.15 K ) ⎝⎛⎜ 2.00 atm ⎞⎟⎠ ⎜⎝⎛ 1 ⎞⎠⎟ = 210 K = −63o C
T1 T2 P1 V1 1.350 atm 2
35. We ignore the weight of the stopper. Initially there is a net force (due to air pressure) on the stopper
of 0, because the pressure is the same both above and below the stopper. With the increase in
temperature, the pressure inside the tube will increase, and so there will be a net upward force given
( )by Fnet = Pin − Pout A. The inside pressure can be expressed in terms of the inside temperature by
means of the ideal gas law for a constant volume and constant mass of gas.
P Vin tube = P0Vtube → Pin = P0 Tin
Tin T0 T0
Fnet = ( Pin )− Pout A = ⎛ P0 Tin − P0 ⎞ A = P0 ⎛ Tin − 1⎟⎞ A →
⎜ T0 ⎟ ⎜ T0 ⎠
⎝ ⎠ ⎝
⎛ Fnet 1⎟⎠⎞ ⎡ (10.0 N) ⎤ = 454 K 181°C
⎜⎝ P0 A ⎢ 1.013 ×105 Pa π (0.0075m)2
( )Tin= + T0 = ⎢⎣ + 1⎥ (273K + 18 K) =
⎦⎥
36. Assume that the nitrogen and carbon dioxide are ideal gases, and that the volume and temperature
are constant for the two gases. From the ideal gas law, the value of P RT is constant. Also note
n= V
that concerning the ideal gas law, the identity of the gas is unimportant, as long as the number of
moles is considered.
P1 = P2 →
n1 n2
⎛ 21.6 kg CO2 ⎞
⎜ ⎟
P2 = P1 n2 44.01×10−3 kg CO2 mol ⎟ = (3.85 atm) ⎝⎛⎜ 28.01 ⎠⎟⎞ = 2.45 atm
n1 = (3.85atm) ⎜ 21.6 kg N2 mol ⎟⎠⎟ 44.01
⎝⎜⎜ 28.01×10−3 kg N2
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540
Chapter 17 Temperature, Thermal Expansion, and the Ideal Gas Law
37. (a) Assume the nitrogen is an ideal gas. The number of moles of nitrogen is found from the atomic
weight, and then the ideal gas law is used to calculate the volume of the gas.
n = ( 28.5 kg) 1 mole N2 = 1017 mol
28.01×10−3 kg
PV = nRT → V = nRT = (1017 mol) (8.314 J moliK) (273 K) = 22.79 m3 ≈ 22.8 m3
P
1.013 ×105 Pa
(b) Hold the volume and temperature constant, and again use the ideal gas law.
n = ( 28.5 kg + 25.0 kg) 1 mole N2 = 1910 mol
28.01×10−3 kg
PV = nRT →
P = nRT = (1910 mol) (8.314 J moliK) (273 K) = 1.90 ×105 Pa = 1.88atm
V
22.79 m3
38. We assume that the mass of air is unchanged, and the volume of air is unchanged (since the tank is
rigid). Use the ideal gas law.
[ ]P1 P2 P2 ⎛ 194 atm ⎞
T2 P1 ⎜⎝ 204 atm ⎠⎟
T1
= → T2 = T1 = (273+ 29) K = 287 K = 14o C
39. Assume the argon is an ideal gas. The number of moles of argon is found from the atomic weight,
and then the ideal gas law is used to find the pressure.
n = (105.0 kg) 1 mole Ar = 2628 mol
39.95 ×10−3 kg
nRT (2628 mol) (8.314 J molik) (293.15 K) 1.69 ×108 Pa
V (38.0 L) 1.00 ×10−3 m3 L
( )PV = nRT → P = = =
This is 1660 atm.
40. Assume that the oxygen and helium are ideal gases, and that the volume and temperature are
constant for the two gases. From the ideal gas law, the value of P RT is constant. Also note that
n= V
concerning the ideal gas law, the identity of the gas is unimportant, as long as the number of moles is
considered. Finally, gauge pressure must be changed to absolute pressure.
P1 = P2 → n2 = n1 P2 = (30.0 kg O2 ) ⎛ 1 mole O2 ⎞ ( 8.00 atm) = 8.152 ×102 moles
n1 n2 P1 ⎜⎝ 32 ×10−3 kg ⎠⎟ ( 9.20 atm)
( )8.152 ×102 moles ⎛ 4.0 ×10−3 kg ⎞ = 3.26 kg He
⎝⎜ 1 mole He ⎠⎟
41. We assume that the gas is ideal, that the amount of gas is constant, and that the volume of the gas is
constant.
P1 = P2 → T2 = T1 P2 = [( 273.15 + 20.0) K]⎝⎜⎛ 2.00 atm ⎞ = 586.3 K = 313.15°C ≈ 313°C
T1 T2 P1 1.00 atm ⎠⎟
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541
Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual
42. Assume that the air is an ideal gas. The pressure and volume are held constant. From the ideal gas
law, the value of PV = nT is held constant.
R
n1T1 = n2T2 → n2 = T1 = (273 + 38) K = 311 = 1.080
n1 T2 (273 + 15) K 288
Thus 8.0 % must be removed.
43. Assume the oxygen is an ideal gas. Since the amount of gas is constant, the value of PV T is
constant.
P1V1 = P2V2 → P2 = P1 V1 T2 = ( 2.45 atm) ⎛ 61.5 L ⎞ (273 + 56.0) K = 3.49 atm
T1 T2 V2 T1 ⎜⎝ 48.8 L ⎟⎠ (273 + 18.0) K
44. Assume the helium is an ideal gas. Since the amount of gas is constant, the value of PV T is
constant. We assume that since the outside air pressure decreases by 30%, the air pressure inside the
balloon will also decrease 30%.
P1V1 = P2V2 → V2 = P1 T2 = ⎛ 1.0atm ⎞ (273 + 5.0) K = 1.4 times the original volume
T1 T2 V1 P2 T1 ⎜⎝ 0.68atm ⎟⎠ (273 + 20.0) K
45. Since the container can withstand a pressure difference of 0.50 atm, we find the temperature for
which the inside pressure has dropped from 1.0 atm to 0.50 atm. We assume the mass of contained
gas and the volume of the container are constant.
P1 = P2 → T2 = T1 P2 = [(273.15 + 18) K]⎝⎜⎛ 0.50 atm ⎞ = 146.6 K ≈ −130°C
T1 T2 P1 1.0 atm ⎠⎟
46. The pressure inside the bag will change to the surrounding air pressure as the volume of the bag
changes. We assume the amount of gas and temperature of the gas are constant. Use the ideal gas
equation.
P1V1 = P2V2 → V2 = V1 P1 = V1 ⎛ 1.0 atm ⎞ = 1.3V1
P2 ⎜⎝ 0.75 atm ⎠⎟
Thus the bag has expanded by 30%.
47. We assume that all of the gas in this problem is at the same temperature. Use the ideal gas equation.
First, find the initial number of moles in the tank at 34 atm, n1. Then find the final number of moles
in the tank at 204 atm, n2. The difference in those two values is the number of moles needed to add
to the tank.
P1V1 = n1RT → n1 = P1V1 = (34 atm) (12 L)
RT
RT
P2V2 = n2RT → n2 = P2V2 = (204 atm) (12 L)
RT
RT
nadd = n2 − n1 = (170atm) (12 L)
RT
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542
Chapter 17 Temperature, Thermal Expansion, and the Ideal Gas Law
Use this number of moles at atmospheric pressure to find the volume of air needed to add, and then
find the time needed to add it.
nadd RT (170atm) (12 L)
Padd
P Vadd add = nadd RT → Vadd = = RT = 2040 L
1atm
2040 L ⎛ 1min ⎞ = 7.0 min
⎜⎝ 290 L ⎟⎠
48. From the ideal gas equation, we have P1V1 = nRT1 and P2V2 = nRT2 , since the amount of gas is
constant. Use these relationships along with the given conditions to find the original pressure and
temperature.
P1V1 = nRT1 ; P2V2 = nRT2 → P1V1 − P2V2 = nRT1 − nRT2 = nR (T1 − T2 ) →
P1V1 − ( P1 + 450 Pa)V2 = nR (T1 − T2 ) → P1 (V1 −V2 ) −V2 (450 Pa) = nR (9.0 K) →
( )P1
= nR (9.0 K) +V2 (450 Pa) = (4.0 mol) (8.314 J moliK) (9.0 K) + 0.018 m3 (450 Pa)
V1 − V2 0.002 m3
= 1.537 ×105 Pa ≈ 1.5 ×105 Pa
( )( )T1
= P1V1 = T1 = 1.537 ×105 Pa 0.020 m3 = 92.4 K ≈ −181°C
nR
(4.0 mol) (8.314 J moliK)
49. We calculate the density of water vapor, with a molecular mass of 18.0 grams per mole, from the
ideal gas law.
PV = nRT → nP →
V = RT
( )m Mn MP (0.0180 kg mol) 1.013 ×105 Pa = 0.588 m3
ρ = V = V = RT = (8.314 J moliK) (373K)
The density from Table 13-1 is 0.598 m3 . Because this gas is very “near” a phase change state
(water can also exist as a liquid at this temperature and pressure), we would not expect it to act like
an ideal gas. It is reasonable to expect that the molecules will have other interactions besides purely
elastic collisions. That is evidenced by the fact that steam can form droplets, indicating an attractive
force between the molecules.
50. The ideal gas law can be used to relate the volume at the surface to the submerged volume of the
bubble. We assume the amount of gas in the bubble doesn’t change as it rises. The pressure at the
submerged location is found from Eq. 13-6b.
PV P V P Vsurface surface
T T = Tsurface
PV = nRT → = nR = constant → submerged submerged →
submerged
V = V PP TT = V P P+ ρ gh TTsurface
submerged submerged surface submerged atm surface
surface submerged atm submerged
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543
Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual
( ) ( ( ))( )=
1.00 cm3 ⎣⎡1.013 ×105 + 1.00 ×103 kg m3 9.80 m s2 (37.0 m)⎤⎦ (273.15+ 18.5) K
1.013 ×105 Pa (273.15 + 5.5) K
= 4.79 cm3
51. At STP, 1 mole of ideal gas occupies 22.4 L.
1 mole ⎛ 6.02×1023 molecules ⎞⎛ 1 L ⎞ = 2.69 ×1025 molecules m3
⎜ ⎟ ⎝⎜ ⎠⎟
22.4 L ⎝ mole ⎠ 10−3 m3
52. We assume that the water is at 4oC so that its density is 1000 kg m3 .
1.000 L ⎛ 10−3m3 ⎞ ⎛ 1000 kg ⎞ ⎛ (15.9994 + 2 1 mol ) × 10−3 kg ⎞ = 55.51 mol
⎝⎜ 1L ⎠⎟ ⎝⎜ 1 m3 ⎠⎟ ⎜ × 1.00794 ⎟
⎝ ⎠
55.51 mol ⎛ 6.022 × 1023 molecules ⎞ = 3.343 ×1025 molecules
⎝⎜ 1 mol ⎠⎟
53. We use Eq. 17-4.
PV = NkT → P= N kT = ⎛ 1 molecule ⎞ ⎛ 1003 cm3 ⎞ ⎜⎝⎛ 1.38 × 10−23 J ⎟⎠⎞ ( 3 K ) = 4 ×10−17 Pa
V ⎝⎜ 1cm3 ⎠⎟ ⎜⎝ m3 ⎟⎠ K
54. (a) Since the average depth of the oceans is very small compared to the radius of the Earth, the
ocean’s volume can be calculated as that of a spherical shell with surface area π4 R2 and a
Earth
thickness Δy . Then use the density of sea water to find the mass, and the molecular weight of
water to find the number of moles.
6.38×106 m 2
( ) ( ) ( )Volume = 0.75
π4 R2 Δy = 0.75(4π ) 3×103 m = 1.15×1018 m3
Earth
1.15 ×1018 m3 ⎛ 1025 kg ⎞ ⎛ 1 mol ⎞ = 6.55 ×1022 moles ≈ 7 ×1022 moles
⎜⎝ m3 ⎟⎠ ⎜⎝ × 10−3 ⎠⎟
18 kg
( )(b) 6.55 ×1022 moles 6.02 ×1023 molecules 1 mol ≈ 4 ×1046 molecules
55. Assume the gas is ideal at those low pressures, and use the ideal gas law.
NP 1 ×10−12 N m2 ⎝⎛⎜ 3 ×108 molecules ⎞ ⎛ 10−6 m3 ⎞
V = kT = 1.38 ×10−23 J K m3 ⎟⎠ ⎜⎝ 1 cm3 ⎟⎠
( )PV = NkT → (273 K) =
= 300 molecules cm3
56. We assume an ideal gas at STP. Example 17-10 shows that the molar volume of this gas is 22.4 L.
We calculate the actual volume of one mole of gas particles, assuming a volume of l 3 , and then find
0
the ratio of the actual volume of the particles to the volume of the gas.
( ()(( )) )( )Vmolecules
V =gas
6.02 ×1023 molecules 3.0 ×10−10 m 3 molecule = 7.3 ×10−4
22.4 L 1×10−3 m3 1L
The molecules take up less than 0.1% of the volume of the gas.
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544
Chapter 17 Temperature, Thermal Expansion, and the Ideal Gas Law
57. We assume that the last breath Galileo took has been spread uniformly throughout the atmosphere
since his death. Calculate the number of molecules in Galileo’s last breath, and divide it by the
volume of the atmosphere, to get “Galileo molecules/m3”. Multiply that factor times the size of a
breath to find the number of Galileo molecules in one of our breaths.
( ( )( ) )PV = NkT
→ N = PV = 1.01×105 Pa 2.0 ×10−3 m3 = 4.9 ×1022 molecules
kT
1.38×10−23 J K (300 K )
6.38×106 m 2
( ) ( )Atmospheric = 5.1×1018 m3
volume = 4π R2 h = 4π 1.0 ×104 m
Earth
Galileo molecules = 4.9 ×1022 molecules = 9.6 ×103 molecules m3
m3 5.8 ×1018 m3
# Galileo molecules = 9.6 ×103 molecules ⎛ 2.0 ×10−3 m3 ⎞ = 19 molecules
breath m3 ⎝⎜ 1 breath ⎟⎠ breath
58. Use Eq. 17-5a for the constant-volume gas thermometer to relate the boiling point to the triple point.
T = (273.16 K) P → Pbp = Tbp = (273.15 + 100) K = 1.3660
Ptp Ptp 273.16 K
273.16 K
59. (a) For the constant-volume gas thermometer, we use Eq. 17-5a.
T = (273.16 K) P → Ptp = P 273.16 K = (187 torr ) 273.16 K K = 71.2 torr
Ptp T
(273.15 + 444.6)
(b) We again use Eq. 17-5a.
T = ( 273.16 K ) P = ( 273.16 K ) 118 torr = 453 K = 180°C
Ptp 71.2 torr
60. From Fig. 17-17, we estimate a temperature of 373.35 K from the oxygen curve at a pressure of 268
torr. The boiling point of water is 373.15 K.
(a) The inaccuracy is ΔT = 373.35 K − 373.15 K = 0.20 K
(b) As a percentage, we have the following.
ΔT (100) = 0.20 K (100) = 0.054%
T 373.15 K
The answers may vary due to differences in reading the graph.
61. Since the volume is constant, the temperature of the gas is proportional to the pressure of the gas.
First we calculate the two temperatures of the different amounts of gas.
T1 ( )= P1 melt ( 273.16 K) 218 = 208.21K
273.16 K P1 tp = 286
T2 ( )= P2 melt ( 273.16 ) 128 = 214.51K
273.16 K P2 tp = K 163
Assume that there is a linear relationship between the melting-point temperature and the triple-point
pressure, as shown in Fig. 17-17. The actual melting point is the y-intercept of that linear
relationship. We use Excel to find that y-intercept. The graph is shown below.
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545
Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual
216T (K)
214
212
210
208 y = -0.0512 x + 222.85
206
150 175 200 225 250 275 300
P tp (torr)
We see the melting temperature is 222.85 K ≈ 223K . The spreadsheet used for this problem can be
found on the Media Manager, with filename “PSE4_ISM_CH17.XLS,” on tab “Problem 17.61.”
62. Since the glass does not expand, the measuring cup will contain 350 mL of hot water. Find the
volume of water after it cools.
( )ΔV = V0βΔT = (350 mL) 210 ×10−6 C° (20°C − 95°C) = −5.5mL
The volume of cool water is about 5.5 mL less than the desired volume of 350 mL.
63. (a) At 36oC, the tape will expand from its calibration, and so will read low .
( )(b) Δl = αΔT = 12 ×10−6 °C (36°C − 15°C) = 2.52 ×10−4 ≈ 2.5 ×10−2 %
l0
64. The net force on each side of the box will be the pressure difference between the inside and outside
of the box, times the area of a side of the box. The outside pressure is 1 atmosphere. The ideal gas
law is used to find the pressure inside the box, assuming that the mass of gas and the volume are
constant.
P = nR = constant → P2 = P1 → P2 = P1 T2 = (1.00 atm) (273 + 185) K = 1.590 atm
T V T2 T1 T1 (273 + 15) K
The area of a side of the box is given by the following.
( )Area = l 2 = ⎡⎣(Volume of )box ⎦⎤1/3 2 = 6.15 ×10−2 m2 2/3 = 1.5581×10−1m2
The net force on a side of the box is the pressure difference times the area.
( ) ( )F = (Δ Pressure) (Area) = (0.590atm) 1.01×105 Pa 1.5581×10−1m2 = 9300 N
65. Assume the helium is an ideal gas. The volume of the cylinder is constant, and we assume that the
temperature of the gas is also constant in the cylinder. From the ideal gas law, PV = nRT , under
these conditions the amount of gas is proportional to the absolute pressure.
PV = nRT → P = RT = constant → P1 = P2 → n2 = P2 = 5atm + 1atm = 6
n V n1 n2 n1 P1 32 atm + 1atm 33
Thus 6 33 = 0.182 ≈ 20% of the original gas remains in the cylinder.
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546
Chapter 17 Temperature, Thermal Expansion, and the Ideal Gas Law
66. When the rod has a length l, then a small (differential) change in temperature will cause a small
(differential) change in length according to Eq. 17-1a, expressed as dl = αl ΔT.
dl l2 dl T2 ln l 2 = α (T2 − T1 )
l l1
= αdT
∫ ∫(a)
dl = αl ΔT → = αΔT → ll 1 → →
T1
l l= eα(T2 −T1)
21
T2
l2 dl T2 ln l 2 T2 ∫ αdT
= αdT = αdT
∫ ∫ ∫(b) → → l2 = l eT1
ll 1 l 1 T1 1
T1
dl = αΔT l2 dl T2 T2
l ll 1
= αdT =
T1 T1
( )→
dl = αl ΔT α0 + bT dT
∫ ∫ ∫(c) → →
( ) ( )ln l 2 → l = l e ( )⎣⎡α0 (T2 −T1 ) + 1 b T22 −T12 ⎦⎤
l1 2
= α0 T2 − T1 + 1 b T22 − T12
2 21
67. Assume that the air in the lungs is an ideal gas, that the amount of gas is constant, and that the
temperature is constant. The ideal gas law then says that the value of PV is constant. The pressure
a distance h below the surface of a fluid is given by Eq. 13-6b, P = P0 + ρ gh , where P0 is
atmospheric pressure and ρ is the density of the fluid. We assume that the diver is in sea water.
→ V = V PP = V P P+ ρ ghsurface
( ) ( )PV = PVsurface submerged submerged submerged submerged atm
surface atm
( )( )= (5.5L) 1.01×105 Pa +
1025 kg m3 9.80 m s2 (8.0 m) 9.9 L
1.01×105 Pa
=
This is obviously very dangerous, to have the lungs attempt to inflate to twice their volume. Thus it
is not advisable to quickly rise to the surface.
68. (a) Assume the pressure and amount of gas are held constant, and so P0V0 = nRT0 and P0V = nRT .
From these two expressions calculate the change in volume and relate it to the change in
temperature.
V = V0 + ΔV → ΔV =V − V0 = nRT − nRT0 = nR ( T )− T0 = V0 ΔT
P0 P0 P0 T0
But ΔV = βV0ΔT , and so ΔV = V0 ΔT = βV0ΔT → 1
T0 β = T0
For T0 = 293 K , β 1 1 = 3.4 ×10−3 K , which agrees well with Table 17-1.
= = 293 K
T0
(b) Assume the temperature and amount of gas are held constant, and so P0V0 = nRT0 = PV . From
these two expressions calculate change in volume and relate it to the change in pressure.
V = V0 + ΔV →
ΔV =V − V0 = nRT0 − nRT0 = nRT0 ⎛ 1 − 1 ⎞ = nRT0 ⎛ P0 − P ⎞ = V0 1 ( −ΔP )
P P0 ⎜ P P0 ⎟ P0 ⎝⎜ P ⎠⎟ P
⎝ ⎠
But from Eq. 12-7, ΔV = −V0 1 ΔP and so ΔV = V0 1 ( −ΔP ) = −V0 1 ΔP → B=P
B P B
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547
Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual
69. To do this problem, the “molecular weight” of air is needed. If we approximate air as 70% N2
(molecular weight 28) and 30% O2 (molecular weight 32), then the average molecular weight is
0.70(28) + 0.30(32) = 29.2.
(a) Treat the air as an ideal gas. Assume that the pressure is 1.00 atm.
( )( )PV = nRT
→ n= PV = 1.013 ×105 Pa 870 m3 = 3.6178 ×104 moles
RT
(8.314 J molik) (293 K)
( ) ( )m = 3.6178 ×104 moles 29.2 ×10−3 kg mol = 1056.4 kg ≈ 1100 kg
(b) Find the mass of air at the lower temperature, and then subtract the mass at the higher
temperature.
( )( )n
= PV = 1.013 ×105 Pa 870 m3 = 4.0305 ×104 moles
RT
(8.314 J molik) (263 K)
( ) ( )m = 4.0305 ×104 moles 29.2 ×10−3 kg mol = 1176.9 kg
The mass entering the house is 1176.9 kg − 1056.4 kg = 120.5kg ≈ 100 kg .
70. We are given that P ∝ 1 for constant temperature and V ∝ T2/3 for constant pressure. We also
V2
assume that V ∝ n for constant pressure and temperature. Combining these relationships gives the
following.
PV 2 = n2RT 4 / 3
R= PV 2 = (1.00atm) (22.4 L)2 = (1.00atm) (22.4 L)2 = 0.283 L2 iatm
n2T 4 / 3 (1.00 mol)2 (273.15 K)4/3 (1.00 mol)2 (273.15 K)4/3 mol2 iK4/3
71. (a) The iron floats in the mercury because ρHg > ρFe . As the substances are heated, the density of
both substances will decrease due to volume expansion. The density of the mercury decreases
more upon heating than the density of the iron, because βHg > βFe . The net effect is that the
densities get closer together, and so relatively more mercury will have to be displaced to hold
up the iron, and the iron will float lower in the mercury.
(b) The fraction of the volume submerged is VHg VFe . Both volumes expand as heated. The
displaced
subscript “displaced” is dropped for convenience.
(( )) (( ))fractional change = VHg V0 Hg 1 + βHgΔT − V0 Hg V0 Fe
VFe − V0 Hg V0 Fe = V0 Fe 1 + βFeΔT 1 + βHgΔT
V V0 Hg 0 Fe = 1 + βFeΔT −1
V V0 Hg 0 Fe
( ) ( )1 + 180 ×10−6 Co 25Co − 1 = 1.0045 − 1 = 3.6 × 10−3 ; % change = 0.36 %
( ) ( )= 1 + 35 ×10−6 25Co 1.000875
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Física para Ciencias e Ingeniería. Volumen I - Douglas C. Giancoli - 4ta edición [Solucionario]
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