CHAPTER
17 THE PRINCIPLE OF
LINEAR SUPERPOSITION AND
INTERFERENCE PHENOMENA
CONCEPTUAL QUESTIONS
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1.
REASONING AND SOLUTION The principle of linear superposition states that when
two or more waves are present simultaneously at the same place, the resultant wave is the
sum of the individual waves. This principle does not imply that two sound waves, passing
through the same place at the same time, always create a louder sound than either wave
alone. The resultant wave pattern depends on the relative phases of the two sound waves
when they meet. If two sound waves arrive at the same place at the same time, and they are
exactly in phase, then the two waves will interfere constructively and create a louder sound
than either wave alone. On the other hand, if two waves arrive at the same place at the same
time, and they are exactly out of phase, destructive interference will occur; the net effect is
a mutual cancellation of the sound. If the two sound waves have the same amplitude and
frequency, they will completely cancel each other and no sound will be heard.
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2.
REASONING AND SOLUTION If you are sitting at the overlap point between the two
speakers in Figure 17.4, the two sound waves reaching you are exactly out of phase. You
hear no sound because of destructive interference. If one of the speakers is suddenly shut
off, then only one sound wave will reach your ears. Since there is no other sound wave to
interfere with this sound wave, you will hear the sound from the single speaker.
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3.
SSM REASONING AND SOLUTION Consider the situation in Figure 17.3. If you walk
along a line that is perpendicular to the line between the speakers and passes through the
overlap point, you will always be equidistant from both speakers. Therefore the sound
waves along this line always overlap exactly in phase. Hence, you will always hear the
same loudness; you will not observe the loudness to change from loud to faint to loud.
On the other hand, if you walk along a line that passes through the overlap point and is
parallel to the line between the speakers, your distance from the two speakers will vary such
that the difference in path lengths traveled by the two waves will vary. At certain points the
path length difference between the two sound waves will be an integer number of
wavelengths; constructive interference will occur at these points, and the sound intensity
will be a maximum. At other points, the path length difference between the two sound
waves will be an odd number of half-wavelengths [(1/2)λ, (3/2)λ, (5/2)λ, etc.]; destructive
interference will occur and the sound intensity will be a minimum. In between the points of
constructive and destructive interference, the waves will be out of phase by varying degrees.
Therefore, as you walk along this line, you will observe the sound intensity to alternate
between faint and loud.
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Chapter 17 Conceptual Questions
4.
873
REASONING AND SOLUTION If the width of the speakers is D, then sound of wavelength λ will diffract more readily if the ratio λ / D is large. Since longer wavelengths
correspond to lower frequencies, we see that lower-frequency sounds diffract more readily
from a given speaker than higher-frequency sounds. Since the frequencies of the sounds of
the female vocalists are higher than the frequencies of the sounds of the rhythmic bass, the
sounds of the female vocalists will not diffract to the same extent as the sounds of the
rhythmic bass. Thus, diffraction allows the bass tones to penetrate the regions to either side
of the stage, but it does not permit the same help to the sounds of the female vocalists.
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5.
REASONING AND SOLUTION When a wave encounters an obstacle or the edges of an
opening, it bends or diffracts around them. The extent of the diffraction depends on the ratio
of the wavelength λ to the size D of the obstacle or opening. If the ratio λ / D is small, little
diffraction occurs. As the ratio λ / D is made larger, the wave diffracts to a greater extent.
In Example 1 in Section 16.2, it is shown that the wavelength of AM radio waves is 244 m,
while the wavelength of FM radio waves is 3.26 m. For a given obstacle, the ratio λ / D
will be greater for AM radio waves; therefore AM radio waves will diffract more readily
around a given obstacle than FM waves.
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6.
REASONING AND SOLUTION A tuning fork has a frequency of 440 Hz. The string of a
violin and this tuning fork, when sounded together, produce a beat frequency of 1 Hz. This
beat frequency is the difference between the frequency of the tuning fork and the frequency
of the violin string. A violin string of frequency 439 Hz, as well as a violin string of 441
Hz, will produce a beat frequency of 1 Hz, when sounded together with the 440 Hz tuning
fork. We conclude that, from these two pieces of information alone, it is not possible to
distinguish between these two possibilities. Therefore, it is not possible to determine the
frequency of the violin string.
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7.
SSM REASONING AND SOLUTION Tuning forks vibrate at 438 Hz and 440 Hz.
When sounded together, a listener hears a beat frequency of 2 Hz, the difference in
frequency between the two forks. The forks are then vibrated underwater with the listener
also underwater. The forks vibrate at 438 and 440 Hz, just as they do in air. The speed of
sound, however, is four times faster in water than in air. From the relation v = λ f , we see
that the wavelength of the sound in water is four times longer than it is in air. Since the
speed of the waves is four times greater than it is in air, the regions of constructive and
destructive interference move past the ears of the listener four times more rapidly than they
do in air. However, the regions of condensation and rarefaction in water are separated by
four times the distance than they are in air. Therefore, at a given point in space, the
amplitude variation of the resultant wave occurs at the same rate that it does in air.
Consequently, the listener hears a beat frequency of 2 Hz, just as he does in air.
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8.
REASONING AND SOLUTION
The frequency of a guitar string is given by
Equation 17.3, f n = n v / ( 2 L ) , where v is the speed of the wave on the string, L is the
distance between the two fixed ends of the guitar string, and n = 1, 2, 3, 4, . . . The speed v
874 THE PRINCIPLE OF LINEAR SUPERPOSITION AND INTERFERENCE PHENOMENA
of the wave is given by Equation 16.2, v = F /(m / L) , where F is the tension in the guitar
string, and m / L is the mass per unit length of the string.
If the tension in a guitar string is doubled, then, according to Equation 16.2, the speed of
the wave on the string will increase by a factor of 2 . According to Equation 17.3, the
frequency of oscillation is directly proportional to the speed of the wave. Therefore, the
frequency of oscillation will not double; rather, it will increase by a factor of 2 .
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9.
REASONING AND SOLUTION A string is attached to a wall and vibrates back and forth
as in Figure 17.18. The vibration frequency and length of the string are fixed. The tension
in the string is changed. From Equation 16.2, v = F /(m / L) , we see that increasing the
tension F results in increasing the speed v of the waves on the string. From the relationship
v = λ f , we see that, since the frequency remains fixed, an increase in the wave speed
results in an increase in the wavelength of the wave.
It is observed that at certain values of the tension, a standing wave pattern develops.
Since the two ends of the string are "fixed," the ends of the string are nodes; thus, the length
L of the string must contain an integer number of half-wavelengths. Therefore, standing
waves will occur at the wavelengths λ n = 2L / n, where n = 1, 2, 3, 4, . . . The largest
possible wavelength that will result in a standing wave pattern occurs when n = 1, and the
wavelength is equal to twice the length of the string. If the tension is increased beyond the
value for which λ = 2L, the string cannot sustain a standing wave pattern.
____________________________________________________________________________________________
A string is being vibrated back and forth as in
10. REASONING AND SOLUTION
Figure 17.18a. The tension in the string is decreased by a factor of four, with the frequency
and the length of the string remaining the same. From Equation 16.2, v = F /(m / L) , we
see that decreasing the tension F by a factor of four results in decreasing the wave speed v
by a factor of 4 or 2. From the relationship v = λ f , we see that, when the frequency f is
fixed, decreasing the wave speed by a factor of 2 results in decreasing the wavelength λ by a
factor of 2. Therefore, when the tension in the string is decreased by a factor of 4, the
wavelength of the resulting standing wave pattern will decrease by a factor of 2. In
Figure 17.18a, the wavelength of the standing wave is equal to twice the length of the string.
Since decreasing the tension reduces the wavelength by a factor of 2, the new wavelength is
equal to the length of the string. With this new wavelength, the pattern shown below
results.
____________________________________________________________________________________________
Chapter 17 Conceptual Questions
875
11. REASONING AND SOLUTION a. Since the bottom end of the rope is free to move, there
is an antinode at the bottom.
b. Since the rope has mass, it has weight. The upper part of the rope supports more of the
weight than the lower part does. Thus, the tension in the rope is greater near the top than
near the bottom. The distance between successive nodes is one-half the wavelength of the
waves traveling up and down the rope. But the wavelength is equal to the speed of the
waves divided by the frequency, according to Equation 16.1. In turn, the speed is
proportional to the square root of the tension, according to Equation 16.2. As a result, the
wave speed is greater near the top than near the bottom and so is the wavelength. Since the
separation between successive nodes is one-half the wavelength, it is also greater near the
top than near the bottom.
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12. REASONING AND SOLUTION
Consider
the tube in Figure 17.23. We suppose that the
air in the tube is replaced with a gas in which
the speed of sound is twice what it is in air. If
the frequency of the tuning fork remains
unchanged, we see from the relation v = λ f
that the wavelength of the sound will be twice
as long in the gas as it is in air. The resulting
pattern is shown in the drawing at the right.
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13. REASONING AND SOLUTION When a concert performer generates a 2093-Hz tone, the
wavelength of the sound is
λ=
v 343 m/s
=
= 0.164 m or 16.4 cm
f 2093 Hz
If there is excessive reflection of the sound that the performer generates, a large amplitude
standing wave can result. The distance between an antinode (maximum loudness) and the
next adjacent node (zero loudness) on a standing wave is one-quarter of a wavelength. For a
standing sound wave with a frequency of 2093 Hz, the distance between an antinode and the
adjacent node is (16.4/4) cm or 4.1 cm. It would be possible, therefore, for a listener to
move a distance of only 4.1 cm and hear the loudness of the tone change from loud to faint.
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14. SSM REASONING AND SOLUTION The speed of sound v in an ideal gas is given by
Equation 16.5, v = γ kT / m , where γ = C P / CV is the ratio of the specific heat capacities,
k is Boltzmann's constant, T is the Kelvin temperature, and m is the mass of a molecule of
the gas. Thus, the speed of sound in an ideal gas is directly proportional to the square root
of the Kelvin temperature T.
876 THE PRINCIPLE OF LINEAR SUPERPOSITION AND INTERFERENCE PHENOMENA
If the instrument is open at both ends, then the frequency of sound produced by the
instrument is given by Equation 17.4, f n = n v / ( 2 L ) , where v is the speed of sound, L is
the distance between the two open ends of the instrument, and n = 1, 2, 3, 4, . . . Similarly,
if the instrument is open at only one end and closed at the other end, the frequency of sound
produced by the instrument is given by Equation 17.5, f n = n v / ( 4 L ) , where v is the
speed of sound, L is the distance between the open and closed ends of the instrument, and
n = 1, 3, 5, 7, . . .
When a wind instrument is brought inside from the cold outdoors, its frequency will
change because the temperature inside is higher than the temperature outside. If we can treat
air as an ideal gas, then according to Equation 16.5, the speed of sound will increase
because T increases. According to both Equations 17.4 and 17.5, the frequency produced by
a wind instrument is directly proportional to v, the speed of sound; therefore, since the speed
of sound increases with increasing temperature, the frequency of the sound produced by the
instrument will also increase. Note that the length L also typically increases with increasing
temperature, as discussed in Section 12.4. However, the change in length is relatively
negligible compared to the change in the speed of sound, as can be verified using data from
Table 12.1.
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15. REASONING AND SOLUTION Tones produced by a typical orchestra are complex
sound waves, and most have fundamental frequencies less than 5000 Hz. Since these tones
are complex sound waves, each one consists of a mixture of the fundamental and the higher
harmonic frequencies. The higher harmonics are integer multiples of the fundamental
frequency. For example, in a tone with a fundamental frequency of 5000 Hz, the fourth
harmonic will have a frequency of 20 000 Hz. Since most orchestra tones have fundamental
frequencies that are less than 5000 Hz, a high-quality stereo system must be able to
reproduce accurately all frequencies up to 20 000 Hz in order to include at least the fourth
harmonic of all tones.
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Chapter 17 Problems
877
CHAPTER 17 THE PRINCIPLE OF
LINEAR SUPERPOSITION AND
INTERFERENCE PHENOMENA
PROBLEMS
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1.
SSM REASONING AND SOLUTION According to the principle of linear superposition,
when two or more waves are present simultaneously at the same place, the resultant wave is
the sum of the individual waves. Therefore, the shape of the string at the indicated times
looks like the following:
t=1s
t=2s
t=3s
t=4s
0
2
4
6
8
10
12 cm
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878 THE PRINCIPLE OF LINEAR SUPERPOSITION AND INTERFERENCE PHENOMENA
2.
REASONING When the difference in path lengths traveled by the two sound waves is a
half-integer number 12 , 1 12 , 2 12 , … of wavelengths, the waves are out of phase and
(
)
destructive interference occurs at the listener. The smallest separation d between the
speakers is when the difference in path lengths is 12 of a wavelength, so d = 12 λ. The
wavelength is, according to Equation 16.1, is equal to the speed v of sound divided by the
frequency f ; λ = v/f .
SOLUTION Substituting λ = v/f into d = 12 λ gives
v
 343 m/s 
d = 12 λ = 12   = 12 
 = 0.700 m
 245 Hz 
 f 
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3.
REASONING AND SOLUTION In a time of t = 1 s, the pulse on the left has moved to the
right a distance of 1 cm, while the pulse on the right has moved to the left a distance of
1 cm. Adding the shapes of these two pulses when t = 1 s reveals that the height of the
resultant pulse is
a.
2 cm at x = 3 cm.
b. 1 cm at x = 4 cm.
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4.
REASONING For destructive interference to occur, the difference in travel distances for
the sound waves must be an integer number of half wavelengths. For larger and larger
distances between speaker B and the observer at C, the difference in travel distances
becomes smaller and smaller. Thus, the largest possible distance between speaker B and the
observer at C occurs when the difference in travel distances is just one half wavelength.
SOLUTION Since the triangle ABC in Figure 17.7 is a right triangle, we can apply the
2
.
( 5.00 m )2 + d BC
Pythagorean theorem to obtain the distance dAC as
Therefore, the
difference in travel distances is
2
− d BC =
( 5.00 m )2 + d BC
λ
2
=
v
2f
where we have used Equation 16.1 to express the wavelength λ as λ = v/f. Solving for the
distance dBC gives
Chapter 17 Problems
( 5.00 m )
2
2
+ d BC
= d BC
2
2
= d BC
+
( 5.00 m )2 + d BC
( 5.00 m )
2
v2
− 2
4f
v
+
2f
d BC v
f
or
+
v2
4f2
( 5.00 m ) −
2
( 5.00 m )
or
2
2
+ d BC

v 
=  d BC +

2f 

( 5.00 m )2 =
d BC v
f
+
879
2
v2
4f2
( 343 m/s )2
2
4 (125 Hz )
=
= 8.42 m
v
343 m/s
f
125 Hz
______________________________________________________________________________
d BC =
5.
SSM REASONING When constructive interference occurs again at point C, the path
length difference is two wavelengths, or ∆s = 2 λ = 3.20 m . Therefore, we can write the
expression for the path length difference as
2
2
sAC – sBC = sAB
+ sBC
– sBC = 3.20 m
This expression can be solved for sAB .
SOLUTION Solving for sAB , we find that
sAB = (3.20 m + 2.40 m) 2 – (2.40 m)2 = 5.06 m
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6.
REASONING The two speakers are vibrating exactly out of phase. This means that the
conditions for constructive and destructive interference are opposite of those that apply
when the speakers vibrate in phase, as they do in Example 1 in the text. Thus, for two wave
sources vibrating exactly out of phase, a difference in path lengths that is zero or an integer
number (1, 2, 3, …) of wavelengths leads to destructive interference; a difference in path
lengths that is a half-integer number
(
1
1
1
, 1 , 2 , ...
2
2
2
) of wavelengths leads to constructive
interference. First, we will determine the wavelength being produced by the speakers.
Then, we will determine the difference in path lengths between the speakers and the
observer and compare the differences to the wavelength in order to decide which type of
interference occurs.
SOLUTION According to Equation 16.1, the wavelength λ is related to the speed v and
frequency f of the sound as follows:
880 THE PRINCIPLE OF LINEAR SUPERPOSITION AND INTERFERENCE PHENOMENA
v 343 m/s
=
= 0.800 m
f
429 Hz
λ=
Since ABC in Figure 17.7 is a right triangle, the Pythagorean theorem applies and the
difference ∆d in the path lengths is given by
2
2
∆d = d AC − d BC = d AB
+ d BC
− d BC
We will now apply this expression for parts (a) and (b).
a. When d BC = 1.15 m , we have
2
2
∆d = d AB
+ d BC
− d BC =
( 2.50 m )2 + (1.15 m )2 − 1.15 m = 1.60 m
Since 1.60 m = 2 ( 0.800 m ) = 2λ , it follows that the interference is destructive (the
speakers vibrate out of phase).
b. When d BC = 2.00 m , we have
2
2
∆d = d AB
+ d BC
− d BC =
( 2.50 m )2 + ( 2.00 m )2 − 2.00 m = 1.20 m
( ) λ , it follows that the interference is
Since 1.20 m = 1.5 ( 0.800 m ) = 1
1
2
constructive (the
speakers vibrate out of phase).
7.
SSM WWW REASONING The geometry of the positions of the loudspeakers and the
listener is shown in the following drawing.
C
d 1 = 1.00 m
d2
y
60.0°
B
A
x1
x2
Chapter 17 Problems
881
The listener at C will hear either a loud sound or no sound, depending upon whether the
interference occurring at C is constructive or destructive. If the listener hears no sound,
destructive interference occurs, so
nλ
d 2 − d1 =
n = 1, 3, 5, …
(1)
2
SOLUTION Since v = λ f , according to Equation 16.1, the wavelength of the tone is
λ=
v 343 m/s
=
= 5.00 m
f 68.6 Hz
Speaker B will be closest to Speaker A when n = 1 in Equation (1) above, so
d2 =
nλ
5.00 m
+ d1 =
+ 1.00 m = 3.50 m
2
2
From the figure above we have that,
x1 = (1.00 m) cos 60.0° = 0.500 m
y = (1.00 m) sin 60.0° = 0.866 m
Then
x22 + y 2 = d 22 = (3.50 m) 2
or
x2 = (3.50 m) 2 − (0.866 m)2 = 3.39 m
Therefore, the closest that speaker A can be to speaker B so that the listener hears no sound
is x1 + x2 = 0.500 m + 3.39 m = 3.89 m .
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8.
REASONING The fact that a loud sound is heard implies constructive interference, which
occurs when the difference in path lengths is an integer number (1, 2, 3, …) of
wavelengths. This difference is 50.5 m − 26.0 m = 24.5 m. Therefore, constructive
interference occurs when 24.5 m = nλ, where n = 1, 2, 3, …. The wavelength is equal to the
speed v of sound divided by the frequency f; λ = v/f (Equation 16.1). Substituting this
relation for λ into 24.5 m = nλ, and solving for the frequency gives
f =
nv
24.5 m
This relation will allow us to find the two lowest frequencies that the listener perceives as
being loud due to constructive interference.
SOLUTION The lowest frequency occurs when n = 1:
882 THE PRINCIPLE OF LINEAR SUPERPOSITION AND INTERFERENCE PHENOMENA
f =
(1)( 343 m/s ) = 14 Hz
nv
=
24.5 m
24.5 m
This frequency lies below 20 Hz, so it cannot be heard by the listener. For n = 2 and n = 3,
the frequencies are 28 and 42 Hz , which are the two lowest frequencies that the listener
perceives as being loud.
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9.
REASONING AND SOLUTION Since v = λ f, the wavelength of the tone is
λ =
v
343 m/s
=
= 4.70 m
f
73.0 Hz
The figure below shows the line between the two speakers and the distances in question.
L
L-x
x
B
A
P
Constructive interference will occur when the difference in the distances traveled by the two
sound waves in reaching point P is an integer number of wavelengths. That is, when
(L – x) – x = nλ
where n is an integer (or zero). Solving for x gives
x=
L − nλ
2
(1)
When n = 0, x = L/2 = (7.80 m)/2 = 3.90 m . This corresponds to the point halfway
between the two speakers. Clearly in this case, each wave has traveled the same distance
and therefore, they will arrive in phase.
When n = 1,
x=
(7.80 m) − (4.70 m)
= 1.55 m
2
Thus, there is a point of constructive interference 1.55 m from speaker A . The points of
constructive interference will occur symmetrically about the center point at L/2, so there is
Chapter 17 Problems
883
also a point of constructive interference 1.55 m from speaker B, that is at the point
7.80 m – 1.55 m = 6.25 m from speaker A .
When n > 1, the values of x obtained from Equation (1) will be negative. These values
correspond to positions of constructive interference that lie to the left of A or to the right of
C. They do not lie on the line between the speakers.
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10. REASONING For a rectangular opening (“single slit”) such as a doorway, the diffraction
angle θ at which the first minimum in the sound intensity occurs is given by sin θ =
λ
D
(Equation 17.1), where λ is the wavelength of the sound and D is the width of the opening.
This relation can be used to find the angle provided we realize that the wavelength λ is
related to the speed v of sound and the frequency f by λ = v/f (Equation 16.1).
SOLUTION
a. Substituting λ = v/f into Equation 17.1 and using D = 0.700 m (only one door is open)
gives
λ
343 m/s
v
=
= 0.807
sin θ = =
θ = sin −1 ( 0.807 ) = 53.8°
D f D ( 607 Hz )( 0.700 m )
b. When both doors are open, D = 2 × 0.700 m and the diffraction angle is
λ
343 m/s
v
=
= 0.404
θ = sin −1 ( 0.404 ) = 23.8°
D f D ( 607 Hz )( 2 × 0.700 m )
______________________________________________________________________________
sin θ =
=
11. SSM REASONING The diffraction angle for the first minimum for a circular opening is
given by Equation 17.2: sin θ = 1.22 λ / D , where D is the diameter of the opening.
SOLUTION
a. Using Equation 16.1, we must first find the wavelength of the 2.0-kHz tone:
λ=
v
343 m/s
=
= 0.17 m
f 2.0 × 10 3 Hz
The diffraction angle for a 2.0-kHz tone is, therefore,


θ = sin –1  1.22 ×
b. The wavelength of a 6.0-kHz tone is
0.17 m 
 =
0.30 m 
44°
884 THE PRINCIPLE OF LINEAR SUPERPOSITION AND INTERFERENCE PHENOMENA
λ=
v
343 m/s
=
= 0.057 m
f 6.0 × 103 Hz
Therefore, if we wish to generate a 6.0-kHz tone whose diffraction angle is as wide as that
for the 2.0-kHz tone in part (a), we will need a speaker of diameter D, where
1.22 λ (1.22)(0.057 m)
=
= 0.10 m
sin θ
sin 44°
______________________________________________________________________________
D=
12. REASONING Equation 17.1 specifies the diffraction angle θ according to sin θ = λ / D ,
where λ is the wavelength of the sound and D is the width of the opening. The wavelength
depends on the speed and frequency of the sound. Since the frequency is the same in winter
and summer, only the speed changes with the temperature. We can account for the effect of
the temperature on the speed by assuming that the air behaves as an ideal gas, for which the
speed of sound is proportional to the square root of the Kelvin temperature.
SOLUTION Equation 17.1 indicates that
sin θ =
λ
D
Into this equation, we substitute λ = v / f (Equation 16.1), where v is the speed of sound
and f is the frequency:
λ v/ f
sin θ = =
D
D
Assuming that air behaves as an ideal gas, we can use v = γ kT / m (Equation 16.5), where
γ is the ratio of the specific heat capacities at constant pressure and constant volume, k is
Boltzmann’s constant, T is the Kelvin temperature, and m is the average mass of the
molecules and atoms of which the air is composed:
sin θ =
v
1 γ kT
=
fD fD m
Applying this result for each temperature gives
sin θsummer =
1 γ kTsummer
fD
m
and
sin θ winter =
Dividing the summer-equation by the winter-equation, we find
1 γ kTwinter
fD
m
Chapter 17 Problems
sin θsummer
sin θ winter
885
1 γ kTsummer
T
fD
m
=
= summer
Twinter
1 γ kTwinter
fD
m
Thus, it follows that
sin θsummer = sin θ winter
Tsummer
Twinter
= sin 15.0°
311 K
= 0.276
273 K
or θsummer = sin −1 ( 0.276 ) = 16.0°
13. REASONING In both cases, Equation 17.1 (sin θ = λ/D) offers a direct solution, since the
width D of the doorway is given and we can use Equation 16.1 (λ = v/f ) to determine the
wavelength λ in terms of the speed v and frequency f.
SOLUTION Using Equations 17.1 and 16.1, we find
sin θ =
λ
D
=
v
Df
Using this result we have


 v 
343 m/s
−1 
 = 5.1°
=
sin

 ( 0.77 m ) 5.0 ×103 Hz 
Df 


θ = sin −1 
a.
(
)


343 m/s
v 
−1 
 = 63°
b.
θ = sin 
 = sin 
2
D
f


( 0.77 m ) 5.0 ×10 Hz 

______________________________________________________________________________
−1 
14. REASONING AND SOLUTION
The figure at the right shows the
geometry of the situation.
The tone will not be heard at seats located 8.7 m
at the first diffraction minimum. This
occurs when
sin θ =
λ
D
=
v
fD
That is, the angle θ is given by
(
)
Diffraction horn
Stage
θ
θ
First row
of seats
C
x
886 THE PRINCIPLE OF LINEAR SUPERPOSITION AND INTERFERENCE PHENOMENA

 v 
343 m/s
−1 
 = 27.2°
 = sin 
4
(1.0
10
Hz)(0.075
m)
×
 f D


θ = sin −1 
From the figure at the right, we see that
tan 27.2° =
x
8.7 m
⇒
x = (8.7 m)(tan 27.2°) = 4.47 m
8.7 m
θ
x
Thus, seats at which the tone cannot be heard are a distance x on either side of the center
seat C. Thus, the distance between the two seats is
2x = 2(4.47 m) = 8.9 m
______________________________________________________________________________
15. REASONING AND SOLUTION At 0 °C the speed of sound is air is given as 331 m/s in
Table 16.1 in the text. This corresponds to a wavelength of
λ1 = v/f = (331 m/s)/(3.00 × 103 Hz) = 0.1103 m
The diffraction angle is given by Equation 17.2 as
 1.22 λ 
–1 1.22 ( 0.1103 m ) 
 = 50.3°
 = sin 
0.175 m
 D 


θ1 = sin –1 
For an ideal gas, the speed of sound is proportional to the square root of the Kelvin
temperature, according to Equation 16.5. Therefore, the speed of sound at 29 °C is
b
v = 331 m / s
g
302 K
= 348 m / s
273 K
The wavelength at this temperature is λ2 = (348 m/s)/(3.00 × 103 Hz) = 0.116 m. This gives
a diffraction angle of θ2 = 54.0°. The change in the diffraction angle is thus
∆θ = 54.0° − 50.3° = 3.7°
______________________________________________________________________________
16. REASONING When two frequencies are sounded simultaneously, the beat frequency
produced is the difference between the two. Thus, knowing the beat frequency between the
tuning fork and one flute tone tells us only the difference between the known frequency and
the tuning-fork frequency. It does not tell us whether the tuning-fork frequency is greater or
smaller than the known frequency. However, two different beat frequencies and two flute
Chapter 17 Problems
887
frequencies are given. Consideration of both beat frequencies will enable us to find the
tuning-fork frequency.
SOLUTION The fact that a 1-Hz beat frequency is heard when the tuning fork is sounded
along with the 262-Hz tone implies that the tuning-fork frequency is either 263 Hz or
261 Hz. We can eliminate one of these values by considering the fact that a 3-Hz beat
frequency is heard when the tuning fork is sounded along with the 266-Hz tone. This
implies that the tuning-fork frequency is either 269 Hz or 263 Hz. Thus, the tuning-fork
frequency must be 263 Hz .
______________________________________________________________________________
17. SSM REASONING The beat frequency of two sound waves is the difference between
the two sound frequencies. From the graphs, we see that the period of the wave in the upper
figure is 0.020 s, so its frequency is f1 = 1/ T1 = 1/(0.020 s)=5.0 × 101 Hz . The frequency of
the wave in the lower figure is f 2 = 1/(0.024 s)=4.2 ×101 Hz .
SOLUTION The beat frequency of the two sound waves is
f beat = f1 − f 2 = 5.0 × 101 Hz – 4.2 × 101 Hz = 8 Hz
______________________________________________________________________________
18. REASONING The time between successive beats is the period of the beat frequency. The
period that corresponds to any frequency is the reciprocal of that frequency. The beat
frequency itself is the magnitude of the difference between the two sound frequencies
produced by the pianos. Each of the piano frequencies can be determined as the speed of
sound divided by the wavelength.
SOLUTION According to Equation 10.5, the period T (the time between successive beats)
that corresponds to the beat frequency fbeat is T = 1/ f beat . The beat frequency is the
magnitude of the difference between the two sound frequencies fA and fB, so that
f beat = f A − f B . With this substitution, the expression for the period becomes
1
T=
f beat
=
1
fA − fB
(1)
The frequencies fA and fB are given by Equation 16.1 as
fA =
v
λA
and
fB =
Substituting these expressions into Equation (1) gives
v
λB
888 THE PRINCIPLE OF LINEAR SUPERPOSITION AND INTERFERENCE PHENOMENA
T=
1
1
1
=
=
= 0.25 s
343 m/s 343 m/s
fA − fB
v
v
−
−
0.769 m 0.776 m
λ
λ
A
B
19. SSM REASONING AND SOLUTION The first case requires that the frequency be either
440 Hz − 5 Hz = 435 Hz
or
440 Hz + 5 Hz = 445 Hz
The second case requires that the frequency be either
436 Hz − 9 Hz = 427 Hz
or
436 Hz + 9 Hz = 445 Hz
The frequency of the tuning fork is 445 Hz .
______________________________________________________________________________
20. REASONING The beat frequency is the difference between two sound frequencies.
Therefore, the original frequency of the guitar string (before it was tightened) was either
3 Hz lower than that of the tuning fork (440.0 Hz − 3 Hz = 337 Hz) or 3 Hz higher (440.0
Hz + 3 Hz = 443 Hz):
443 Hz
440.0 Hz
437 Hz
} 3-Hz beat frequency
} 3-Hz beat frequency
To determine which of these frequencies is the correct one (437 or 443 Hz), we will use the
information that the beat frequency decreases when the guitar string is tightened
SOLUTION When the guitar string is tightened, its frequency of vibration (either 437 or
443 Hz) increases. As the drawing below shows, when the 437-Hz frequency increases, it
becomes closer to 440.0 Hz, so the beat frequency decreases. When the 443-Hz frequency
increases, it becomes farther from 440.0 Hz, so the beat frequency increases. Since the
problem states that the beat frequency decreases, the original frequency of the guitar string
was 437 Hz .
443 Hz
440.0 Hz
}
Beat frequency increases
} Beat frequency decreases
437 Hz
Tuning
fork
Original
string
Tightened
string
______________________________________________________________________________
Chapter 17 Problems
889
21. REASONING The beat frequency is the difference between the frequency of the sound
wave traveling in seawater and that generated by the 440.0-Hz tuning fork. We can use
Equation 16.1, v = f λ, to find the frequency of the sound wave, provided we can determine
the speed v of sound in seawater. This can be accomplished by using Equation 16.6.
SOLUTION
The speed of sound in a liquid is given by Equation 16.6 as v = Bad / ρ ,
where Bad is the adiabatic bulk modulus and ρ is the density of the liquid. The frequency of
the sound traveling in sea water is
Bad
f =
v
λ
=
ρ
=
λ
2.31 × 109 Pa
1025 kg/m3
= 448 Hz
3.35 m
The beat frequency is
448 Hz − 440.0 Hz = 8 Hz
______________________________________________________________________________
22. REASONING AND SOLUTION
a. We know that fx − fy = 8 Hz. Also, fx is either 395 Hz or 389 Hz and fy is either 397 Hz
or 387 Hz. The fact that fx > fy requires that
f y = 387 Hz
and, therefore,
f x = 395 Hz
b. Now we know that fy − fx = 8 Hz and fx is either 395 Hz or 389 Hz and fy is either
397 Hz or 387 Hz. The fact that fx < fy requires that
f y = 397 Hz
and, therefore,
f x = 389 Hz
______________________________________________________________________________
23. SSM REASONING The fundamental frequency f1 is given by Equation 17.3 with n = 1 :
f1 = v /(2 L) . Since values for f1 and L are given in the problem statement, we can use this
expression to find the speed of the waves on the cello string. Once the speed is known, the
tension F in the cello string can be found by using Equation 16.2, v = F /(m / L) .
SOLUTION Combining Equations 17.3 and 16.2 yields
2 L f1 =
F
m/ L
890 THE PRINCIPLE OF LINEAR SUPERPOSITION AND INTERFERENCE PHENOMENA
Solving for F, we find that the tension in the cello string is
F = 4 L2 f12 (m / L) = 4(0.800 m) 2 (65.4 Hz)2 (1.56 ×10 –2 kg/m) = 171 N
______________________________________________________________________________
24. REASONING The frequencies fn of the standing waves on a string fixed at both ends are
 v 
given by Equation 17.3 as f n = n 
 , where n is an integer that specifies the harmonic
 2L 
number, v is the speed of the traveling waves that make up the standing waves, and L is the
length of the string. For the second harmonic, n = 2.
SOLUTION The frequency f2 of the second harmonic is
 140 m/s 
 v 
2
f2 = n 
=
2

 = 5.0 × 10 Hz

2
0.28
m
(
)
 2L 


______________________________________________________________________________
25. REASONING The time it takes for a wave to travel the length L of the string is t = L/v,
where v is the speed of the wave. The speed can be obtained since the fundamental
frequency is known, and Equation 17.3 (with n = 1) gives the fundamental frequency as
f1 = v/(2L). The length is not needed, since it can be eliminated algebraically between this
expression and the expression for the time.
SOLUTION Solving Equation 17.3 for the speed gives v = 2Lf1. With this result for the
speed, the time for a wave to travel the length of the string is
L
L
1
1
=
=
=
= 1.95 ×10−3 s
v 2 L f1 2 f1 2 ( 256 Hz )
______________________________________________________________________________
t=
26. REASONING Equation 17.3 (with n = 1) gives the fundamental frequency as f1 = v/(2L),
where L is the wire’s length and v is the wave speed on the wire. The speed is given by
F
Equation 16.2 as v =
, where F is the tension and m is the mass of the wire.
m/ L
SOLUTION Using Equations 17.3 and 16.2, we obtain
v
=
f1 =
2L
F
m/ L = 1 F = 1
2L
2 mL 2
(
160 N
)
6.0 × 10−3 kg ( 0.41 m )
= 130 Hz
______________________________________________________________________________
Chapter 17 Problems
891
27. SSM REASONING According to Equation 17.3, the fundamental (n = 1) frequency of a
string fixed at both ends is related to the wave speed v by f1 = v / 2 L , where L is the length
of the string. Thus, the speed of the wave is v = 2Lf1. Combining this with Equation 16.2,
v = F /(m / L) , we have, after some rearranging,
F
= 4 f12 (m / L)
2
L
Since the strings have the same tension and the same lengths between their fixed ends, we
have
2
2
f1E
(m / L) E = f1G
( m / L)G
where the symbols “E” and “G” represent the E and G strings on the violin. This equation
can be solved for the linear density of the G string.
SOLUTION The linear density of the string is
( m / L) G =
2
f1E
2
f1G
 f
(m / L) E =  1E
 f
 1G
2
(
2

 (m / L) E


)
 659.3 Hz 
–4
–3
=
 3.47 × 10 kg/m = 3.93 × 10 kg/m
 196.0 Hz 
______________________________________________________________________________
28. REASONING The series of natural frequencies for a wire fixed at both ends is given by
f n = nv / ( 2 L ) (Equation 17.3), where the harmonic number n takes on the integer values 1,
2, 3, etc.. This equation can be solved for n. The natural frequency fn is the lowest
frequency that the human ear can detect, and the length L of the wire is given. The speed v
at which waves travel on the wire can be obtained from the given values for the tension and
the wire’s linear density.
SOLUTION According to Equation 17.3, the harmonic number n is
f 2L
(1)
n= n
v
F
(Equation 16.2), where F is the tension and m/L is the linear
The speed v is v =
m/ L
density. Substituting this expression into Equation (1) gives
n=
fn 2L
v
=
fn 2L
F
m/ L
= fn 2L
m/ L
0.0140 kg/m
= ( 20.0 Hz ) 2 ( 7.60 m )
= 2
F
323 N
892 THE PRINCIPLE OF LINEAR SUPERPOSITION AND INTERFERENCE PHENOMENA
29. REASONING A standing wave is composed of two oppositely traveling waves. The speed
F
v of these waves is given by v =
(Equation 16.2), where F is the tension in the string
m/ L
and m/L is its linear density (mass per unit length). Both F and m/L are given in the
statement of the problem. The wavelength λ of the waves can be obtained by visually
inspecting the standing wave pattern. The frequency of the waves is related to the speed of
the waves and their wavelength by f = v/λ (Equation 16.1).
SOLUTION
a. The speed of the waves is
v=
F
280 N
=
= 180 m/s
m/ L
8.5 × 10−3 kg/m
b. Two loops of any standing wave comprise one
wavelength. Since the string is 1.8 m long and consists
of three loops (see the drawing), the wavelength is
λ=
2
3
(1.8 m ) = 1.2 m
1.8 m
λ
c. The frequency of the waves is
180 m/s
= 150 Hz
λ 1.2 m
______________________________________________________________________________
f =
v
=
30. REASONING The frequencies fn of the standing waves allowed on a string fixed at both
 v 
ends are given by Equation 17.3 as f n = n 
 , where n is an integer that specifies the
 2L 
harmonic number, v is the speed of the traveling waves that make up the standing waves,
and L is the length of the string. The speed v is related to the tension F in the string and the
F
linear density m/L via v =
(Equation 16.2). Therefore, the frequencies of the
m/ L
standing waves can be written as

F

 v 
m/ L
fn = n 
 = n
 2L
 2L 

 n
F
=
 2L m / L
Chapter 17 Problems
893
The tension F in each string is provided by the weight W (either WA or WB) that hangs from
the right end, so F = W. Thus, the expression for fn becomes f n =
n
W
. We will use
2L m / L
this relation to find the weight WB.
A
WA
B
WB
SOLUTION String A has one loop so n = 1, and the frequency f1A of this standing wave is
1 WA
. String B has two loops so n = 2, and the frequency f 2B of this standing
2L m / L
WB
2
wave is f 2B =
. We are given that the two frequencies are equal, so
2L m / L
f1A =
WB
1 WA
2
=
2L m / L 2L m / L
f1A
f 2B
Solving this expression for WB gives
WB = 14 WA = 14 ( 44 N ) = 11 N
______________________________________________________________________________
31. SSM REASONING We can find the extra length that the D-tuner adds to the E-string by
calculating the length of the D-string and then subtracting from it the length of the E string.
For standing waves on a string that is fixed at both ends, Equation 17.3 gives the
frequencies as f n = n(v / 2 L) . The ratio of the fundamental frequency of the D-string to that
of
the
E-string is
f D v /(2 LD ) LE
=
=
f E v /(2 LE ) LD
This expression can be solved for the length LD of the D-string in terms of quantities given
in the problem statement.
894 THE PRINCIPLE OF LINEAR SUPERPOSITION AND INTERFERENCE PHENOMENA
SOLUTION The length of the D-string is
 f 
 41.2 Hz 
LD = LE  E  = (0.628 m) 
 = 0.705 m
 36.7 Hz 
 fD 
The length of the E-string is extended by the D-tuner by an amount
LD − LE = 0.705 m − 0.628 m = 0.077 m
______________________________________________________________________________
32. REASONING AND SOLUTION We are given f j = 12 2 f j –1 .
a. The length of the unfretted string is L0 = v/(2f0) and the length of the string when it is
pushed against fret 1 is L1 = v/(2f1). The distance between the frets is
L0 – L1 =
 v 
f   v 
v
v
1 
–
=
1– 0  = 

  1 – 12 






2 f 0 2 f1  2 f 0  
f1   2 f 0  
2
1 

= L0 1 – 12  = ( 0.628 m )( 0.0561) = 0.0352 m
2

b. The frequencies corresponding to the sixth and seventh frets are f 6 =
f7 =
(12 2 )
7
(12 2 )
6
f 0 and
f 0 . The distance between fret 6 and fret 7 is
L6 – L7 =
v
v
–
2 f6 2 f7
v
=
2
(12 2 )

= L0 

6
v
–
f0
2
(12 2 )
7
 v 
=

f0  2 f0  

1
–
1
(12 2 ) (12 2 )
6
7




 = ( 0.628 m )( 0.0397 ) = 0.0249 m
12
12
2
2 

______________________________________________________________________________
33. SSM WWW
1
–
1
( ) ( )
6
REASONING
7
The beat frequency produced when the piano and the
other instrument sound the note (three octaves higher than middle C) is f beat = f − f 0 ,
where f is the frequency of the piano and f 0 is the frequency of the other instrument
Chapter 17 Problems
895
( f 0 = 2093 Hz ). We can find f by considering the temperature effects and the mechanical
effects that occur when the temperature drops from 25.0 °C to 20.0 °C.
SOLUTION The fundamental frequency f0 of the wire at 25.0 °C is related to the tension
F0 in the wire by
f0 =
v
=
2 L0
F0 /(m / L)
(1)
2 L0
where Equations 17.3 and 16.2 have been combined.
The amount ∆L by which the piano wire attempts to contract is (see Equation 12.2)
∆L = α L0 ∆T , where α is the coefficient of linear expansion of the wire, L0 is its length at
25.0 °C, and ∆T is the amount by which the temperature drops. Since the wire is prevented
from contracting, there must be a stretching force exerted at each end of the wire.
According to Equation 10.17, the magnitude of this force is
 ∆L 
∆F = Y 
A
 L 
 0
where Y is the Young's modulus of the wire, and A is its cross-sectional area. Combining
this relation with Equation 12.2, we have
 α L ∆T
∆F = Y  0
 L
0


 A = α ( ∆T )Y A

Thus, the frequency f at the lower temperature is
f =
( F0 + ∆F ) /(m / L)
v
=
=
2 L0
2 L0
 F0 + α ( ∆T )Y A /(m / L)
2 L0
(2)
Using Equations (1) and (2), we find that the frequency f is
f = f0
f = ( 2093 Hz )
 F0 + α ( ∆T )Y A /(m / L)
F + α ( ∆T )Y A
= f0 0
F0
F0 /(m / L)
818.0 N + (12 ×10 –6 /C°)(5.0 C°)(2.0 × 1011 N/m 2 ) (7.85 × 10 –7 m 2 )
818.0 N
= 2105 Hz
Therefore, the beat frequency is 2105 Hz − 2093 Hz = 12 Hz .
______________________________________________________________________________
896 THE PRINCIPLE OF LINEAR SUPERPOSITION AND INTERFERENCE PHENOMENA
34. REASONING The frequency of a pipe open at both ends is given by Equation 17.4 as
 v 
fn = n 
 , where n is an integer specifying the harmonic number, v is the speed of
 2L 
sound, and L is the length of the pipe. This relation can be used to find L, since all the other
variables are known.
SOLUTION Solving the equation above for L, and recognizing that n = 3 for the third
harmonic, we have
 343 m/s 
 v 
L = n 
 = 3 
 = 1.96 m
f
2
2
262
Hz
(
)


 n

______________________________________________________________________________
35. SSM REASONING AND SOLUTION
The distance between one node and an adjacent
antinode is λ/4. Thus, we must first determine the wavelength of the standing wave. A tube
open at only one end can develop standing waves only at the odd harmonic frequencies.
Thus, for a tube of length L producing sound at the third harmonic (n = 3), L = 3(λ/4).
Therefore, the wavelength of the standing wave is
λ = 43 L =
4
3
b1.5 mg = 2.0 m
and the distance between one node and the adjacent antinode is λ/4 = 0.50 m .
____________________________________________________________________________________________
36. REASONING AND SOLUTION We know that L = v/(2f ). For 20.0 Hz
L = (343 m/s)/[2(20.0 Hz)] = 8.6 m
For 20.0 kHz
L = (343 m/s)/[2(20.0 × 103 Hz)] = 8.6 ×10−3 m
______________________________________________________________________________
37. REASONING The fundamental frequency f1A of air column A, which is open at both
 v 
ends, is given by Equation 17.4 with n = 1: f1A = (1) 
 , where v is the speed of sound
 2 LA 
in air and LA is the length of the air column. Similarly, the fundamental frequency f1B of air
column B, which is open at only one end, can be expressed using Equation 17.5 with n = 1:
 v 
f1B = (1) 
 . These two relations will allow us to determine the length of air column B.
4
L
 B
Chapter 17 Problems
897
SOLUTION Since the fundamental frequencies of the two air columns are the same
f1A = f1B , so that
 v
 2 LA
(1) 
f1A

 v 
 = (1) 


 4 LB 
or
LB = 12 LA =
1
2
( 0.70 m ) =
0.35 m
f1B
______________________________________________________________________________
38. REASONING For a tube open at only one end, the series of natural frequencies is given by
f n = nv / ( 4 L ) (Equation 17.5), where n has the values 1, 3, 5, etc., v is the speed of sound,
and L is the tube length. We will apply this expression to both the air-filled and the heliumfilled tube in order to determine the desired ratio.
SOLUTION According to Equation 17.5, we have
f n, air =
nvair
4L
and
f n, helium =
nvhelium
4L
Dividing the expression for helium by the expression for air, we find that
nvhelium
f n, helium
vhelium 1.00 ×103 m/s
L
4
=
=
=
= 2.92
nvair
f n, air
vair
343 m/s
4L
39. SSM REASONING The natural frequencies of a tube open at only one end are given by
 v 
Equation 17.5 as f n = n 
 , where n is any odd integer (n = 1, 3, 5, …), v is the speed of
 4L 
sound, and L is the length of the tube. We can use this relation to find the value for n for the
450-Hz sound and to determine the length of the pipe.
SOLUTION
 v 
a. The frequency fn of the 450-Hz sound is given by 450 Hz = n 
 . Likewise, the
 4L 
 v 
frequency of the next higher harmonic is 750 Hz = ( n + 2 ) 
 , because n is an odd
 4L 
integer and this means that the value of n for the next higher harmonic must be n + 2.
Taking the ratio of these two relations gives
 v 
n + 2) 
(

750 Hz
 4L  = n + 2
=
450 Hz
n
 v 
n

 4L 
898 THE PRINCIPLE OF LINEAR SUPERPOSITION AND INTERFERENCE PHENOMENA
Solving this equation for n gives n = 3 .
 v 
b. Solving the equation 450 Hz = n 
 for L and using n = 3, we find that the length of
 4L 
the tube is
 343 m/s 
 v 
L = n 
 = 3 
 = 0.57 m
f
4
4
450
Hz
(
)


 n

______________________________________________________________________________
40. REASONING AND SOLUTION The distance between the nodes of the standing wave is
L = λ /2 = v/(2f ). The man travels this distance in a time
t = 1/(3.0 Hz) = 0.33 s
His speed is then
vlistener = L/t = v/(2f t ) = (343 m/s)/[2(0.33 s)(440 Hz)] = 1.2 m/s
______________________________________________________________________________
41. REASONING The well is open at the top and closed at the bottom, so it can be
approximated as a column of air that is open at only one end. According to Equation 17.5,
the natural frequencies for such an air column are
 v 
fn = n 

 4L 
where n = 1, 3, 5, …
The depth L of the well can be calculated from the speed of sound, v = 343 m/s, and a
knowledge of the natural frequencies fn.
SOLUTION We know that two of the natural frequencies are 42 and 70.0 Hz. The ratio of
these two frequencies is
70.0 Hz 5
=
42 Hz
3
Therefore, the value of n for each frequency is n = 3 for the 42-Hz sound, and
n = 5 for the 70.0-Hz sound. Using n = 3, for example, the depth of the well is
nv 3 ( 343 m/s )
=
= 6.1 m
4 f3
4( 42 Hz )
______________________________________________________________________________
L=
42. REASONING We will make use of the series of natural frequencies (including the first
overtone frequency) given by f n = nv / ( 2 L ) (Equation 17.4), for a tube open at both ends.
In this expression, n takes on the integer values 1, 2, 3, etc. and has the value of n = 2 for the
first overtone frequency that is given. We can solve Equation 17.4 for L, but must deal with
the fact that no value is given for the speed v of sound in gas B. To obtain the necessary
Chapter 17 Problems
899
value, we will use the fact that both gases are ideal gases and utilize the speed given for gas
A and the masses of the two types of molecules.
SOLUTION According to Equation 17.4 as applied to gas B, we have f n = nvB / ( 2 L ) ,
which can be solved for L to show that
nv
L= B
(1)
2 fn
Since both gases are ideal gases, the speed is given by v = γ kT / m (Equation 16.5), where
γ is the ratio of the specific heat capacities at constant pressure and constant volume, k is
Boltzmann’s constant, T is the Kelvin temperature, and m is the mass of a molecule of the
gas. Noting that γ and T are the same for each gas, we can apply this expression to both
gases:
γ kT
γ kT
and vB =
vA =
mA
mB
Dividing the expression for gas B by that for gas A gives
γ kT
vB
vA
=
mB
γ kT
=
mA
mB
or
vB = vA
mA
mB
mA
Substituting this result for vB into Equation (1), we find that
L=
nvB nvA
=
2 fn 2 fn
mA 2 ( 259 m/s ) 7.31×10−26 kg
=
= 0.557 m
mB
2 ( 386 Hz ) 1.06 × 10−25 kg
43. SSM WWW REASONING
According to Equation 11.4, the absolute pressure at the
bottom of the mercury is P = Patm + ρ gh , where the height h of the mercury column is the
original length L0 of the air column minus the shortened length L. Hence,
P = Patm + ρ g ( L0 − L)
SOLUTION From Equation 17.5, the fundamental (n = 1) frequency f1 of the shortened
tube is f1 = 1(v/4L), where L is the length of the air column in the tube. Likewise, the
frequency f3 of the third (n = 3) harmonic in the original tube is f3 = 3(v/4L0), where L0 is
the length of the air column in the original tube. Since f1 = f3, we have that
900 THE PRINCIPLE OF LINEAR SUPERPOSITION AND INTERFERENCE PHENOMENA
 v 
 v 
1

 = 3 
 4L 
 4 L0 
or
L =
1L
3 0
The pressure at the bottom of the mercury is
P = Patm + ρ g
( 23 L0 )
= 1.01× 105 Pa+(13 600 kg/m3 )(9.80 m/s2 )
( 23 × 0.75 m ) =
1.68 × 105 Pa
______________________________________________________________________________
44. REASONING AND SOLUTION The original tube has a fundamental given by f = v/(4L),
so that its length is L = v/(4f ). The cut tube that has one end closed has a length of
Lc = v/(4fc), while the cut tube that has both ends open has a length Lo = v/(2fo).
We know that L = Lc + Lo. Substituting the expressions for the lengths and solving for f
gives
fo fc
( 425 Hz )( 675 Hz ) = 162 Hz
=
f =
2 f c + f o 2 ( 675 Hz ) + 425 Hz
______________________________________________________________________________
45. REASONING AND SOLUTION Since the wavelength is twice the distance between two
successive nodes, we can use Equation 16.1 and see that
v = λ f = (2L)f = 2(0.30 m)(4.0 Hz) = 2.4 m/s
______________________________________________________________________________
46. REASONING Equation 17.5 (with n = 1) gives the fundamental frequency as f1 = v/(4L),
where L is the length of the auditory canal and v is the speed of sound.
SOLUTION Using Equation 17.5, we obtain
v
343 m/s
=
= 3.0 ×103 Hz
4 L 4 ( 0.029 m )
______________________________________________________________________________
f1 =
47. SSM REASONING The tones from the two speakers will produce destructive
interference with the smallest frequency when the path length difference at C is one-half of
a wavelength. From Figure 17.7, we see that the path length difference is ∆s = sAC – sBC .
From Example 1, we know that sAC = 4.00 m , and from Figure 17.7, sBC = 2.40 m .
Therefore, the path length difference is ∆s = 4.00 m – 2.40 m = 1.60 m .
Chapter 17 Problems
901
SOLUTION Thus, destructive interference will occur when
λ
2
= 1.60 m
or
λ = 3.20 m
This corresponds to a frequency of
v
343 m / s
= 107 Hz
λ
3.20 m
______________________________________________________________________________
f =
=
48. REASONING AND SOLUTION Two ultrasonic sound waves combine and form a beat
frequency that is in the range of human hearing. We know that the frequency range of
human hearing is from 20 Hz to 20 kHz. The frequency of one of the ultrasonic waves is
70 kHz. The beat frequency is the difference between the two sound frequencies. The
smallest possible value for the ultrasonic frequency can be found by subtracting the upper
limit of human hearing from the value of 70 kHz. The largest possible value for the
ultrasonic frequency can be determined by adding the upper limit of human hearing to the
value of 70 kHz.
a. The smallest possible frequency of the other ultrasonic wave is
f = 70 kHz – 20 kHz = 50 kHz
which results in a beat frequency of 70 kHz – 50 kHz = 20 kHz .
b. The largest possible frequency for the other wave is
f = 70 kHz + 20 kHz = 90 kHz
which results in a beat frequency of 90 kHz – 70 kHz = 20 kHz .
______________________________________________________________________________
49. REASONING AND SOLUTION
a. For a string fixed at both ends the fundamental frequency is f1 = v/(2L) so fn = nf1.
f 2 = 800 Hz,
f3 = 1200 Hz,
f 4 = 1600 Hz
b. For a pipe with both ends open the fundamental frequency is f1 = v/(2L) so fn = nf1.
f 2 = 800 Hz,
f3 = 1200 Hz,
f 4 = 1600 Hz
c. For a pipe open at one end only the fundamental frequency is f1 = v/(4L) so fn = nf1 with
n odd.
f3 = 1200 Hz, f5 = 2000 Hz, f 7 = 2800 Hz
______________________________________________________________________________
902 THE PRINCIPLE OF LINEAR SUPERPOSITION AND INTERFERENCE PHENOMENA
50. REASONING AND SOLUTION The shape of the string looks like
t=1s
t=2s
t=3s
t=4s
0
2
4
6
8
10
______________________________________________________________________________
51. SSM REASONING For standing waves on a string that is clamped at both ends,
Equations 17.3 and 16.2 indicate that the standing wave frequencies are
 v 
fn = n 

 2L 
where
v=
F
m/ L
Combining these two expressions, we have, with n = 1 for the fundamental frequency,
Chapter 17 Problems
f1 =
903
1
F
2L m / L
This expression can be used to find the ratio of the two fundamental frequencies.
SOLUTION The ratio of the two fundamental frequencies is
f old
f new
1 Fold
= 2L m / L =
1 Fnew
2L m / L
Fold
Fnew
Since Fnew = 4 Fold , we have
f new = f old
Fnew
Fold
= f old
4 Fold
Fold
= f old 4 = (55.0 Hz) (2) =
1.10 × 102 Hz
______________________________________________________________________________
52. REASONING Each string has a node at each end, so the frequency of vibration is given by
Equation 17.3 as fn = nv/(2L), where n = 1, 2, 3, … The speed v of the wave can be
determined from Equation 16.2 as v = F / ( m / L ) . We will use these two relations to find
the lowest frequency that permits standing waves in both strings with a node at the junction.
SOLUTION
Since the frequency of the left string is equal to the frequency of the right string, we can
write
F
F
nright
nleft
( m / L )right
( m / L )left
=
2Lleft
2L right
Substituting in the data given in the problem yields
nleft
190.0 N
190.0 N
nright
−2
6.00 × 10 kg/m
1.50 × 10−2 kg/m
=
2 ( 3.75 m )
2 (1.25 m )
This expression gives nleft = 6nright. Letting nleft = 6 and nright = 1, the frequency of the left
string (which is also equal to the frequency of the right string) is
(6)
f6 =
190.0 N
6.00 × 10−2 kg/m
=
2 ( 3.75 m )
45.0 Hz
904 THE PRINCIPLE OF LINEAR SUPERPOSITION AND INTERFERENCE PHENOMENA
______________________________________________________________________________
53. REASONING When the difference
1− 2
in path lengths traveled by the two sound
( 12 , 1 12 , 2 12 , …)
waves is a half-integer number
of wavelengths, destructive interference
occurs at the listener. When the difference in path lengths is zero or an integer number
(1, 2, 3, …) of wavelengths, constructive interference occurs. Therefore, we will divide the
distance 1 − 2 by the wavelength of the sound to determine if constructive or destructive
interference occurs. The wavelength is, according to Equation 16.1, the speed v of sound
divided by the frequency f ; λ = v/f .
SOLUTION
a. The distances 1 and 2 can be
determined by applying the Pythagorean
theorem to the two right triangles in the
drawing:
1
=
( 2.200 m )2 + (1.813 m )2
= 2.851 m
2
=
( 2.200 m )2 + (1.187 m )2
= 2.500 m
Therefore,
1− 2=
0.351 m.
1.813 m
1.187 m
2.200 m
2
1
P
The
v 343 m/s
=
= 0.234 m . Dividing the distance
f 1466 Hz
the wavelength λ gives the number of wavelengths in this distance:
wavelength of the sound is λ =
Number of wavelengths =
1− 2
λ
=
1− 2
by
0.351 m
= 1.5
0.233 m
( )
Since the number of wavelengths is a half-integer number 1 12 , destructive interference
occurs at the listener.
v 343 m/s
=
= 0.351 m . Dividing the distance
f
977 Hz
by the wavelength λ gives the number of wavelengths in that distance:
b. The wavelength of the sound is now λ =
1− 2
Number of wavelengths =
1− 2
λ
=
0.351 m
=1
0.351 m
Since the number of wavelengths is an integer number (1) , constructive interference
occurs at the listener.
______________________________________________________________________________
Chapter 17 Problems
905
54. REASONING AND SOLUTION The speed of the speakers is
vs = 2π r/t = 2π (9.01 m)/(20.0 s) = 2.83 m/s
The sound that an observer hears coming from the right speaker is Doppler shifted to a new
frequency given by Equation 16.11 as
f OR =
fs
1–vs /v
=
100.0 Hz
= 100.83 Hz
1– ( 2.83 m/s ) / ( 343.00 m/s ) 
The sound that an observer hears coming from the left speaker is shifted to a new frequency
given by Equation 16.12 as
f OL =
fs
1 + vs /v
=
100.0 Hz
= 99.18 Hz
1 + ( 2.83 m/s ) / ( 343.00 m/s ) 
The beat frequency heard by the observer is then
100.83 Hz − 99.18 Hz = 1.7 Hz
______________________________________________________________________________
55. SSM WWW REASONING The natural frequencies of the cord are, according to
Equation 17.3,
f n = nv / ( 2 L ) , where n = 1, 2, 3, .... The speed v of the waves on the cord
is, according to Equation 16.2, v = F / ( m / L ) , where F is the tension in the cord.
Combining these two expressions, we have
2
nv
n
F
fn =
=
2L 2L m / L
 fn 2L 
F

 =
m/ L
 n 
or
Applying Newton's second law of motion, ΣF = ma, to the forces that act on the block and
are parallel to the incline gives
F – Mg sin θ = Ma = 0
or
F = Mg sin θ
where Mg sin θ is the component of the block's weight that is parallel to the incline.
Substituting this value for the tension into the equation above gives
2
 fn 2L 
Mg sin θ

 =
m/ L
 n 
This expression can be solved for the angle θ and evaluated at the various harmonics. The
answer can be chosen from the resulting choices.
906 THE PRINCIPLE OF LINEAR SUPERPOSITION AND INTERFERENCE PHENOMENA
SOLUTION Solving this result for sin θ shows that
(m / L) 
sin θ =
2
fn 2L 
1.20 × 10 –2 kg/m  (165 Hz ) 2 ( 0.600 m ) 
3.20
=
= 2




Mg  n 
n
n
(15.0 kg ) 9.80 m/s2 

2
(
)
Thus, we have
 3.20 

 n2 
θ = sin –1 
Evaluating this for the harmonics corresponding to the range of n from n = 2 to n = 4 , we
have
 3.20 
θ = sin –1  2  = 53.1° for n = 2
 2 
 3.20 
 = 20.8° for n = 3
 32 
θ = sin –1 
 3.20 
 = 11.5° for n = 4
 42 
θ = sin –1 
The angles between 15.0° and 90.0° are θ = 20.8 ° and θ = 53.1° .
______________________________________________________________________________
56. CONCEPT QUESTIONS a. In drawing 1 the two speakers are equidistant from the
observer. Since each wave travels the same distance in reaching the observer, the difference
in travel-distances is zero, and constructive interference will occur. It will occur for any
frequency. Different frequencies will correspond to different wavelengths, but the path
difference will always be zero. Condensations will always meet condensations and
rarefactions will always meet rarefactions at the observation point.
b. Destructive interference occurs only when the difference in travel distances for the two
waves is an odd integer number n of half-wavelengths. Only certain frequencies, therefore,
will be consistent with this requirement.
SOLUTION Frequency and wavelength are related by Equation 16.1 (λ = v/f). Using this
equation together with the requirement for destructive interference in drawing 2, we have
L2 + L2 − L = n
Difference in travel
distances
λ
2
=n
v
2f
where n = 1, 3, 5, ...
Here we have used the Pythagorean theorem to determine the length of the diagonal of the
square. Solving for the frequency f gives
Chapter 17 Problems
f =
2
(
nv
907
)
2 −1 L
The problem asks for the minimum frequency, so n = 1, and we obtain
f =
2
v
(
)
2 −1 L
=
2
(
343 m/s
)
2 − 1 ( 0.75 m )
= 550 Hz
______________________________________________________________________________
57. CONCEPT QUESTIONS a. The diffraction angle θ is determined by the ratio of the
wavelength λ to the diameter D, according to Equation 17.2 (sin θ = 1.22 λ/D).
b. The wavelength is related to the frequency f and the speed v of the wave by
Equation 16.1 (λ = v/f ).
SOLUTION Using Equations 17.2 and 16.1, we have
sin θ = 1.22
λ
D
= 1.22
v
Df
Since the speed of sound is a constant, this result indicates that the diffraction angle will be
the same for each of the three speakers, provided that the diameter D times the frequency f
has the same value. Thus, we pair the diameter and the frequency as follows
Diameter × Frequency
(0.050 m)(12.0 × 103 Hz) = 6.0 × 102 m/s
(0.10 m)(6.0 × 103 Hz) = 6.0 × 102 m/s
(0.15 m)(4.0 × 103 Hz) = 6.0 × 102 m/s
The common value of the diffraction angle, then, is
v 
−1 1.22 ( 343 m/s ) 
 = 44°
 = sin 
2
Df 

 6.0 ×10 m / s 
______________________________________________________________________________

θ = sin −1  1.22
58. CONCEPT QUESTIONS a. The bystander hears a frequency from the moving horn that is
greater than the emitted frequency fs. This is because of the Doppler effect (Section 16.9).
The bystander is the observer of the sound wave emitted by the horn. Since the horn is
moving toward the observer, more condensations and rarefactions of the wave arrive at the
observer’s ear per second than would otherwise be the case. More cycles per second means
that the observed frequency is greater than the emitted frequency.
908 THE PRINCIPLE OF LINEAR SUPERPOSITION AND INTERFERENCE PHENOMENA
b. The bystander hears a frequency from the stationary horn that is equal to the frequency fs
produced by the horn. Since the horn is stationary, there is no Doppler effect.
c. Yes. Because the two frequencies heard by the bystander are different, he hears a beat
frequency that is the difference between the two.
SOLUTION According to Equation 16.11, the frequency that the bystander hears from the
moving horn is
 1 
f o = fs 
 1 − v / v 
s


where vs is the speed of the moving horn and v is the speed of sound. The beat frequency
heard by the bystander is fo – fs, so we find that
 1 
 1

− fs = fs 
− 1
f o − fs = fs 

 1− v / v 
 1− v / v 
s
s






1
= ( 395 Hz ) 
− 1 = 14 Hz
1 − (12.0 m/s ) / ( 343 m/s ) 
______________________________________________________________________________
59. CONCEPT QUESTIONS a. The harmonic frequencies are integer multiples of the
fundamental frequency. Therefore, for wire A, the fundamental is one half of 660 Hz, or
330 Hz. Similarly, for wire B, the fundamental is one third of 660 Hz, or 220 Hz. Thus, the
fundamental frequency of wire A is greater than that for wire B.
b. The fundamental frequency is related to the length L of the wire and the speed v at which
individual waves travel back and forth on the wire by Equation 17.3 with n = 1: f1 = v/(2L).
c. According to Equation 17.3, the fundamental frequency is proportional to the speed.
Therefore, since the fundamental frequency of wire A is greater, the speed must be greater
on wire A than on wire B.
SOLUTION Using Equation 17.3 with n = 1, we find
v
or v = 2 L f1
f1 =
2L
Wire A
v = 2 (1.2 m )( 330 Hz ) = 790 m/s
Wire B
v = 2 (1.2 m )( 220 Hz ) = 530 m/s
As expected, the speed for wire A is greater.
______________________________________________________________________________
Chapter 17 Problems
909
60. CONCEPT QUESTIONS a. The speed v at which individual waves travel on the wire
F
related to the tension F according to Equation 16.2: v =
, where m/L is the mass per
m/ L
unit length of the wire.
b. The tension in the wire in Part 2 less than the tension in Part 1. The reason is related to
Archimedes’ principle (Equation 11.6). This principle indicates that when an object is
immersed in a fluid, the fluid exerts an upward buoyant force on the object. In Part 2 the
upward buoyant force from the mercury supports part of the block’s weight, thus reducing
the amount of the weight that the wire must support. As a result, the tension in the wire is
less than in Part 1.
c. Since the tension F is less in Part 2, the speed v is also less. The fundamental frequency
of the wire is given by Equation 17.3 with n = 1: f1 = v/(2L). Since v is less, the
fundamental frequency of the wire is less in Part 2 than in Part 1.
SOLUTION Using Equations 17.3 and 16.2, we can obtain the fundamental frequency of
the wire as follows:
v
1
F
(1)
f1 =
=
2L 2L m / L
In Part 1 of the drawing, the tension F balances the weight of the block, keeping it from
falling. The weight of the block is its mass times the acceleration due to gravity. The mass,
according to Equation 11.1 is the density ρcopper times the volume V of the block. Thus, the
tension in Part 1 is
F = ( mass ) g = ρcopperVg
Part 1 tension
In Part 2 of the drawing, the tension is reduced from this amount by the amount of the
upward buoyant force. According to Archimedes’ principle, the buoyant force is the weight
of the liquid mercury displaced by the block. Since half of the block’s volume is immersed,
the volume of mercury displaced is V/2. The weight of this mercury is the mass times the
acceleration due to gravity. Once again, according to Equation 11.1, the mass is the density
ρmercury times the volume, which is V/2. Thus, the tension in Part 2 is
Part 2 tension
F = ρcopperVg − ρ mercury (V / 2 ) g
With these two values for the tension we can apply Equation (1) to both parts of the drawing
and obtain
f1 =
Part 1
Part 2
f1 =
1
2L
1
2L
ρcopperVg
m/ L
ρcopperVg − ρ mercury (V / 2 ) g
m/ L
910 THE PRINCIPLE OF LINEAR SUPERPOSITION AND INTERFERENCE PHENOMENA
Dividing the Part 2 by the Part 1 result, gives
f1, Part 2
f1, Part 1
1
= 2L
=
ρcopperVg − ρ mercury (V / 2 ) g
1
2L
m/ L
ρcopperVg
=
ρcopper − 12 ρmercury
ρcopper
m/ L
(
8890 kg/m3 − 12 13 600 kg/m3
8890 kg/m
3
)=
0.485
As expected, the fundamental frequency is less in Part 2 than Part 1.
______________________________________________________________________________
61. CONCEPT QUESTIONS a. At the end of the tube where the tuning fork is, there is an
antinode, because the gas molecules there are free to vibrate. At the plunger, there is a
node, because the gas molecules there are not free to vibrate.
b. Since there is an antinode at one end of the tube and a node at the other, the smallest
value of L occurs when the length of the tube is one quarter of a wavelength.
SOLUTION Since the smallest value for L is a quarter of a wavelength, we have L = λ/4 or
λ = 4L. According to Equation 16.1, the speed of sound is
b
gb
g
v = fλ = f 4 L = 485 Hz 4 0.264 m = 512 m / s
______________________________________________________________________________
62. CONCEPT QUESTIONS
a. The phrase “the speakers are vibrating out of phase” means that when the diaphragm of
one speaker is moving outward, the diaphragm of the other speaker is moving inward. In
other words, when one speaker is creating a condensation, the other is creating a rarefaction.
b. The sound waves reaching the listener would exhibit destructive interference. Both sound
waves travel the same distance from the speakers to the listener. Since the speakers are
vibrating out of phase, whenever a condensation from one speaker reaches the listener, it is
met by a rarefaction from the other, and vice versa. Therefore, the two sound waves
experience destructive interference, and the listener hears no sound.
c. When the listener begins to move sideways, the distance between the listener and each
speaker is no longer the same. Consequently, the sound waves no longer produce destructive
interference, and the sound intensity begins to increase.
Chapter 17 Problems
SOLUTION The two speakers are vibrating out of phase.
Therefore, when the difference in path lengths 1 − 2
traveled by the two sounds is one-half a wavelength, or
1
1 − 2 = 2 λ , constructive interference occurs. Note that this
condition is different than that for two speakers vibrating in
phase. The frequency f of the sound is equal to the speed v of
sound divided by the wavelength λ; f = v/λ (Equation 16.1).
Thus, we have that
1− 2
= 12 λ =
v
2f
or
f =
v
2( 1 −
2
Midpoint
4.00 m
)
911
0.92 m
1
1.50 m
2
1.50 m
The distances 1 and 2 can be determined by applying the Pythagorean theorem to the
right triangles in the drawing:
=
( 4.00m )2 + (1.50 m + 0.92 m )2
=
( 4.00 m )2 + (1.50 m
1
2
= 4.68 m
− 0.92 m ) = 4.04 m
2
The frequency of the sound is
f =
v
2( 1 −
2
)
=
343 m/s
= 270 Hz
2 ( 4.68 m − 4.04 m )
______________________________________________________________________________
63. CONCEPT QUESTIONS
a. The waves on the longer string have the same speed as those on the shorter string. The
speed v of a transverse wave on a string is given by v = F / ( m / L ) (Equation 16.2), where
F is the tension in the string and m/L is the mass per unit length (or linear density). Since F
and m/L are the same for both strings, the speed of the waves is the same.
b. From the drawing it is evident that both strings are vibrating at their fundamental
frequencies, and the longer string will vibrate at a lower frequency. The reason is that the
fundamental frequency of vibration (n = 1) for a string fixed at each end is given by
Equation 17.3 as f1 = v/(2L). Since the speed v is the same for both strings, but L is greater
for the longer string, the longer string vibrates at the lower frequency.
c. The beat frequency will increase. The beat frequency is equal to the higher frequency of
the shorter string minus the lower frequency of the longer string. If the longer string is
increased in length, its frequency decreases even more, so the difference between the
frequencies (the beat frequency) increases.
912 THE PRINCIPLE OF LINEAR SUPERPOSITION AND INTERFERENCE PHENOMENA
0.57 cm
SOLUTION
The beat frequency is the
frequency of the shorter string minus the
frequency of the longer string; fshorter − flonger .
We are given that fshorter = 225 Hz.
L
According to Equation 17.3 with n = 1, we
have flonger = v/(2Llonger), where Llonger is the
length of the longer string; in other words, we
have Llonger = L + 0.0057 m. Thus,
f longer =
v
2Llonger
=
v
2 ( L + 0.0057 m )
According to our answer to Concept Question (a), the speed v of the waves on the longer
string is the same as those on the shorter string, so v = 41.8 m/s. The length L of the shorter
string can be obtained directly from Equation 17.3:
L=
v
41.8 m/s
=
= 0.0929 m
2 f1 2 ( 225 Hz )
Substituting this number back into the expression for flonger yields
f longer =
v
41.8 m/s
=
= 212 Hz
2 ( L + 0.0057 m ) 2 ( 0.0929 m + 0.0057 m )
The beat frequency is fshorter − flonger = 225 Hz − 212 Hz = 13 Hz .
______________________________________________________________________________