Week 13 homework
IMPORTANT NOTE ABOUT WEBASSIGN:
In the WebAssign versions of these problems, various details have been changed, so that
the answers will come out differently. The method to find the solution is the same, but
you will need to repeat part of the calculation to find out what your answer should have
been.
WebAssign Problem 1: The drawing shows a loudspeaker A and point C, where a
listener is positioned. A second loud-speaker B is located somewhere to the right of A.
Both speakers vibrate in phase and are playing a 68.6-Hz tone. The speed of sound is 343
m/s. What is the closest to speaker A that speaker B can be located, so that the listener
hears no sound?
REASONING The geometry of the positions of the loudspeakers and the listener is
shown in the following drawing.
C
d 1 = 1 .0 0 m
d2
y
6 0 .0 °
A
B
x1
x2
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, K
(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
nλ
5.00 m
+ d1 =
+ 1.00 m = 3.50 m
2
2
d2 =
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 .
WebAssign Problem 2: Speakers A and B are vibrating in phase. They are directly
facing each other, are 7.80 m apart, and are each playing a 73.0-Hz tone. The speed of
sound is 343 m/s. On the line between the speakers there are three points where
constructive interference occurs. What are the distances of these three points from
speaker A?
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 . 9 0 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 . 5 5 m f r o m s p e a k e r A . The
points of constructive interference will occur symmetrically about the center point at
L/2, so there is also a point of constructive interference 1.55 m from speaker B, that
is at the point
7.80 m – 1.55 m = 6 . 2 5 m f r o m s p e a k e r 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.
WebAssign Problem 3: A row of seats is parallel to a stage at a distance of 8.7 m from
it. At the center and front of the stage is a diffraction horn loudspeaker. This speaker
sends out its sound through an opening that is like a small doorway with a width D of
7.5 cm. The speaker is playing a tone that has a frequency of 1.0 × 104 Hz. The speed
of sound is 343 m/s. What is the distance between two seats, located near the center
of the row, at which the tone cannot be heard?
REASONING AND SOLUTION
The figure at the right shows the
geometry of the situation.
D i ffra c ti o n h o rn
The tone will not be heard at seats located 8 . 7 m
at the first diffraction minimum. This
occurs when
sin θ =
λ
v
=
D fD
S ta g e
θ
θ
F irst ro w
o f se a ts
C
x
That is, the angle θ is given by

 v 
343 m/s
−1
θ = sin − 1 
 = 27.2°
 = sin 
4
 f D
 (1.0 × 10 Hz)(0.075 m) 
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
WebAssign Problem 4: Two pure tones are sounded together. The drawing shows the
pressure variations of the two sound waves, measured with respect to atmospheric
pressure. What is the beat frequency?
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
WebAssign Problem 5: A string has a linear density of 8.5 × 10–3 kg/m and is under a
tension of 280 N. The string is 1.8 m long, is fixed at both ends, and is vibrating in the
standing wave pattern shown in the drawing. Determine the (a) speed, (b) wavelength,
and (c) frequency of the traveling waves that make up the standing wave.
REASONING A standing wave is composed of two oppositely traveling waves. The
F
speed v of these waves is given by v =
(Equation 16.2), where F is the
m/ L
tension in the string 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
=
m/ L
280 N
= 180 m/s
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 =
c.
2
3
( 1.8 m ) =
1.2 m
The frequency of the waves is
v 180 m/s
f = =
= 150 Hz
λ
1.2 m
1.8 m
λ
WebAssign Problem 6: Divers working in underwater chambers at great depths must
deal with the danger of nitrogen narcosis (the “bends”), in which nitrogen dissolves
into the blood at toxic levels. One way to avoid this danger is for divers to breathe a
mixture containing only helium and oxygen. Helium, however, has the effect of
giving the voice a high-pitched quality, like that of Donald Duck’s voice. To see
why this occurs, assume for simplicity that the voice is generated by the vocal cords
vibrating above a gas-filled cylindrical tube that is open only at one end. The quality
of the voice depends on the harmonic frequencies generated by the tube; larger
frequencies lead to higher-pitched voices. Consider two such tubes at 20 °C. One is
filled with air, in which the speed of sound is 343 m/s. The other is filled with
helium, in which the speed of sound is 1.00 × 103 m/s. To see the effect of helium on
voice quality, calculate the ratio of the nth natural frequency of the helium-filled tube
to the nth natural frequency of the air-filled tube.
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 airfilled and the helium-filled 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
3
f n, helium
4 L = vhelium = 1.00 × 10 m/s = 2.92
=
nvair
f n, air
vair
343 m/s
4L
WebAssign Problem 7: Two ultrasonic sound waves combine and form a beat frequency
that is in the range of human hearing. The frequency of one of the ultrasonic waves
is 70 kHz. What is (a) the smallest possible and (b) the largest possible value for the
frequency of the other ultrasonic wave?
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 kH z – 20 kH z =
50 kH z
which results in a beat frequency of 7 0 k H z – 5 0 k H z = 2 0 k H z .
b. The largest possible frequency for the other wave is
f = 70 kH z + 20 kH z =
90 kH z
which results in a beat frequency of 9 0 k H z – 7 0 k H z = 2 0 k H z .
WebAssign Problem 8: The fundamental frequency of a vibrating system is 400 Hz. For
each of the following systems, give the three lowest frequencies (excluding the
fundamental) at which standing waves can occur: (a) a string fixed at both ends, (b) a
cylindrical pipe with both ends open, and (c) a cylindrical pipe with only one end
open.
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
WebAssign Problem 9: The A string on a string bass is tuned to vibrate at a fundamental
frequency of 55.0 Hz. If the tension in the string were increased by a factor of four,
what would be the new fundamental frequency?
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
frequency,
f1 =
for the fundamental
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
Practice conceptual problems:
1. Does the principle of linear superposition imply that two sound waves, passing
through the same place at the same time, always create a louder sound than either wave
alone? Explain.
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.
6. 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. From these two pieces of
information alone, is it possible to determine the exact frequency of the violin string?
Explain.
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.
10. A string is vibrating back and forth as in 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. Draw the new standing wave pattern that develops on the string. Give your
reasoning.
REASONING AND SOLUTION A string is being vibrated back and forth as in
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.
13. Standing waves can ruin the acoustics of a concert hall if there is excessive reflection
of the sound waves that the performers generate. For example, suppose a performer
generates a 2093-Hz tone. If a large-amplitude standing wave is present, it is possible for
a listener to move a distance of only 4.1 cm and hear the loudness of the tone change
from loud to faint. Account for this observation in terms of standing waves, pointing out
why the distance is 4.1 cm.
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 onequarter 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.