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Chapter 22. Sound Physics, 6 Edition
Chapter 22. Sound
Speed of Sound Waves
22-1. Young's modulus for steel is 2.07 x 10
11 Pa and its density is 7800 kg/m
3
. Compute the
speed of sound in a steel rod.
11
3
2.07 x 10 Pa
7800 kg/m
Y
v
(^) ; v = 5150 m/s
22-2. A 3-m length of copper rod has a density 8800 kg/m
3 , and Young's modulus for copper is
1.17 x 10
11 Pa. How much time will it take for sound to travel from one end of the rod to
the other?
11
3
1.17 x 10 Pa
8800 kg/m
Y
v
; v = 3646 m/s
2(3 m)
;
3646 m/s
s s
v t
t v
; t = 1.65 ms
22-3. What is the speed of sound in air (M = 29 g/mol and = 1.4) on a day when the
temperature is 30
0 C? Use the approximation formula to check this result.
0 0 (1.4)(8.314 J/kg K)(273 30 )K
0.029 kg/mol
RT
v
M
(^) ; v = 349 m/s
0
0
m/s 331 m/s 0.6 (30 C);
C
v
v = 349 m/s
22-4. The speed of longitudinal waves in a certain metal rod of density 7850 kg/m
3 is measured
to be 3380 m/s. What is the Young's modulus for the metal?
2 ; ;
Y Y
v v
(^) Y = v
2 = (7850 kg/m
3 )(3380 m/s)
2 ; Y = 8.97 x 10
10 Pa
Chapter 22. Sound Physics, 6 Edition
22-5. If the frequency of the waves in Problem 22-4 is 312 Hz, what is the wavelength?
3380 m/s
;
312 Hz
v
v f
f
(^) ; = 10.8 m
22-6. Compare the theoretical speeds of sound in hydrogen (M = 2.0 g/mol, = 1.4) with
helium (M = 4.0 g/mol, = 1.66) at 0
0 C.
1.4)(8.314 J/mol K)(273 K)
0.002 kg/mol
H
RT
v
M
; vH = 1260 m/s
1.66)(8.314 J/mol K)(273 K)
0.004 kg/mol
He
RT
v
M
; vHe = 971 m/s
971 m/s
1260 m/s
He
H
v
v
(^) ; v He =^ 0.771^ vH
*22-7. A sound wave is sent from a ship to the ocean floor, where it is reflected and returned. If
the round trip takes 0.6 s, how deep is the ocean floor? Consider the bulk modulus for sea
water to be 2.1 x 10
9 Pa and its density to be 1030 kg/m
3 .
9
3
2.1 x 10 Pa
1030 kg/m
B
v
; v = 1428 m/s
h = vt = (1328 m/s)(0.3 s); h = 428 m
Vibrating Air Columns
22-8. Find the fundamental frequency and the first three overtones for a 20-cm pipe at 20
0 C if
the pipe is open at both ends. v = 331 m/s + (0.6 )(
0 ) = 343 m/s.
1
(1)(343 m/s) ; 858 Hz
2 (2)(0.20 m)
n
nv f f
l
(^) f 1 = 858 Hz
(First overtone, n = 2) fn = nf 1 ; f 2 = 2(857.5 Hz) = 1715 Hz
22-8 (Cont.) (2nd overtone, n = 3) fn = nf 1 ; f 2 = 3(857.5 Hz) = 2573 Hz
Chapter 22. Sound Physics, 6 Edition
*22-14. In a resonance experiment, the air in a closed tube of variable length is found to resonate
with a tuning fork when the air column is first 6 cm and then 18 cm long. What is the
frequency of the tuning fork if the temperature is 20
0 C? [ v = 343 m/s at 20
0 C_._ ]
The distance between adjacent nodes of resonance is one-half of a wavelength.
343 m/s
18 cm - 6 cm; 24 cm;
2 0.24 m
v
f
; f = 1430 Hz
*22-15. A closed pipe and an open pipe are each 3 m long. Compare the wavelength of the fourth
overtone for each pipe at 20
0 C.? ( Only odd harmonics allowed for closed pipe .)
For an open pipe, the fourth overtone is the fifth harmonic, n = 5.
Open (fourth overtone): 5
2 2(3 m)
; ;
5
n
l
n
5 = 1.20 m
For closed pipe, the fourth overtone is the ninth harmonic, n = 9.
Closed (fourth overtone): (^) 9
4 4(3 m) ; ;
9
n
l
n
9 = 1.33 m
Sound Intensity and Intensity Level
22-16. What is the intensity level in decibels of a sound whose intensity is 4 x 10
2 ?
-5 2
-12 2
0
4 x 10 W/m 10 log 10 log
1 x 10 W/m
I
I
(^) ; = 76.0 dB
22-17. The intensity of a sound is 6 x 10
2
. What is the intensity level?
-8 2
-12 2
0
6 x 10 W/m 10 log 10 log
1 x 10 W/m
I
I
(^) ; = 47.8 dB
Chapter 22. Sound Physics, 6 Edition
22-18. A 60 dB sound is measured at a particular distance from a whistle. What is the intensity
of this sound in W/m
2 ?
6
0 0
10log 60 dB; 10 ;
I I
I I
I = (
6 )(1 x 10
2 ); I = 1 x 10
2
*22-19. What is the intensity of a 40 dB sound?
4
0 0
10log 40 dB; 10 ;
I I
I I
I = (
4 )(1 x 10
2 ); I = 1 x 10
2
*22-20. Compute the intensities for sounds of 10 dB, 20 dB, and 30 dB.
1
0 0
10log 10 dB; 10 ;
I I
I I
I = (
1 )(1 x 10
2 ); I = 1 x 10
2
2
0 0
10log 20 dB; 10 ;
I I
I I
I = (
2 )(1 x 10
2 ); I = 1 x 10
2
3
0 0
10 log 30 dB; 10 ;
I I
I I
I = (
3 )(1 x 10
2 ); I = 1 x 10
2
22-21. Compute the intensity levels for sounds of 1 x 10
2 , 2 x 10
2 , and
3 x 10
2 .
-6 2
-12 2
0
1 x 10 W/m 10 log 10 log
1 x 10 W/m
I
I
(^) ; = 60.0 dB
-6 2
-12 2
0
2 x 10 W/m
10 log 10 log
1 x 10 W/m
I
I
(^) ; = 63.0 dB
-6 2
-12 2
0
3 x 10 W/m
10 log 10 log
1 x 10 W/m
I
I
(^) ; = 64.8 dB
Chapter 22. Sound Physics, 6 Edition
The Doppler Effect
Assume that the speed of sound in 343 m/s for all of these problems.
22-26. A stationary source of sound emits a signal at a frequency of 290 Hz. What are the
frequencies heard by an observer (a) moving toward the source at 20 m/s and (b) moving
away from the source at 20 m/s? ( Approach = +, recede = -)
0
0
343 m/s + 20 m/s
290 Hz
343 m/s - 0
s
s
V v
f f
V v
; f 0 = 307 Hz
0
0
343 m/s + (-20 m/s)
290 Hz
343 m/s - 0
s
s
V v
f f
V v
; f 0 = 273 Hz
22-27. A car blowing a 560-Hz horn moves at a speed of 15 m/s as it first approaches a
stationary listener and then moves away from a stationary listener at the same speed.
What are the frequencies heard by the listener? ( Approach = +, recede = -)
0 0
343 m/s + 0 560 Hz
343 m/s - 15 m/s
s
s
V v f f
V v
; f 0 = 586 Hz
0
0
343 m/s + 0
360 Hz
343 m/s - (-15 m/s)
s
s
V v
f f
V v
; f 0 = 537 Hz
22-28. A person stranded in a car blows a 400-Hz horn. What frequencies are heard by the driver
of a car passing at a speed of 60 km/h? ( Approach = +, recede = -)
0
km 1000 m 1 h
60 16.7 m/s
h 1 km 3600 s
v
Approaching:
0
0
343 m/s + 16.7 m/s
400 Hz
343 m/s - 0
s
s
V v
f f
V v
; f 0 = 419 Hz
At same point as car there is no change: f 0 = 400 Hz
Leaving:
0 0
343 m/s + (-16.7 m/s) 400 Hz
343 m/s - 0
s
s
V v f f
V v
; f 0 = 381 Hz
Chapter 22. Sound Physics, 6 Edition
22-29. A train moving at 20 m/s blows a 300-Hz whistle as it passes a stationary observer. What
are the frequencies heard by the observer as the train passes?
Approaching:
0 0
343 m/s + 0 300 Hz
343 m/s - 20 m/s
s
s
V v f f
V v
; f 0 = 319 Hz
When at the same position there is no change: f 0 = 300 Hz
Leaving:
0
0
343 m/s + 0
300 Hz
343 m/s - (-20 m/s)
s
s
V v
f f
V v
; f 0 = 283 Hz
*22-30. A child riding a bicycle north at 6 m/s hears a 600 Hz siren from a police car heading
south at 15 m/s. What is the frequency heard by the child? (Approaches are +)
Approaching:
0
0
343 m/s + 6 m/s
600 Hz
343 m/s - 15 m/s
s
s
V v
f f
V v
; f 0 = 638 Hz
*22-31. An ambulance moves northward at 15 m/s. Its siren has a frequency of 600 Hz at rest. A
car heads south at 20 m/s toward the ambulance. What frequencies are heard by the car
driver before and after they pass? ( Approach = +, recede = -)
Before passing:
0
0
343 m/s + 20 m/s
600 Hz
343 m/s - 15 m/s
s
s
V v
f f
V v
; f 0 = 664 Hz
After passing:
0
0
343 m/s + (-20 m/s)
600 Hz
343 m/s - (-15 m/s)
s
s
V v
f f
V v
; f 0 = 541 Hz
*22-32. A truck traveling at 24 m/s overtakes a car traveling at 10 m/s in the same direction. The
trucker blows a 600-Hz horn. What frequency is heard by the car driver?
The car is moving away, so v 0 = - 10 m/s ; Truck is approaching, so vs = + 24 m/s
0 0
343 m/s + (-10 m/s) 600 Hz
343 m/s - (+24 m/s)
s
s
V v f f
V v
^ ^
; f 0 = 626 Hz
Chapter 22. Sound Physics, 6 Edition
22-37. A 40-g string 2 m in length vibrates in three loops. The tension in the string is 270 N.
What is the wavelength? What is the frequency?
2 2(2 m)
;
3
l
n
= 1.33 m
(270 N)(2 m)
0.040 kg
Fl v
m
v = 116 ft/s;
116 m/s
1.33 m
v f
(^) ; f = 87.1 Hz
22-38. How many beats per second are heard when two tuning forks of 256 Hz and 259 Hz are
sounded together?
f’- f = 259 Hz – 256 Hz = 3 beats/s
*22-39. What is the length of a closed pipe if the frequency of its second overtone is 900 Hz on a
day when the temperature is 20C?
The second overtone for a closed pipe occurs when n = 5, and v = 343 m/s.
5(343 m/s) ;
4 4(900 Hz)
n
nv f l
l
;^ l =^ 47.6 cm
*22-40. The fundamental frequency for an open pipe is 360 Hz. If one end of this pipe is closed,
what will be the new fundamental frequency? ( n = 1 for fundamental )
We will first find the length of an open pipe that has a frequency of 360 Hz
1(343 m/s)
;
2 2(360 Hz)
n
nv
f l
l
(^) ; l = 47.6 cm
Now, take this length for a closed pipe to find new frequency:
1
(1) (343 m/s)
;
4 4(0.476 m)
v
f f
l
; f = 180 Hz
Chapter 22. Sound Physics, 6 Edition
*22-41. A 60-cm steel rod is clamped at one end as shown in Fig. 23-13a. Sketch the
fundamental and the first overtone for these boundary conditions. What are the
wavelengths in each case?
Boundary conditions = node at right, antinode to left:
1 = 4l = 4(0.60 m); 1 = 2.40 m
First overtone adds one node, 1
st ovt. = 3rd harmonic.
3
4 4(0.60 m)
l
; 3 = 0.800 m
*22-42. The 60-cm rod in Fig. 22-13b is now clamped at its midpoint. What are the
wavelengths for the fundamental and first overtone?
Boundary conditions: A node must be at the center,
and an antinode must be at each end in both cases.
Fundamental: 1
2(0.60 m);
1
l
1 = 1.20 m
First overtone is first possibility after the fundamental. Because of clamp at midpoint,
First overtone:
2 2(0.60 m)
l (^) = 0.400 m
*22-43. The velocity of sound in a steel rod is 5060 m/s. What is the length of a steel rod
mounted as shown in Fig. 22-13a if the fundamental frequency of vibration for the rod
is 3000 Hz?
5060 m/s
;
3000 Hz
v
v f
f
(^) ; = 1.69 m
For fundamental: 1 = 4 l ;
1.69 m
l
(^) ; l = 42.2 cm
First overtone
Fundamental
First overtone
Fundamental
Fundamental
Chapter 22. Sound Physics, 6 Edition
*22-47. What is the difference in intensity levels (dB) for two sounds whose intensities are
2 x 10
2 and 0.90 W/m
2 ? [ See solution for Prob. 22-44 above. ]
2 2 2 2 1 2 1
0 0 1
10 log 10log ; 10 log
I I I
I I I
2
(^2 1) -5 2
0.90 W/m 10 log
2 x 10 W/m
; = 46.5 dB
Critical Thinking Questions
*22-48. By inhaling helium gas, one can raise the frequency of the voice considerably. For air
M = 29 g/mol and = 1.4; for helium M = 4.0 g/mol and = 1.66. At a temperature
of 27
0 C, you sing a “C” note at 256 Hz. What is the frequency that will be heard if you
inhale helium gas and other parameters are unchanged? Notice that both v and f were
increased. How do you explain this in view of the fact that v = f . Discuss.
1.4)(8.314 J/mol K)(300 K)
0.029 kg/mol
air
RT
v
M
(^) ; vH = 347 m/s
1.66)(8.314 J/mol K)(300 K)
0.004 kg/mol
He
RT
v
M
; vHe = 1017 m/s
The fundamental wavelength is a property of the boundary conditions which don’t
change. Therefore, the frequencies are directly proportional to the velocities.
1017 m/s
2.93;
347 m/s
He
air
f
f
(^) f He =^ 2.93^ fair =^ 2.93(256 Hz);^ fHe =^ 750 Hz
The ratio of v/f is constant and equal to the wavelength in each case.
Chapter 22. Sound Physics, 6 Edition
*22-49. A toy whistle is made out of a piece of sugar cane that is 8 cm long. It is essentially an
open pipe from the air inlet to the far end. Now suppose a hole is bored at the midpoint
so that the finger can alternately close and open the hole. (a) If the velocity of sound is
340m/s, what are the two possible fundamental frequencies obtained by closing and
opening the hole at the center of the cane? (b) What is the fundamental frequency if the
finger covers the hole and the far end is plugged?
An opening forces an antinode to occur at that point.
(a) With the finger closing the hole, the fundamental is:
= 2l = 2(0.08 m); 1 = 1.60 m
1
340 m/s
1.60 m
f ; f 1 = 212 Hz
Now, with the finger removed, the fundamental is: = l = 0.08 m, so that
1
340 m/s
0.80 m
f (^) ; f 1 = 425 m
(b) Now if the far end is plugged and the hole is covered, f 1 = 4l:
= 4(0.08 m) = 3.2 m;
340 m/s
3.20 m
f (^) ; f = 106 Hz
*22-50. A tuning fork of frequency 512 Hz is moved away from an observer and toward a flat
wall with a speed of 3 m/s. The speed of sound in the air is 340 m/s. What is the
apparent frequency of the unreflected sound? What is the apparent frequency of the
reflected sound? How many beats are heard each second?
For unrelfected sound, vs = + 3 m/s; for reflected sound, vs = -3 m/s
Unreflected:
0
0
340 m/s + 0
(512 Hz)
340 m/s -(+3 m/s)
s
s
V v
f f
V v
f 0 = 517 Hz
0.08 m
Chapter 22. Sound Physics, 6 Edition
resonance positions occur at 17, 51, and 85 cm from the top of the tube. What is the
velocity of sound in the air? What is the approximate temperature in the room?
The distance between adjacent resonance points must be equal to one-half of a
wavelength. Therefore, the wavelength of the sound must be:
= 2(51 cm – 17 cm) = 68.0 cm
v = f = (512 Hz)(0.680 m); v = 348 m/s
2
m/s
331 m/s 0.6 348 m/s
C
v t
t = 28.
0 C
*21-53. What is the difference in intensity levels of two sounds, one being twice the intensity of
the other?
Assume I 2 = 2I 1 :
2 1
2 1
1 1
10 log 10 log
I I
I I
(^) ; = 3.01 dB
