CHAPTER 17
SOUND AND OTHER
WAVE PHENOMENA
ActivPhysics
can help with these problems:
Activities 10.3, 10.4, 10.5, 10.6, 10.8, 10.9
Sections 171 and 172:
Sound Waves and the
Speed of Sound in Gases
Problem
1. Show that the quantity
p
P/ρ
has the units of
speed.
Solution
The units of pressure (force per unit area) divided by
density (mass per unit volume) are (N
/
m
2
)
/
(kg
/
m
3
) =
(N
/
kg)(m
3
/
m
2
) = (m
/
s
2
)m = (m
/
s)
2
,
or those of
speed squared.
Problem
2. Dimensional analysis alone suggests that the sound
speed in a gas should be given roughly by
p
P/ρ.
By how much would an estimate based on this
simple analysis be in error for a gas with
γ
=
7
5
?
Solution
The fractional difference between
p
P/ρ
and
p
γP/ρ
is Δ
v/v
= (
p
P/ρ

p
γP/ρ
)
/
p
γP/ρ
=
p
1
/γ

1
.
For
γ
= 7
/
5
,
this is
p
5
/
7

1 =

0
.
155
≈ 
15% (i.e.,
the rough value is 15% smaller than Equation 171).
Problem
3. Find the wavelength, period, angular frequency,
and wave number of a 1.0kHz sound wave in air
under the conditions of Example 171.
Solution
The value of the speed of sound in air from
Example 171 was 343 m/s. Therefore (a)
λ
=
v/f
=
(343 m
/
s)
/
(1 kHz) = 34
.
3 cm; (b)
T
= 1
/f
= 1 ms;
(c)
ω
= 2
πf
= 6
.
28
×
10
3
s

1
; and (d)
k
=
ω/v
=
2
π/λ
= 18
.
3 m

1
.
(See Equations 161, 3, 4, and 6.)
Problem
4. (a) Determine an approximate value for the speed
of sound in miles per second. (b) Suppose you see a
lightning flash and, 10 s later, hear the thunder.
How many miles away did the flash occur? Neglect
the travel time for the light (why?)
Solution
(a) From Example 171,
v
= (343 m
/
s)(1 mi
÷
1609 m) = 0
.
213 mi
/
s
.
(b) If it took the sound of the
thunder roughly 10 s to reach you, its source was
about
vt
= (0
.
213 mi
/
s)(10 s) = 2
.
13 mi away. (The
speed of light in air, approximately 186,000 mi/s, is so
much bigger than the speed of sound that the light
travel time is negligible compared to sound travel
times.)
Problem
5. Timers in sprint races start their watches when
they see smoke from the starting gun, not when
they hear the sound (Fig. 1725). Why? How much
error would be introduced by timing a 100m race
from the sound of the shot?
Solution
The sound of the starting gun takes (100 m)
÷
(340 m
/
s) = 0
.
294 s to reach the finish line. An error
of this magnitude is significant in short races, where
world records are measured in hundredths of a second.
(This problem is almost the same as Problem 2,
Chapter 2.)
Problem
6. The factor
γ
for nitrogen dioxide (NO
2
) is 1.29.
Find the sound speed in NO
2
at a pressure of 4
.
8
×
10
4
N
/
m
2
and density 0
.
35 kg
/
m
3
.
Solution
The given data, substituted into Equation 171, gives:
v
=
q
1
.
29(4
.
8
×
10
4
N
/
m
2
)
/
(0
.
35 kg
/
m
3
) = 421 m
/
s
.
Problem
7. At standard atmospheric pressure (1
.
0
×
10
5
N
/
m
2
)
,
what density of air would make the sound speed
1.0 km/s?
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CHAPTER 17
Solution
Solving for the density in Equation 171, we find:
ρ
=
γP/v
2
= 1
.
4(1
.
0
×
10
5
N
/
m
2
)
/
(10
3
m
/
s)
2
= 0
.
14 kg
/
m
3
.
Problem
8. The Sun’s outer atmosphere, or corona, has about
10
8
electrons and an equal number of protons per
cubic centimeter. The pressure of this electron
proton gas is about 3
×
10

3
N
/
m
2
,
and it behaves
like a monatomic gas with
γ
=
5
3
.
To one significant
figure, what is the sound speed in the corona?
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 Spring '10
 SMITH
 Math, Frequency, Wavelength

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