16.1:
a)
→
→
=
=
=
A
p
p
m
f
v
then
,
1000
if
)
b
.
344
.
0
)
Hz
100
(
s)
m
344
(
λ
0
0
1000
A
Therefore, the amplitude is
m.
10
2
.
1
5

×
max
max
increasing
,
Since
c)
p
BkA
p
=
while keeping
A
constant requires decreasing
k
, and increasing
π
, by the same factor.
Therefore the new wavelength is
Hz.
50
m,
9
.
6
)
20
(
m)
688
.
0
(
m
9
.
6
s
m
344
new
=
=
=
f
16.2:
m.
10
21
.
3
or
,
12
Hz)
1000
(
Pa)
9
10
2
.
2
(
2
)
s
m
1480
(
Pa)
2
10
0
.
3
(
2
max

×

×
=
×
=
=
A
A
π
πBf
v
p
The much higher bulk modulus
increases both the needed pressure amplitude and the speed, but the speed is proportional
to the square root of the bulk modulus. The overall effect is that for such a large bulk
modulus, large pressure amplitudes are needed to produce a given displacement.
16.3:
From Eq. (16.5),
.
2
λ
2
max
v
f
πBA
BA
π
BkA
p
=
=
=
a)
Pa.
7.78
s)
m
(344
Hz)
150
(
m)
10
00
.
2
(
Pa)
10
42
.
1
(
2
5
5
=
×
×

π
b)
Pa.
778
Pa
7.78
100
c)
Pa.
77.8
Pa
78
.
7
10
=
×
=
×
The amplitude at
Hz
1500
exceeds the pain threshold, and at
Hz
000
,
15
the sound
would be unbearable.
16.4:
The values from Example 16.8 are
Hz,
1000
Pa,
10
16
.
3
4
=
×
=
f
B
m.
10
2
.
1
8

×
=
A
Using Example 16.5,
,
s
m
295
s
m
344
293K
K
216
=
=
v
so the pressure
amplitude of this wave is
Pa).
10
16
.
3
(
2
4
max
×
=
=
=
A
v
πf
B
BkA
p
Pa.
10
8.1
m)
10
2
.
1
(
s
m
295
Hz)
(1000
2
3
8


×
=
×
π
This is
0.27
Pa)
10
(3.0
Pa)
10
1
.
8
(
2
3
=
×
×


times smaller than the pressure amplitude at sea level (Example 161), so pressure
amplitude decreases with altitude for constant frequency and displacement amplitude.
16.5:
a) Using Equation (16.7),
[
]
×
=
=
=
2
2
2
s)
400
m)(
8
(
so
,
)
λ
(
B
f
ρ
v
B
Pa.
10
33
.
1
)
m
kg
1300
(
10
3
×
=
b)
Using Equation (16.8),
[
]
2
4
2
2
s
10
(3.9
m)
5
.
1
(
)
(

×
=
=
=
ρ
t
L
ρ
v
Y
Pa.
10
47
.
9
)
m
kg
6400
(
10
3
×
=
×
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16.6:
a) The time for the wave to travel to Caracas was
s
579
s
39
min
9
=
and the
speed was
s
m
10
085
.
1
4
×
(keeping an extra figure). Similarly, the time for the wave to
travel to Kevo was 680 s for a speed of
s,
m
10
278
.
1
4
×
and the time to travel to Vienna
was 767 s for a speed of
s.
m
10
258
.
1
4
×
The average speed for these three
measurements is
s.
m
10
21
.
1
4
×
Due to variations in density, or reflections (a subject
addressed in later chapters), not all waves travel in straight lines with constant speeds.
b) From Eq. (16.7),
,
2
ρ
v
B
=
and using the given value of
3
3
m
kg
10
3
.
3
×
=
ρ
and the
speeds found in part (a), the values for the bulk modulus are, respectively,
Pa.
10
2
.
5
and
Pa
10
5.4
Pa,
10
9
.
3
11
11
11
×
×
×
These are larger, by a factor of 2 or 3, than
the largest values in Table (111).
16.7:
Use
s
m
1482
water
=
v
at
C,
20
°
as given in Table
(
29
1
.
16
The sound wave travels
in water for the same time as the wave travels a distance
m
8
.
20
m
20
.
1
m
0
.
22
=

in air,
and so the depth of the diver is
(
29
(
29
m.
6
.
89
s
m
344
s
m
1482
m
8
.
20
m
8
.
20
air
water
=
=
v
v
This is the depth of the diver; the distance from the horn is
m.
8
.
90
16.8:
a), b), c) Using Eq.
(
29
,
10
.
16
(
29
(
29
(
29
(
29
s
m
10
32
.
1
mol
kg
10
02
.
2
K
15
.
300
K
mol
J
3145
.
8
41
.
1
3
3
2
H
×
=
×
⋅
=

v
(
29
(
29
(
29
(
29
s
m
10
02
.
1
mol
kg
10
00
.
4
K
15
.
300
K
mol
J
3145
.
8
67
.
1
3
3
e
H
×
=
×
⋅
=

v
(
29
(
29
(
29
(
29
.
s
m
323
mol
kg
10
9
.
39
K
15
.
300
K
mol
J
3145
.
8
67
.
1
3
Ar
=
×
⋅
=

v
d) Repeating the calculation of Example 16.5 at
K
15
.
300
=
T
gives
,
s
m
348
air
=
v
and
so
air
He
air
H
94
.
2
,
80
.
3
2
v
v
v
v
=
=
and
.
928
.
0
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 Spring '08
 Hickman

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