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164
Chapter 16
(c)
25
2
area = (5.00 mm)
2.5 10 m .
−
=×
Part (a):
12
2
5
2
17
(1 10
W/m )(2.5 10 m )
2.5 10
J/s.
−−
−
××
=
×
Part (b):
32
5
2
8
(3.2 10 W/m )(2.5 10 m )
8.0 10 J/s.
−
=
×
EVALUATE:
For faint sounds the displacement and pressure variation amplitudes are very small. Intensities for
audible sounds vary over a very wide range.
16.15.
IDENTIFY:
Apply Eq.(16.12) and solve for
A
.
/ ,
vf
λ
=
with
/.
vB
ρ
=
SET UP:
2.
f
ω
π
=
For air,
5
1.42 10 Pa.
B
EXECUTE:
(a)
The amplitude is
62
11
2
39
2
22
(
3
.
0
0
1
0
W
m
)
9.44 10
m.
(1000 kg m )(2.18 10 Pa)(2 (3400 Hz))
Ι
A
ρΒω
π
−
−
×
==
=
×
×
The wavelength is
93
(2.18 10 Pa) (1000 kg m )
0.434 m.
3400 Hz
B
ρ
v
ff
×
=
=
(b)
Repeating the above with
5
1.42 10 Pa
B
and the density of air gives
9
5.66 10 m and
0.100 m.
A
−
=
EVALUATE:
(c)
The amplitude is larger in air, by a factor of about 60. For a given frequency, the much less dense
air molecules must have a larger amplitude to transfer the same amount of energy.
16.16.
IDENTIFY
and
SET UP:
Use Eq.(16.7) to eliminate either
v
or
B
in
2
max
.
2
vp
I
B
=
EXECUTE:
From Eq. (19.21),
2
.
=
Using Eq.(16.7) to eliminate
( )
max
max
,2
2
.
vI
B
ρ
pB
p
ρ
B
Using Eq. (16.7) to eliminate
B
,
222
max
max
2(
)
2
.
Iv
p
v
ρ
p
ρ
v
EVALUATE:
The equation in this form shows the dependence of
I
on the density of the material in which the
wave propagates.
16.17.
IDENTIFY
and
SET UP:
Apply Eqs.(16.5), (16.11) and (16.15).
EXECUTE:
(a)
2
(2 rad)(150 Hz)
942.5 rad/s
f
ππ
=
2
2
942.5 rad/s
2.74 rad/m
344 m/s
f
k
vv
ππ ω
=
=
=
5
1.42 10 Pa
B
(Example 16.1)
Then
56
max
(1.42 10 Pa)(2.74 rad/m)(5.00 10 m) 1.95 Pa.
k
A
−
×
×
=
(b)
Eq.(16.11):
2
1
2
IB
k
A
=
2
3
2
1
2
(942.5 rad/s)(1.42 10 Pa)(27.4 rad/m)(5.00 10 m)
4.58 10 W/m .
I
×
=
×
(c)
Eq.(16.15):
0
(10 dB)log( /
),
II
β
=
with
12
2
0
11
0 W
/m.
I
−
( )
1
(10 dB)log (4.58 10 W/m )/(1 10
W/m )
96.6 dB.
×
=
EVALUATE:
Even though the displacement amplitude is very small, this is a very intense sound. Compare the
sound intensity level to the values in Table 16.2.
16.18.
IDENTIFY:
Apply
0
(10 dB)log( /
).
=
In part (b), use Eq.(16.14) to calculate
I
from the information that is
given.
SET UP:
12
2
0
10
W/m .
I
−
=
From Table 16.1 the speed of sound in air at 20.0 C
°
is 344 m/s. The density of air at
that temperature is
3
1.20kg m .
EXECUTE:
(a)
2
12
2
0.500 W/m
(10 dB) log
57 dB.
10
W/m
β
μ
−
⎛⎞
⎜⎟
⎝⎠
(b)
2
52
max
3
(0.150 N m )
2.73 10 W m .
2
2(1.20 kg m )(344 m s)
p
I
ρ
v
−
=
×
Using this in Equation (16.15),
12
2
2.73 10 W m
(10 dB) log
74.4 dB.
10
W m
β
−
−
×
EVALUATE:
As expected, the sound intensity is larger for the jack hammer.
16.19.
IDENTIFY:
Use Eq.(16.13) to relate
I
and
max
.
p
0
(10 dB)log( /
).
=
Eq.(16.4) says the pressure amplitude and
displacement amplitude are related by
max
2
.
f
p
BkA
B
A
v
SET UP:
At 20 C
°
the bulk modulus for air is
5
×
and
344 m/s.
v
=
12
2
0
I
−
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View Full DocumentSound and Hearing
165
EXECUTE:
(a)
25
2
12
2
max
5
(344 m/s)(6.0 10 Pa)
4.4 10
W/m
22
(
1
.
4
2
1
0
P
a
)
vp
I
B
−
−
×
==
=
×
×
(b)
12
2
12
2
4.4 10
W/m
(10 dB)log
6.4 dB
11
0 W
/m
β
−
−
⎛⎞
×
⎜⎟
×
⎝⎠
(c)
5
11
max
5
5.8 10
m
2
2 (400 Hz)(1.42 10 Pa)
vp
A
fB
ππ
−
−
×
=
×
×
EVALUATE:
This is a very faint sound and the displacement and pressure amplitudes are very small. Note that the
displacement amplitude depends on the frequency but the pressure amplitude does not.
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This note was uploaded on 07/16/2011 for the course PHY 2053 taught by Professor Buchler during the Spring '06 term at University of Florida.
 Spring '06
 Buchler
 Physics

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