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CHAPTER 16 – Sound
5. If we let
L
1
represent the thickness of the top layer, the total transit time is
t
= (
L
1
/
v
1
) + [(
L
–
L
1
)/
v
2
)];
4.5 s = [
L
1
/(331 m/s)] + [(1500 m –
L
1
)/(343 m/s)], which gives
L
1
=
1200 m
.
Thus the bottom layer is 1500 m – 1200 m =
300 m
.
7.
We find the displacement amplitude from
Δ
P
M
= 2
π
ρ
vD
M
f
.
(
a
) For the frequency of 100 Hz, we have
3
.
0
×
10
–3
Pa = 2
π
(1.29 kg/m
3
)(331 m/s)
D
M
(100 Hz), which gives
D
M
=
1.1
×
10
–8
m
.
(
b
) For the frequency of 10 kHz, we have
3
.
0
×
10
–3
Pa = 2
π
(1.29 kg/m
3
)(331 m/s)
D
M
(10
×
10
3
Hz), which gives
D
M
=
1.1
×
10
–10
m
.
11. (
a
) We find the intensity level from
β
= 10 log
10
(
I
/
I
0
) = 10 log
10
[(8.5
×
10
–8
W/m
2
)/(10
–12
W/m
2
)] =
49 dB
.
(
b
) We find the intensity from
= 10 log
10
(
I
/
I
0
);
25 dB = 10 log(
I
/10
–12
W/m
2
), which gives
I
=
3.2
×
10
–10
W/m
2
.
21. The intensity with one firecracker is onehalf that of two firecrackers.
Thus the change in intensity level is
2
–
1
= 10 log
10
(
I
2
/
I
1
);
2
– 90 dB = 10 log
10
(
½
), which gives
2
=
87 dB
.
25. (
a
) We find the intensity of the sound from
= 10 log
10
(
I
/
I
0
);
130 dB = 10 log(
I
/10
–12
W/m
2
), which gives
I
= 10 W/m
2
.
The power output of the speaker is
P
=
IA
=
I
4
π
r
2
= (10 W/m
2
)4
π
(3.4 m)
2
=
1.5
×
10
3
W
.
(
b
) We find the intensity of the sound from
= 10 log
10
(
I
/
I
0
);
90 dB = 10 log(
I
/10
–12
W/m
2
), which gives
I
= 1.0
×
10
–3
W/m
2
.
We find the distance from
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 Spring '08
 HKMIET

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