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
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '08
 HKMIET
 Hertz, m/s, Hz

Click to edit the document details