homework 11 – KIM, JI – Due: Nov 14 2007, 4:00 am
1
Question 1, chap 17, sect 2.
part 1 of 1
10 points
An explosive charge is detonated at a height
of several kilometers in the atmosphere.
At
a
distance
of
488
m
from
the
explosion
the acoustic pressure reaches a maximum of
14
.
1 N
/
m
2
.
The density of the air is 1
.
4 kg
/
m
3
and the
speed of sound is 343 m
/
s and sound waves
in air are absorbed at a rate of approximately
9
.
52 dB
/
km.
What will be the sound level at a distance
2
.
53 km from the explosion?
Consider the
atmosphere to be homogeneous over the dis
tances considered.
Correct answer:
79
.
426
dB (tolerance
±
1
%).
Explanation:
Let :
λ
β
= 9
.
52 dB
/
km
,
L
1
= 488 m
,
P
m
= 14
.
1 N
/
m
2
,
ρ
= 1
.
4 kg
/
m
3
,
v
= 343 m
/
s
,
and
L
2
= 2
.
53 km
.
The intensity at
L
1
is
I
1
=
P
2
m
2
ρ v
=
(14
.
1 N
/
m
2
)
2
2 (1
.
4 kg
/
m
3
) (343 m
/
s)
= 0
.
207007 W
/
m
2
.
Since the intensity is inversely proportional
to the distance squared,
I
2
I
1
=
L
2
1
L
2
2
I
2
=
I
1
L
2
1
L
2
2
= (0
.
207007 W
/
m
2
)
(488 m)
2
(2
.
53 km)
2
= 0
.
00770167 W
/
m
2
.
The sound level at
L
2
is
β
2
= 10 log
parenleftbigg
I
2
I
0
parenrightbigg
= 10 log
parenleftbigg
0
.
00770167 W
/
m
2
1
×
10
−
12
W
/
m
2
parenrightbigg
= 98
.
8658 dB
,
so allowing for absorption of the wave over
the distance traveled,
β
′
2
=
β
2

λ
β
(
L
2

L
1
)
= 98
.
8658 dB

(9
.
52 dB
/
km) (2
.
53 km

0
.
488 km)
= 79
.
426 dB
.
Question 2, chap 17, sect 2.
part 1 of 1
10 points
A rock group is playing in a bar.
Sound
emerging from the door spreads uniformly in
all directions. The intensity level of the music
is 102 dB at a distance of 7
.
22 m from the
door.
At what distance is the music just barely
audible to a person with a normal threshold
of hearing? Disregard absorption.
Correct answer: 908944 m (tolerance
±
1 %).
Explanation:
Let :
r
= 7
.
22 m
β
= 102 dB
,
and
I
0
= 1
×
10
−
12
dB
.
I
∝
1
r
2
,
so
I
I
0
=
R
2
0
r
2
.
The decibel level is
β
= 10 log
parenleftbigg
I
I
0
parenrightbigg
= 10 log
parenleftbigg
R
2
0
r
2
parenrightbigg
= 20 log
parenleftbigg
R
0
r
parenrightbigg
,
so
R
0
r
= 10
β/
20
R
0
=
r
10
β/
20
= (7
.
22 m) 10
(102 dB)
/
(20 dB)
=
908944 m
.
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homework 11 – KIM, JI – Due: Nov 14 2007, 4:00 am
2
Question 3, chap 17, sect 2.
part 1 of 1
10 points
A source of sound (1000 Hz) emits uni
formly in all directions. An observer 4
.
18 m
from the source measures a sound level of
62
.
3 dB.
Calculate the average power output of the
source.
Correct answer: 372
.
874
μ
W (tolerance
±
1
%).
Explanation:
Let :
β
= 62
.
3 dB
,
I
0
= 1
×
10
−
12
W
/
m
2
,
and
L
= 4
.
18 m
.
The intensity of the sound is
I
=
I
0
10
β/
10
= (1
×
10
−
12
W
/
m
2
) 10
(62
.
3 dB)
/
(10 dB)
= 1
.
69824
×
10
−
6
W
/
m
2
so the average power output is
P
= 4
π L
2
I
= 4
π
(4
.
18 m)
2
(1
.
69824
×
10
−
6
W
/
m
2
)
= 0
.
000372874 W
=
372
.
874
μ
W
.
Question 4, chap 17, sect 2.
part 1 of 1
10 points
Two sources have sound levels of 65
.
8 dB
and 114 dB.
What is their combined intensity?
Correct answer: 0
.
251192
W
/
m
2
(tolerance
±
1 %).
Explanation:
Let :
β
1
= 65
.
8 dB
and
β
2
= 114 dB
.
The intensities of the sources are
I
1
=
I
0
10
β
1
/
10
= 1
×
10
−
12
W
/
m
2
10
(65
.
8 dB)
/
(10 dB)
= 3
.
80189
×
10
−
6
W
/
m
2
I
2
=
I
0
10
β
2
/
10
= 1
×
10
−
12
W
/
m
2
10
(114 dB)
/
(10 dB)
= 0
.
251189 W
/
m
2
.
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 Fall '08
 Turner
 Physics, Charge, Work, Correct Answer, m/s, Pressure measurement

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