Nguyen, Thanh – Homework 20 – Due: Nov 16 2007, 7:00 pm – Inst: D Weathers
1
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printout
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have
9
questions.
Multiplechoice questions may continue on
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before answering.
The due time is Central
time.
001
(part 1 of 1) 10 points
It is found that a 5
.
2 m segment of a long
string contains 3 complete wavelengths and
has a mass of 170 g. At one point it is vibrat
ing sinusoidally with a frequency of 51 Hz and
a peak to valley displacement of 0
.
95 m.
What power is being transmitted along the
string?
Correct answer: 33477
.
7 W.
Explanation:
Let :
l
= 5
.
2 m
,
λ
=
5
.
2 m
3
,
m
= 170 g
,
f
= 51 Hz
,
and
A
=
0
.
95 m
2
.
Since
k
≡
2
π
λ
,
ω
≡
2
π f
,
v
≡
ω
k
,
P
=
1
2
μ ω
2
A
2
v
=
1
2
‡
m
l
·
(2
π f
)
2
A
2
λ f
=
1
2
170 g
5
.
2 m
¶
[2
π
(51 Hz)]
2
×
0
.
95 m
2
¶
2
5
.
2 m
3
¶
(51 Hz)
=
33477
.
7 W
.
keywords:
002
(part 1 of 2) 10 points
A point source emits sound waves with an
average power output of 0
.
907 W.
Find the intensity 2
.
25 m from the source.
Correct answer: 0
.
0142571 W
/
m
2
.
Explanation:
A point source emits energy in the form of
a spherical waves.
At a distance
R
from
the source, the power is distributed over the
surface area of a sphere, 4
πR
2
. Therefore, the
intensity at a distance
R
= 2
.
25 m from the
source is
I
=
P
av
4
π R
2
= 0
.
0142571 W
/
m
2
.
003
(part 2 of 2) 10 points
Find the distance at which the sound reduces
to a level of 69 dB.
Correct answer: 95
.
3232 m.
Explanation:
We can find the intensity at the
β
= 69 dB
level by using the equation
β
= 10 log
I
I
0
¶
with
I
0
= 1
×
10

12
W
/
m
2
:
69 dB = 10 log
I
1
×
10

12
W
/
m
2
¶
therefore
I
= 10
β/
10
I
0
= 10
6
.
9
(1
×
10

12
W
/
m
2
)
= 7
.
94328
×
10

6
W
/
m
2
.
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 Winter '08
 Weathers
 Work, Light, Frequency, Wavelength, Thanh

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