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strong (ams4684) – HW13 – ditmire – (58216)
1
This printout should have 21 questions.
Multiplechoice questions may continue on
the next column or page – fnd all choices
beFore answering.
001
10.0 points
IF the Frequency oF sound is doubled, how will
its speed change?
How will its wavelength
change?
1.
Both speed and wavelength will double.
2.
Its speed will halve, and its wavelength
will double.
3.
Its speed will double, and its wavelength
will not change.
4.
Its speed will not change, and its wave
length will halve.
correct
5.
There will be no change in its speed and
wavelength.
Explanation:
v
=
fλ
= (2
f
)
±
1
2
λ
²
The speed oF sound is constant in a given
medium. IF the Frequency oF sound is doubled,
its speed will not change, but its wavelength
must be “compressed” to halF. The speed oF
sound depends only on the medium through
which it travels, not on its Frequency, wave
length, or intensity (until the intensity is so
great that a shock wave results.)
002
10.0 points
A stereo speaker represented by
P
in the fg
ure emits sound waves with a power output oF
102.3 W.
P
x
13
.
2m
What is the intensity oF the sound waves at
point
x
, 13.2 m away?
Correct answer: 0
.
0467216 W
/
m
2
.
Explanation:
Let :
P
= 102
.
3 W
and
r
= 13
.
2m
.
I
=
P
4
π r
2
=
102
.
3W
4
π
(13
.
2 m)
2
=
0
.
0467216 W
/
m
2
.
003
10.0 points
What is the time lapse between seeing a
lightning strike and hearing the thunder iF the
lightning ±ash is 31 km away? The speed oF
lightwaves in air is 3
×
10
8
m
/
s and the speed
oF sound waves in air is 333 m
/
s.
Correct answer: 93
.
093 s.
Explanation:
Let :
c
=3
×
10
8
m
/
s
,
v
= 333 m
/
s
,
and
d
= 31 km
.
The time For the light to travel 31 km is
t
1
=
d
c
and the time For the sound to travel 31 km is
t
2
=
d
v
2
,
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View Full Document strong (ams4684) – HW13 – ditmire – (58216)
2
so the time lapse is
Δ
t
=
t
2

t
1
=
d
±
1
v
2

1
c
²
= (31 km)
±
1
333 m
/
s

1
3
×
10
8
m
/
s
²
=
93
.
093 s
.
004
10.0 points
At a maximum level of loudness, the power
output of a 75piece orchestra radiated as
sound is 72.0 W.
What is the intensity of these sound waves
to a listener who is sitting 28.8 m from the
orchestra?
Correct answer: 0
.
00690777 W
/
m
2
.
Explanation:
Let :
P
= 72
.
0 W
and
r
= 28
.
8m
.
I
=
P
4
π r
2
=
72 W
4
π
(28
.
8 m)
2
=
0
.
00690777 W
/
m
2
.
005
10.0 points
The speed of sound in water is 1432 m
/
s. A
sonar signal is sent from a ship at a point just
below the water surface and 2
.
25 s later the
reFected signal is detected.
How deep is the ocean beneath the ship?
Correct answer: 1611 m.
Explanation:
Let :
v
= 1432 m
/
s
and
t
=2
.
25 s
.
The sound has to travel to the ocean Foor
and back, so
2
d
=
vt
d
=
vt
2
=
(1432 m
/
s) (2
.
25 s)
2
=
1611 m
.
006
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This note was uploaded on 11/03/2010 for the course PHYSICS 303 taught by Professor Shih during the Spring '10 term at University of Texas at Austin.
 Spring '10
 SHIH

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