pokharel (pp7242) – Sound – walther – (16180)
1
This printout should have 24 questions.
Multiplechoice questions may continue on
the next column or page – fnd all choices
beFore answering.
Good luck, this is your home work on Ch.
12
C± Walther
001
10.0 points
Consider the Following:
A) the speed oF the source
B) the speed oF the observer
C) the loudness oF the sound
In the Doppler e²ect For sound waves, which
Factors a²ect the Frequency that the observer
hears?
1.
A and B only
correct
2.
C only
3.
B and C only
4.
None oF these.
5.
A, B, and C
6.
A only
7.
A and C only
8.
B only
Explanation:
In the Doppler e²ect oF sound waves, both
the speed oF the source and the speed oF the
observer a²ect the Frequency that the observer
hears. (The directions oF the movement are
Factors, too.)
On the other hand, the loudness oF the
sound (the intensity oF the sound wave) is
irrelevant.
002
10.0 points
A small vibrating object on the surFace oF a
ripple tank is the source oF waves oF Frequency
20 Hz and speed 60 cm/s. The source
S
is
moving to the right with speed 20 cm/s, as
shown.
S
C
A
B
D
At which oF the labeled points will the Fre
quency measured by a stationary observer be
greatest?
1.
It will be the same For all Four points.
2.
B
3.
C
4.
A
correct
5.
D
Explanation:
The Doppler e²ect For stationary observers
is
f
′
=
v
sound
v
sound
−
v
source
f
0
.
Since the source is moving directly toward
point
A
, the measured Frequency at point
A
by a stationary observer will be greatest.
003
(part 1 oF 2) 10.0 points
An
ambulance
is traveling south at
62
.
9 m
/
s, away From a car that is traveling
north at 30
.
5 m
/
s.
The ambulance driver
hears his siren at a Frequency oF 535 Hz.
Ambulance
62
.
9 m
/
s
30
.
5 m
/
s
Car
What wavelength does a person who is
standing between the car and the ambulance
detect From the sound oF the ambulance’s
siren? The velocity oF sound in air is 343 m
/
s.
Correct answer: 0
.
758692 m.
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2
Explanation:
Let :
v
amb
= 62
.
9 m
/
s
,
f
0
= 535 Hz
,
and
v
= 343 m
/
s
.
From the Doppler shift,
f
′
=
v
sound
±
v
observer
v
sound
∓
v
source
f .
The ambulance moves away from the car,
so
f
′
=
v
v
+
v
amb
f
0
and
v
=
f
′
λ
′
λ
′
=
v
f
′
=
v
+
v
amb
f
0
=
343 m
/
s + 62
.
9 m
/
s
535 Hz
=
0
.
758692 m
.
The wave form is stretched out, which re
sults in a longer wavelength.
004
(part 2 of 2) 10.0 points
At what frequency does the driver of the car
hear the ambulance’s siren?
Correct answer: 411
.
893 Hz.
Explanation:
Let :
v
car
= 30
.
5 m
/
s
.
The car moves away from the ambulance
and the ambulance moves away from the car,
so
f
′
=
v
−
v
car
v
+
v
amb
f
0
=
p
343 m
/
s
−
30
.
5 m
/
s
343 m
/
s + 62
.
9 m
/
s
P
(535 Hz)
=
411
.
893 Hz
.
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 Fall '09
 KLEINMAN
 Work, Wavelength, Decibel, Correct Answer, superposed waves

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