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JEE Advanced 2017 Paper 1
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© examsnet.com
Question : 11
Total: 54
A stationary source emits sound of frequency
f
0
=
492
Hz. The sound is reflected by a large car approaching the source with a speed of
2
m
s
−
1
.The reflected signal is received by the source and superposed with the original. What will be the beat frequency of the resulting signal in Hz? (Given that the speed of sound in air is
330
m
s
−
1
and the car reflects the sound at the frequency it has received).
Your Answer:
Validate
Solution:
f
0
•
–
–
–
–
–
▸
2
m
/
s
◂
–
–
–
•
S
v
=
0
O
Frequency received by car
=
[
330
+
2
330
]
492
=
332
330
×
492
=
495
Hz
Frequency received by source
=
330
330
−
2
×
495
=
498
Hz
Original frequency
f
1
=
492
Hz
Final frequency
f
2
=
498
Hz
Beat frequency
=
|
f
1
−
f
2
|
=
6
Hz
© examsnet.com
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