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JEE Advanced 2021 Paper 2
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Section:
Physics
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© examsnet.com
Question : 2
Total: 57
A source, approaching with speed
u
towards the open end of a stationary pipe of length
L
, is emitting a sound of frequency
f
s
−
. The farther end of the pipe is closed. The speed of sound in air is
v
and
f
0
is the fundamental frequency of the pipe. For which of the following combination(s) of
u
and
f
s
, will the sound reaching the pipe lead to a resonance ?
[JEE Adv 2021 P2]
u
=
0.8
v
and
f
s
=
f
0
u
=
0.8
v
and
f
s
=
2
f
0
u
=
0.8
v
and
f
s
=
0.5
f
0
u
=
0.5
v
and
f
s
=
1.5
f
0
Validate
Solution:
From Doppler's effect
f
=
f
s
(
v
v
−
u
)
As the pipe is closed at one end
∴
For resonance,
f
=
f
0
(
v
v
−
u
)
=
n
f
0
where
n
=
odd integer
(a) For,
u
=
0.8
v
and
f
s
=
f
0
,
f
=
f
0
(
v
v
−
0.8
v
)
=
5
f
0
(b) For,
u
=
0.8
v
and
f
s
=
2
f
0
,
f
=
2
f
0
(
v
v
−
0.8
v
)
=
10
f
0
(c) For,
u
=
0.8
v
and
f
s
=
0.5
f
0
f
=
0.5
f
0
(
v
v
−
0.8
v
)
=
2.5
f
0
(d) For
u
=
0.5
v
and
f
s
=
1.5
f
0
f
=
1.5
f
0
(
v
v
−
0.5
v
)
=
3
f
0
© examsnet.com
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