JEE Advance - Physics (2021 - Paper 2 Online - No. 2)
A source, approaching with speed u towards the open end of a stationary pipe of length L, is emitting a sound of frequency fs. The farther end of the pipe is closed. The speed of sound in air is v and f0 is the fundamental frequency of the pipe. For which of the following combination(s) of u and fs, will the sound reaching the pipe lead to a resonance?
u = 0.8v and fs = f0
u = 0.8v and fs = 2f0
u = 0.8v and fs = 0.5f0
u = 0.5v and fs = 1.5f0
Explanation
Natural frequency of closed pipe,
f = (2n + 1)f0
f0 is fundamental frequency
n = 0, 1, 2 .......
Frequency of source received by pipe,
$$f' = {f_s}\left[ {{{v - 0} \over {v - u}}} \right]$$
For resonance,
$$f' = f$$
$${f_s}\left[ {{v \over {v - u}}} \right] = (2n + 1){f_0}$$
If u = 0.8v, fs = f0
$$f' = {v \over {0.2v}}{f_0} = 5{f_0}$$
For n = 2 pipe can be in resonance
Hence, option (a) is correct.
If u = 0.8v, fs = 2f0
$$f' = {v \over {0.2v}} \times 2{f_0} = 10{f_0}$$
If u = 0.8v, fs = 0.5f0
$$f' = {v \over {0.2v}} \times 0.5{f_0} = 2.5{f_0}$$
Not possible.
If u = 0.5v, fs = 1.5f0
$$f' = {v \over {0.5v}} \times 1.5{f_0} = 3{f_0}$$
For n = 1 f = 3f0
Pipe can be in resonance.
Hence, option (d) is correct.
f = (2n + 1)f0
f0 is fundamental frequency
n = 0, 1, 2 .......
Frequency of source received by pipe,
$$f' = {f_s}\left[ {{{v - 0} \over {v - u}}} \right]$$
For resonance,
$$f' = f$$
$${f_s}\left[ {{v \over {v - u}}} \right] = (2n + 1){f_0}$$
If u = 0.8v, fs = f0
$$f' = {v \over {0.2v}}{f_0} = 5{f_0}$$
For n = 2 pipe can be in resonance
Hence, option (a) is correct.
If u = 0.8v, fs = 2f0
$$f' = {v \over {0.2v}} \times 2{f_0} = 10{f_0}$$
If u = 0.8v, fs = 0.5f0
$$f' = {v \over {0.2v}} \times 0.5{f_0} = 2.5{f_0}$$
Not possible.
If u = 0.5v, fs = 1.5f0
$$f' = {v \over {0.5v}} \times 1.5{f_0} = 3{f_0}$$
For n = 1 f = 3f0
Pipe can be in resonance.
Hence, option (d) is correct.
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