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Question

A receiver and a source of sonic oscillations of frequency 2000 Hz are located on the x-axis. The source swings harmonically along x-axis with circular frequency ω and amplitude 50 cm. At what approximate value of ω will the frequency bandwidth (difference of maximum and minimum frequency) registered by stationary receiver be equal to 200 Hz? (The velocity of sound in air is 340 m/s and assume that the velocity of source is small compared to the velocity of sound)

A
20 rad/s
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B
34 rad/s
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C
48 rad/s
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D
62 rad/s
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Solution

The correct option is B 34 rad/s
We know
fapp=f0(v±v0v±vs) fmaxfmin=f0(vvvs)f0(vv+vs)=f0v(v+vsv+vsv2v2s)
v>>vsfmaxfmin=fo(2vsv)
200=2vsvf0=2×12×w340×2000
Hence ω=34rads

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