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Question

A stationary observer receives a sound from a source of frequency 2000 Hz moving with constant velocity. The apparent frequency varies with time as shown in the figure.

The value of fmax is (2300 - 10x) Hz. Then the value of x is(Take speed of sound = 300 m/s and neglect the time taken by sound to reach the stationary observer)

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Solution

We know that apparent frequency,
fm=(vvo)f(vvscosθ)
where v,vo,vs are velocities of sound, observer and source respectively.

Maximum apparent frequency is at θ minimum i.e., 0,
fmax=(v0)fvvscos0=vf(vvs).

Minimum apparent frequency is at θ maximum i.e., 180, fmin=(v0)fvvscos180=vf(v+vs).

1fmax=[2f1fmin]=[2200011800]

fmax=2250=230050=230010(5) thus x = 5.

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