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Question

A stationary observer receives a sound from a sound of frequency v0 moving with a constant velocity vs=30 m/s. The apparent frequency varies with time as shown in figure. Velocity of sound v = 300 m/s. Then which of the following is incorrect?


A
The minimum value of apparent frequency is 889 Hz.
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B
The natural frequency of source is 1000 Hz.
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C
The frequency-time curve corresponds to a source moving at an angle to the stationary observer.
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D
The maximum value of apparent frequency is 1111 Hz.
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Solution

The correct option is A The minimum value of apparent frequency is 889 Hz.
This frequency–time curve corresponds to a source moving at an angle to a stationary observer.

In the region SN, the source is moving towards the observer, i.e., the apparent frequency
n=n0(vvvs cosθ)
n=n0(30030030 cosθ)
When θ=π2. i.e., at N,
n=n0=1000 Hz, i.e., natural frequency of source. In the region NS’ the source is moving away from the observer, i.e., apparent frequency
n=n0(30030030 cosθ)
When θ=0, i.e., cos θ=1,
nmax=n0vvvs=(1000 Hz)(300 m/s)(300 m/s30 m/s)
=109×1000 Hz=1111 Hz
nmin=n0vv+vs=1000×300330=909 Hz

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