A source of sound with natural frequency `f_0=1800Hz` moves uniformly along a straight line separated from a stationary observer by a distance `l=250m`. The velocity of the source is equal to `eta=0.80` fraction of the velocity of the sound.
Q. Find the frequency of osund received by the observer at the moment when the source gets closest to him.

Options

A. 2000Hz
B. 6000Hz
C. 3000Hz
D. 5000Hz

Answer

C

Explanation

Let the source be moving along the strainght line AC and observer be located at O, as shown. Let the velocity of sound in air be v. The velocity of source is `etav`.
Let the sound wave received by the observer at the moment when the source is closest to the observer (at C) be emitted by the source when it was at point A.
Therefore, by the time source travels from A to C, the sound wave travels from A to O. If this interval is t,`AC=etact` and `AO=vt`. Velocity of approach of source when it is at A.
`v_S=(etav)costheta=(etav)((AC)/(AO))=etav((etavt)/(vt))=eta^2v`
When the sound wave emitted by the source at A reaches the staionary observer at O, it will receive the frequency
`f=f_0((v)/(v-v_S))=f_0((v)/(v-n^2v))`
`=(f_0)/(1-eta^2)=(1800)/(1=(0.8)^2)=5000Hz`