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Question: A car whistling a frequency of \(f\) is approaching a stationary wall with velocity \(v\). The veloc...

A car whistling a frequency of ff is approaching a stationary wall with velocity vv. The velocity of sound in air is cc. The number of beats heard by the driver per second is
A. cfc−v\dfrac{cf}{c-v}
B. 2vfc+v\dfrac{2vf}{c+v}
C. cfc+v\dfrac{cf}{c+v}
D. 2vfc−v\dfrac{2vf}{c-v}

Explanation

Solution

When a source of wave is moving with respect to an observer, there is an apparent shift in the frequency of the wave as observed by the observer. This effect of apparent change in frequency of the wave is known as Doppler’s effect. The apparent shift in frequency can be calculated by using Doppler shift formula.
Number of beats per second is equivalent to the difference in frequency between the two interfering waves.

Formula used: Doppler shift formula, f′=(v±vov±vs)f0f'=\left( \dfrac{v\pm {{v}_{o}}}{v\pm {{v}_{s}}} \right){{f}_{0}}

Complete step by step answer:
Doppler effect, also known as Doppler shift, is the phenomenon that is observed when the source of a wave is moving with respect to an observer.
If f0{{f}_{0}} and v are the frequency and velocity of the wave produced by the source, then the apparent frequency is given by
f′=(v±vov±vs)f0f'=\left( \dfrac{v\pm {{v}_{o}}}{v\pm {{v}_{s}}} \right){{f}_{0}}
vo{{v}_{o}} is the velocity of observer and is taken positive if the observer moves towards the source and negative if it moves away from the source
vs{{v}_{s}} is the velocity of source and it is positive if the source moves away from the observer and negative if it moves in the opposite direction.
In this question, velocity of sound is cc, frequency of sound wave produced is ff and velocity of observer and source is +v+v and −v-v respectively. Therefore,
f′=c+vc−vff'=\dfrac{c+v}{c-v}f
Number of beats per second is equivalent to the difference in frequency between the two interfering waves. Therefore, number of beats heard by driver is
f′−f0=f(c+vc−v−1)=2vfc−vf'-{{f}_{0}}=f\left( \dfrac{c+v}{c-v}-1 \right)=\dfrac{2vf}{c-v}

So, the correct answer is “Option A”.

Note: If the motion of the source of the wave relative to the observer is towards it then Doppler shift is positive i.e. apparent frequency is higher than actual frequency. If the source moves in the opposite direction with respect to the observer, there is a downward shift in observed frequency.
It must be noted here that Doppler shift is not due to real shift in frequency of the source.