Question

When a source of sound approaches a stationary observer with velocity $v_s$, then the apparent frequency observed by the observer will be $(v$ = velocity of sound$)$ :

• A. $\frac{v_s}{v-v_s}n$
• B. $\frac{v-v_s}{v}n$
• C. $\frac{v+v_s}{v}n$
• D. $\frac{v}{v-v_s}n$

Answer 1

The correct option is D $\frac{v}{v-v_s}n$

Let,
The distance between source and observer is $x$.

The frequency of the vibration of source is $n$

Hence, interval of compression pulses from source is $T=\frac{1}{n}$

Distance traveled by source in time $T$ is $v_{s}T$

Distance of emission of second compression is $x-v_{s}T$

So, time taken by first pulse to reach the observer is $t_1=\frac{x}{v}$

Time taken by next pulse to reach the observer is $t_2=T+\frac{x-v_{s}T}{v}$

Therefore time interval between constructive pulses detected by observer is

$T^{\prime }=t_2-t_1$

$T^{\prime }=T+\frac{x-v_{s}T}{v}-\frac{x}{v}$

$T^{\prime }=(1-\frac{v_s}{v})T=\frac{v-v_s}{v}T$

The apparent frequency detected by the observer is

$n^{\prime }=\frac{1}{T^{\prime }}$

$n^{\prime }=\frac{v}{v-v_s}n$