Question

The frequency changes by 10% as a sound source
approaches a stationary observer with constant
speed Vs What would be percentage change in
the frequency as the source recedes the observer
with same speed (Vs < V)?
(1) 10.5%
(2) 8.5%
(3) 4.5%
(4) 1.5%
​

Answer 1

The percentage change in  the frequency is f' − f / f = − 8.5 %

Explanation:

Case I: (Source approaches a stationary observer)

f'  =  (v / v−vs)f

f' / f = v / v−vs

f' / f − 1 = v / v−vs − 1

f' − f / f = vs / v − vs

0.1 = vs / v−vs

vs = 0.1 v − 0.1 vs

1.1 vs = 0.1 v

v = 11 vs

Case II: (Source recedes the stationary observer)

f' = (v / v+vs) f

f' / f = v / v + vs

f' / f − 1 = v / v+vs − 1

f' − f / f = − vs / v + vs

f' − f / f =−vs / 11 vs + vs

f' − f / f = − 8.5 %

Thus the percentage change in  the frequency is f' − f / f = − 8.5 %

Also learn more

Two sound sources of frequency 360 Hz each,  one moving toward observer, while second moving  away from observer with same speed 5.5 m/s, then  number of beats produced per second is (v = 330  m/s)

(1) 2  (2) 18  (3) 12  (4) 25​

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