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%
Answers
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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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