Question

# The apparent frequency of the whistle of an engine changes by the ratio 5/3 as the engine passes a stationary observer. If the velocity of sound is 340 m$$s^{-1}$$, then the velocity of the engine is

A
340 ms1
B
170 ms1
C
85 ms1
D
42.5 ms1

Solution

## The correct option is D 85 m$$s^{-1}$$Given : Velocity of sound     $$v_{sound} = 340 m/s$$Let the frequency of the whistle be $$n$$ and velocity of the engine be $$v$$.Let the apparent frequencies heard by the observer be  $$n'$$  and $$n''$$ before and after the source passes the observer.   Doppler effect when source   $$(S')$$   approaches the stationary observer:$$n' = n \bigg[\dfrac{v_{sound} }{v_{sound} - v_{source}}\bigg]$$ $$= n \bigg[\dfrac{340}{340 - v}\bigg]$$Doppler effect when source  $$S$$   moves away from the stationary observer:$$n'' = n \bigg[\dfrac{v_{sound} }{v_{sound} + v_{source}}\bigg]$$ $$= n \bigg[\dfrac{340}{340 + v}\bigg]$$Now      $$\dfrac{n'}{n''} = \dfrac{5}{3}$$       (Given)$$\implies \dfrac{340 + v}{340 - v} = \dfrac{5}{3}$$Thus   $$v = 85 m/s$$Physics

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