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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
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B
170 ms1
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C
85 ms1
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D
42.5 ms1
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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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