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Question

An observer moves towards a stationary source of sound, with a velocity one-fifth of the velocity of sound. What is the percentage increase in the apparent frequency?


A
Zero
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B
0.5%
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C
5%
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D
20%
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Solution

The correct option is D $$20\%$$
Let velocity of sound in air be $$v$$.  
$$\Rightarrow $$ Velocity of observer, $$v_{observer}  = \dfrac{v}{5}$$
Let $$\nu$$ be the original frequency of the sound source and $$\nu'$$ be the apparent frequency heard by the moving observer.
Using Doppler effect when observer is moving towards the stationary source:
$$\nu' = \nu \bigg[\dfrac{v_{sound} + v_{observer}}{v_{sound}} \bigg]$$ $$= \nu \bigg[\dfrac{v +\frac{v}{5}}{v} \bigg]   = \dfrac{6}{5}\nu$$
Percentage increase in frequency $$ = \dfrac{\nu' - \nu}{\nu} \times 100   = \dfrac{\frac{6}{5}\nu - \nu}{\nu} \times 100   = 20$$ %

Physics

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