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  • How to solve this problem- - Oscillations and Waves - JEE Main

A whistle producing sound waves of frequencies 9500 Hz and above is approaching a stationary person with speed \nu ms-1 The velocity of sound in air is 300 ms-1. If the person can hear frequencies upto a maximum of 10000 Hz , the maximum value of \nu upto which he can hear the whistle is

  • Option 1)

    30ms^{-1}

  • Option 2)

    15\sqrt{2}ms^{-1}

  • Option 3)

    15/\sqrt{2}ms^{-1}

  • Option 4)

    15 ms^{-1}

 

Answers (1)

best_answer

As we learnt in

Frequency of sound when observer is stationary and source is moving towards observer -

\nu {}'= \nu _{0}.\frac{C}{C-V_{\Delta }}
 

- wherein

C= speed of sound

V_{\Delta }= speed of source

\nu _{0}= original frequency

\nu {}'= apparent frequency

 

 

 

\frac{\upsilon '}{\upsilon }=\frac{v_{s}}{v_{s}-v}

Where v_{s} is the velocity of sound in air.

\frac{10000}{9500}=\frac{300}{300-v}

\Rightarrow (300-v)=285\Rightarrow v=15\; m/s

Correct option is 4.


Option 1)

30ms^{-1}

This is an incorrect option.

Option 2)

15\sqrt{2}ms^{-1}

This is an incorrect option.

Option 3)

15/\sqrt{2}ms^{-1}

This is an incorrect option.

Option 4)

15 ms^{-1}

This is the correct option.

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