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A source of sound of frequency 90 vib/sec is coming towards a stationary observer with a speed equal to \frac{1}{10}^{th} the speed of sound. What will be the frequency heard by the observer.

  • Option 1)

    80 vib/sec

  • Option 2)

    90 vib/sec

  • Option 3)

    100 vib/sec

  • Option 4)

    120 vib/sec

 

Answers (1)

best_answer

n' = \frac{v}{v - v_{s}} . n \Rightarrow \frac{v}{v - v/10} . n \Rightarrow n' = \frac{10}{9} X n

n' = \frac{10}{10} X 90 = 100 vib/s

 

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

\nu {}'= \nu _{0}.\frac{C}{C-V_{s}}
 

- wherein

C= speed of sound

V_{s}= speed of source

\nu _{0}= original frequency

\nu {}'= apparent frequency

 

 

 


Option 1)

80 vib/sec

This is incorrect.

Option 2)

90 vib/sec

This is incorrect.

Option 3)

100 vib/sec

This is correct.

Option 4)

120 vib/sec

This is incorrect.

Posted by

divya.saini

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