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  • Stuck here, help me understand: A source of sound is moving with constant velocity of 20 m/s emitting a note of frequency 1000 Hz. The ratio of frequencies observed by a stationary observer while the source is approaching him and after it crosses him

A source of sound is moving with constant velocity of 20 m/s emitting a note of frequency 1000 Hz. The ratio of frequencies observed by a stationary observer while the source is approaching him and after it crosses him will be (speed of sound is 340 m/s)

  • Option 1)

    9:8

  • Option 2)

    8:9

  • Option 3)

    1:1

  • Option 4)

    9:10

 

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_{s}}
 

- wherein

C= speed of sound

V_{s}= speed of source

\nu _{0}= original frequency

\nu {}'= apparent frequency

 

 When source is approaching, frequency heard will be \nu_{1}=\nu_{0}.\frac{340}{340+20}

Frequency heard when it crosses the observer:\nu_{2}=\nu_{0}.\frac{340}{340-20}

\therefore Ratio of frequencies = \frac{\nu_{1}}{\nu_{2}}

                                     =\frac{\nu_{0}- \frac{340}{360}}{\nu_{0}.\frac{340}{320}}=\frac{8}{9}

 


Option 1)

9:8

This option is correct.

Option 2)

8:9

This option is incorrect.

Option 3)

1:1

This option is incorrect.

Option 4)

9:10

This option is incorrect.

Posted by

Aadil

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