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A source of sound is travelling with velocity 30 m/s towards a stationary observer. If actual frequency of source is 1000 Hz and wind velocity is 20 m/s in a direction 60\degree with the direction of motion of source, then the apparent frequency heard by observer if speed of sound is 340 m/s

  • Option 1)

    1011Hz

  • Option 2)

    1000Hz

  • Option 3)

    1094Hz

  • Option 4)

    1086Hz

 

Answers (1)

v^{'}= \left [ \frac{c+v_{w}}{c+v_{w}- v_{s}} \right ] v_{0}

v^{'}= \left [ \frac{340+20\cos 60\degree}{340+20\cos 60\degree- 30} \right ] 1000 = 1094Hz

 

Frequency of sound when source and observer are moving towards each other -

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

- wherein

C= Speed of sound

V_{0}= Speed of observer

V_{s}= speed of source

\nu _{0 }= Original frequency

\nu {}'= apparent frequency

 

 

 


Option 1)

1011Hz

This is incorrect

Option 2)

1000Hz

This is incorrect

Option 3)

1094Hz

This is correct

Option 4)

1086Hz

This is incorrect

Posted by

Vakul

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