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

  • Option 1)

    5%

  • Option 2)

    20%

  • Option 3)

    0%

  • Option 4)

    0.5%

 

Answers (2)
V Vakul

n' = \left [ \frac{v+v_{0}}{v} \right ]n\Rightarrow \left [ \frac{v+v/5}{v} \right ]n = 6/5n

\Rightarrow 1.2n

Increament in frequency = 0.2

50% increament \frac{0.2n}{n}\times 100= 20 \%

 

Frequency when observer is stationary and source is moving away from 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)

5%

This is incorrect

Option 2)

20%

This is correct

Option 3)

0%

This is incorrect

Option 4)

0.5%

This is incorrect

V Vakul

n' = \left [ \frac{v+v_{0}}{v} \right ]n\Rightarrow \left [ \frac{v+v/5}{v} \right ]n = 6/5n

\Rightarrow 1.2n

Increament in frequency = 0.2

50% increament \frac{0.2n}{n}\times 100= 20 \%

 

Frequency when observer is stationary and source is moving away from 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)

5%

This is incorrect

Option 2)

20%

This is correct

Option 3)

0%

This is incorrect

Option 4)

0.5%

This is incorrect

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