A submarine (A) travelling at 18km/hr is being chased along the line of its velocity by another submarine (B) travelling at 27km/hr. B sends a sonar signal of 500Hz to detect A and receives a reflected sound of frequency v. The value of v is close to : 

(Speed of sound in water =1500ms^{-1}

 

 

  • Option 1)

    504Hz

  • Option 2)

    499Hz

  • Option 3)

    502Hz

  • Option 4)

    507Hz

 

Answers (1)
S solutionqc

 

frequency of sound when source is moving towards the observer and observer is moving away from source -

\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

 

 

We have given that 

                                                            

V_{B}=\frac{5}{18}\times 27                                                                  V_{A}=\frac{5}{18}\times 18=5m/s

Frequency received by A, f_A=f_{0}\left ( \frac{1500-5}{1500-7.5} \right )

Frequency of the reflected sound is heard or received by B=f_A\left ( \frac{1500+7.5}{1500+5} \right )

                                                                                               =f_{0}\left ( \frac{1500-5}{1500-7.5} \right )\left ( \frac{1500+7.5}{1500+5} \right )

=500\times \frac{1495}{1492.5}\times \frac{1507.5}{1505}

=502Hz


Option 1)

504Hz

Option 2)

499Hz

Option 3)

502Hz

Option 4)

507Hz

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