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  • How to solve this problem- A train is moving on a straight track with speed 20 ms -1. It is blowing its whistle at the frequency of 1000 Hz. The percentage change in the frequency heard by a person standing near the track as the train passes him is (

 A train is moving on a straight track with speed 20 ms -1. It is blowing its whistle at the frequency of 1000 Hz. The percentage change in the frequency heard by a person standing near the track as the train passes him is (speed of sound = 320 ms-1) close to :

 

  • Option 1)

    6%

  • Option 2)

    12 %

  • Option 3)

    18 %

  • Option 4)

    24 %

 

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_{\Delta }}
 

- wherein

C= speed of sound

V_{\Delta }= speed of source

\nu _{0}= original frequency

\nu {}'= apparent frequency

 

 

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

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

- wherein

C= speed of sound

V_{\Delta }= speed of source

\nu _{0}= original frequency

\nu {}'= apparent frequency

 

 Frequency of sound emitted by train \upsilon=1000\ Hz

Speed of train \upsilon_{s}=20\ ms^{-1}

Speed of sound \upsilon=320\ ms^{-1}

Observer is stationary 

\upsilon_{1}=\left(\frac{v}{v-v_{s}} \right )\upsilon=\left(\frac{320}{320-20} \right )\times 1000=\frac{32000}{3}\ Hz

Frequency heard by Person as train moves away from him 

\upsilon_{2}=\left(\frac{v}{v+v_{s}} \right )\upsilon=\left(\frac{320}{320+20} \right )\times 1000=\frac{32000}{34}\ Hz

\therefore Percentage change in frequency =\left(\frac{\upsilon_{1}-\upsilon_{2}}{\upsilon_{1}} \right )\times100

    =\frac{\left(\frac{3200}{34}-\frac{3200}{3} \right )}{\frac{3200}{3}}\times100

    \simeq 12 %

Correct option is 2.

 

 


Option 1)

6%

This is an incorrect option.

Option 2)

12 %

This is the correct option.

Option 3)

18 %

This is an incorrect option.

Option 4)

24 %

This is an incorrect option.

Posted by

prateek

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