Question

A train is moving on a straight track with speed $20m{{s}^{-1}}$. It is blowing its whistle at the frequency of 1000Hz. The percentage change in the frequency heard by a person standing near the track as the train passes him is (speed of sound $=320m{{s}^{-1}}$) close to:
A. 12%
B. 18%
C. 24%
D. 60%

Answer 1

Hint: As a first step you could note down all the given quantities from the question. Then, you could recall the expression for observed frequency due to Doppler shift. Then, you could find the observed frequency for both cases when the source is receding and approaching the observer. Then you could accordingly find the percentage change in frequency heard.

Formula used:
Expression for Doppler shift,
${{f}_{0}}={{f}_{s}}\left( \dfrac{v\pm {{v}_{0}}}{v\pm {{v}_{s}}} \right)$

Complete Step by step solution:
In the question, we are given a train that is moving on a track at a speed of $20m{{s}^{-1}}$ that is blowing its whistle at a frequency of 1000Hz. We are supposed to find the percentage change in frequency that is heard by a person standing near the railway track as the train passes him. We are given the speed of sound as $320m{{s}^{-1}}$.

This question is an example of Doppler shift. Doppler shift is nothing but the change in frequency of a wave that is observed due to the relative motion between the source and the receiver.

The Doppler shift so observed is given by,
${{f}_{0}}={{f}_{s}}\left( \dfrac{v\pm {{v}_{0}}}{v\pm {{v}_{s}}} \right)$
Where,
${{f}_{0}}$ is the observed frequency,
${{f}_{s}}$ is actual frequency of the sound wave
$v$ is the speed of sound waves
${{v}_{0}}$ is the velocity of the receiver
${{v}_{s}}$ is the velocity of the source

The signs are assigned depending on whether the source and receiver are approaching or receding each other.

In the given case receiver is at rest,
${{v}_{0}}=0$
Speed of sound is given as,
$v=320m{{s}^{-1}}$

Actual frequency at which the whistle is being blown is,
${{f}_{s}}=1000Hz$
Speed of the train is given as,
${{v}_{s}}=20m{{s}^{-1}}$

When the train (source) is approaching the observed frequency would be,
${{f}_{0}}={{f}_{s}}\left( \dfrac{v}{v-{{v}_{s}}} \right)$
$\Rightarrow {{f}_{0}}=1000\left( \dfrac{320}{320-20} \right)$
$\Rightarrow {{f}_{0}}=1000\times \dfrac{320}{300}$
$\therefore {{f}_{0}}=1066.67Hz$

When the train is receding,
${{f}_{0}}'={{f}_{s}}\left( \dfrac{v}{v+{{v}_{s}}} \right)$
$\Rightarrow {{f}_{0}}'=1000\left( \dfrac{320}{320+20} \right)$
$\Rightarrow {{f}_{0}}'=1000\times \dfrac{320}{340}$
$\therefore {{f}_{0}}'=941.18Hz$

Now the percentage change in observed frequency would be,
$\Delta f=\dfrac{{{f}_{0}}-{{f}_{0}}'}{{{f}_{0}}}\times 100$
$\Rightarrow \Delta f=\dfrac{1066.67-941.18}{1066.67}\times 100$
$\Rightarrow \Delta f=0.12\times 100$
$\therefore \Delta f=12$%.

Therefore, we found the percentage change in the frequency heard by a person standing near the track as the train passes him to be 12%.

Hence, option A is found to be the correct answer.

Note:
You may be wondering how the signs are assigned in the formula given in the solution. All you have to remember so as to avoid getting confused is that ${{v}_{0}}$ is taken positive if the receiver moves towards the source and it negative if the receiver moves away from the source and ${{v}_{s}}$ is taken positive is the source is moving away from the receiver and negative if it moves towards receiver.