Question

A policeman blows a whistle with a frequency of $ 500Hz $ A car approaches him with a velocity of $ 15m/s $ . The change in frequency as heard by the driver of the car as he passes the policeman is (Given, speed of sound in air is $ 300m/s $ .)
A. $ 25Hz $
B. $ 50Hz $
C. $ 100Hz $
D. $ 150Hz $

Answer 1

Hint: Doppler Effect can be considered as a property of sound waves. It is a change in sound wave frequency caused by the relative movement of either source, listener or both .Use the equation of Doppler Effect to find the respective frequency heard by the listener when the car is travelling 15m/s and find their difference.

Formula Used:
$ \nu ' = \dfrac{{c + {v_0}}}{{c + {v_s}}}\nu $ Here $ \nu ' $ is the frequency that the listener hears and $ \nu $ is the frequency of the sound, $ c $ is the velocity of sound in the medium it is travelling in m/s, $ {v_0} $ is the velocity of listener in m/s and $ {v_s} $ is the velocity of source in m/s. Note that $ {v_o} $ and $ {v_s} $ have to be taken $ + ve $ when source and listener are travelling towards each other and $ - ve $ when they are moving away from each other.

Complete step-by-step answer:
We know that the frequency of a sound changes when either source or the listener is moving. This phenomenon is called the Doppler Effect and the relation between apparent and real frequency is:
 $ \nu ' = \dfrac{{c + {v_0}}}{{c - {v_s}}}\nu $
Here $ c $ is the speed of sound, $ {v_o} $ is the speed of the observer and $ {v_s} $ is the speed of source. Note that $ {v_o} $ and $ {v_s} $ have to be taken $ + ve $ when source and listener are travelling towards each other and $ - ve $ when they are moving away from each other.

From (1), we can clearly see that an observer receding from a source would experience a decrease in frequency and someone approaching the source experiences an increase in frequency. (i.e.: apparent frequency for approach and that for retreat are different).
Here in the question, the observer is passing by the source. This means the she/he is approaching the source for some time and then receding away from it. The question asks us to find the difference between apparent frequency during approach and that during retreat.

Let’s first calculate the apparent frequency for approach.
 $ {v_s} = 0 $ $ \left( \because \right. $ Source is stationary $ \left. {} \right) $
 $ {v_0} $ =+15m/s $ \left( \because \right. $ approach $ \left. {} \right) $
 $ c = 300m/s $
 $ \nu = 500Hz $
 $ {\nu _a}^\prime = \dfrac{{c + {v_0}}}{{c - {v_s}}}\nu $
 $ {\nu _a}^\prime = \dfrac{{300 + 15}}{{300}}500Hz $

For retreat:
 $ {v_s} = 0 $ $ (\because $ Source is stationary)
  $ {v_0} $ =-15m/s $ (\because $ approach)
 $ c = 300m/s $
 $ \nu = 500Hz $
 $ {\nu _r}^\prime = \dfrac{{c + {v_0}}}{{c - {v_s}}}\nu $
 $ {\nu _r}^\prime = \dfrac{{300 - 15}}{{300}}500Hz $
We can now find the difference in frequency: $ {\nu _a}^\prime - {\nu _r}^\prime = \dfrac{{300 + 15}}{{300}}500Hz - \dfrac{{300 - 15}}{{300}}500Hz $
 $ = \dfrac{{30}}{{300}}500Hz $
 $ = 50Hz $
So as the observer passes by the source, frequency of the whistle changes by 50Hz.

Additional information:
Doppler Effect occurs for every wave including electromagnetic waves. Equation (1) governs the Doppler Effect in case of all waves except electromagnetic waves.
For electromagnetic waves, effects of special Relativity have to be taken into account and the relation is:
$ \nu ' = \sqrt {\dfrac{{c + {v_0}}}{{c - {v_s}}}} \nu $
So the speed of moving objects should be comparable with speed of light to observe the shift in frequency.

Note: We should remember that the Doppler Effect is not relative. That means the source moving and observer being stationary is different from the observer moving with same speed and source stationary.
Also note that here, the difference between apparent and real frequency is not asked. We are to find the difference between two apparent frequencies.