Question

Two sirens situated one kilometre apart are producing sound of frequency 330 Hz. An observer starts moving from one siren to the other with a speed of 2m/s. If the speed of sound be 330m/s, the beat frequency heard by the observer is
A. 8
B. 4
C. 6
D. 1

Answer 1

Hint:When there is relative motion between the source of sound and the observer, the frequency of the sound heard by the observer is different from the actual frequency of the sound. Use the formula for the apparent frequency when the observer is moving towards or away from the stationary source of sound and find the apparent frequencies of both the sirens. Then use the formula for beat frequency.
Formula used:
${{f}^{'}}=f\left( \dfrac{v+{{v}_{o}}}{v} \right)$
${{f}^{'}}=f\left( \dfrac{v-{{v}_{o}}}{v} \right)$
$beats=\left| {{f}_{1}}-{{f}_{2}} \right|$

Complete answer:
When there is relative motion between the source of sound and the observer, the frequency of the sound heard by the observer is different from the actual frequency of the sound (i.e. the frequency of the sound that is heard when the observer and the source are at rest).
When the observer is moving towards a stationary source of sound, the apparent frequency is given as ${{f}^{'}}=f\left( \dfrac{v+{{v}_{o}}}{v} \right)$ …. (i),
where f is the actual frequency, v is the speed of the sound and ${{v}_{o}}$ is the speed of the observer.
If the observer is away from a stationary source of sound, the apparent frequency is given as ${{f}^{'}}=f\left( \dfrac{v-{{v}_{o}}}{v} \right)$ …. (ii),
The apparent frequency of the sound of the source towards which the observer is moving is ${{f}_{1}}$.
In this case, f = 330Hz, v = 330m/s and ${{v}_{o}}$ = 2m/s.
Substitute the values in (i).
${{f}_{1}}=330\left( \dfrac{330+2}{330} \right)=330\left( \dfrac{332}{330} \right)=332Hz$.
The apparent frequency of the sound of the source from which the observer is moving away is ${{f}_{2}}$.
In this case also, f = 330Hz, v = 330m/s and ${{v}_{o}}$ = 2m/s.
Substitute the values in (i).
${{f}_{2}}=330\left( \dfrac{330-2}{330} \right)=330\left( \dfrac{328}{330} \right)=328Hz$.
Beat frequency is defined as the difference in the frequencies of two sounds.
i.e. $beats=\left| {{f}_{1}}-{{f}_{2}} \right|$.
$\Rightarrow beats=\left| 332-328 \right|=4Hz$.
Therefore, the beat frequency heard by the observer is 4Hz.

So, the correct answer is “Option B”.

Note:
From this solution, we can understand that when the observer is moving towards the stationary source of sound, the frequency of the sound heard by the observer is more than the actual frequency of the sound.
And when the observer is moving away from the stationary source of sound, the frequency of the sound heard by the observer is less than the actual frequency of the sound.