Question

A boy blowing a whistle, is running away from a wall towards an observer with a speed of 1m/s. The frequency of whistles is 680 Hz. The number of beats heard per second by the observer will be (given: v = 340 m/s):
A. zero
B. 2
C. 4
D. 8

Answer 1

Hint: Beats are the interference pattern between two sources of slightly different frequencies. Beats can only be observed by human ears when the two sources have almost the same frequency. According to Doppler’s effect, when a source emitting some frequency moves with respect to initial position, a change in the frequency is observed, called apparent frequency.

Formula used:
$\nu_{apparent}=\nu_{emitted}\left({\dfrac{v_{\circ}+v_{observer}}{v_{\circ}-v_{source}}}\right)$, $Beat\ frequency (\beta)=|\nu_1-\nu_2|$where $\nu_{\circ}$is the speed of sound in air.

Complete step-by-step answer:
First of all, we will calculate what is the frequency observed by the observer when the voice comes after reflection from the wall.
The frequency incident on the wall is given by:
$\nu_{apparent}=\nu_{emitted}\left({\dfrac{v_{\circ}+v_{observer}}{v_{\circ}-v_{source}}}\right)$
Here $v_\circ = speed\ of\ sound\ = 340\ m/s, v_{source} = 1\ m/s\ and\ v_{observer} (wall\ in\ this\ case)= 0\ m/s$
So, $\nu_{apparent}=680\left({\dfrac{340+0}{340+1}}\right) = \dfrac{680\times 340}{341}\ Hz$
Now since both wall and observer are stationary hence it is the frequency which is heard by the observer (coming from the wall).
Now, frequency as listened by the observer directly coming from the boy:
$\nu_{apparent}=\nu_{emitted}\left({\dfrac{v_{\circ}+v_{observer}}{v_{\circ}-v_{source}}}\right)$
Here, $v_{source} = 1\ m/s\ and\ v_{observer} (stationary\ observer\ in\ this\ case)= 0\ m/s$
Thus, $\nu_{apparent}=680\left({\dfrac{340+0}{340-1}}\right) = \dfrac{680\times 340}{339}$
Now, as beats is the difference of two frequencies listened, therefore:
$Beat\ frequency (\beta)=|\nu_1-\nu_2|$
$\beta=|\dfrac{680\times 340}{339}-\dfrac{680\times 340}{341}| = 680\times 340\left( \dfrac{1}{339} - \dfrac{1}{341}\right) = \dfrac{2\times 680\times 340} {339\times 341}\approx 4$
Hence the observer will hear 4 beats per second.

So, the correct answer is “Option C”.

Additional Information: In this question, we have first assumed wall as an observer. This is because when the sound waves fall on a solid boundary such that they then get reflected from it, it starts acting as if it is the original source.

Note: In the Doppler’s effect formula, as the source is at rest, we didn’t alter the denominator term. But if the sources also started moving, we have to apply the logic that if the source is approaching the observer, there must be an increment in frequency. And so we have to decrease the denominator to increase the net term and vice-versa.