Question

A source of sound $S$ of frequency $500Hz$ situated between a stationary observer $O$ and a wall $W$ , moves towards the wall with a speed of $2m/s$ . If the velocity of sound is $332m/s$ , then the number of beats per second heard by the observer is (approximately)

Answer 1

Formula used: $n_1=n\left(\frac{v}{v+v_s}\right)$

$n_2=n\left(\frac{v}{v-v_s}\right),$

Beat frequency $=n_2-n_1$


Given, $v=$ speed of sound $=332m/s$
$n=$ frequency of source $=500Hz$
$v_s=$ velocity of source $2m/s$

For direct sound, source is moving away from the observer. So, the frequency heard,

$\Rightarrow n_1=n\left(\frac{v}{v+v_s}\right)$

$\Rightarrow n_1=500\left(\frac{332}{332+2}\right)$

$\Rightarrow n_1=500\left(\frac{332}{334}\right)$

$\Rightarrow n_1=500\times 0.994$

$\Rightarrow n_1=497Hz$

The other sound is echo, reaching the observer from the wall and the image of source formed by reflection at the wall. This image is approaching the observer in the direction of sound,

Hence for reflected sound, frequency heard by the observer is,

$\Rightarrow n_2=n\left(\frac{v}{v-v_s}\right)$

$\Rightarrow n_2=500\left(\frac{332}{332-2}\right)$

$\Rightarrow n_2=500\left(\frac{332}{330}\right)$

$\Rightarrow n_2=500\times 1.006$

$\Rightarrow n_2=503Hz$

$\Rightarrow \text{Beat frequency}=n_2-n_1$

$\Rightarrow \text{Beat frequency}=503-497$

$\Rightarrow =\text{Beat frequency}=6Hz$

Final answer: 6 Hz