Question

A source of sound of frequency $384Hz$ is moving towards a wall with a velocity of $5m{s}^{-1}$. Velocity of sound is $330m{s}^{-1}$. The number of beats heard by an observer standing behind the source is nearly:

• A. $\frac{384\times 330}{325}-384$
• B. $384-\frac{384\times 330}{325}$
• C. $\frac{384\times 330}{325}-\frac{384\times 330}{335}$
• D. $\frac{384\times 330}{325}-\frac{384\times 330}{325}$

Answer 1

The correct option is C $\frac{384\times 330}{325}-\frac{384\times 330}{335}$

Here source is going away from the observer.
Apparent frequency ${\nu }_1=\frac{v}{v+u}{\nu }_0$, where v = 330 m/s, ${\nu }_0=384\text{Hz}$ and u = 5\ \text{m/s.}
Thus ${\nu }_1=\frac{330\times 384}{335}$
Apparent frequency at wall ${\nu }_{wall}=\frac{v}{v-u}{\nu }_0=\frac{330\times 384}{325}$.
This is frequency does not change because the observer does not have any relative velocity with respect to the wall.
Number of beats per second = $\frac{330\times 384}{325}-\frac{330\times 384}{335}$
Ans: C