Question

There is a source of sound (a tuning fork) moving towards a reflecting wall will a speed of $30 \, m \, s^{- 1}$ . The velocity of the sound in air is $330 \, m \, s^{- 1}$ and the frequency of the tuning fork is $600 \, Hz$ . The observer is between the source and the wall and is moving with some velocity $V_{1}$ towards the wall. The beat frequency heard by the observer is $4 \, Hz$ . If the tuning fork is waxed, the frequency beats heard by the observer becomes $3 \, Hz$ . If the new frequency of tuning fork is $k$ then $k/50$ is

Answer 1

Apparent frequency of source $=n_{0}\left[\frac{c - v_{1}}{c - v_{2}}\right]=n^{′}$

Apparent frequency of reflected sound $=n_{0}\left[\frac{c + v_{1}}{C - V}\right]=n^{′ ′}$
Beat frequency $=n^{′ ′}-n^{′}=\frac{2 n_{0} V_{1}}{C - V_{2}}=\frac{2 \times 600 \times V_{1}}{330 - 30}=4Hz$
$\therefore \, \, V_{1}= \, 1ms^{- 1}$

When the tuning fork is waxed let its frequency become $k$
Beat frequency $=\frac{2 k v_{1}}{C - V_{2}}=\frac{2 \times k \times 1}{330 \times 30} \, =3\text{H}\text{z}$
$\therefore \, \, \, k=450\text{H}\text{z}=$ new frequency of tuning fork.
$k/50=9$