Question

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

Answer 1

Answer: 9

Apparent frequency of source $=n_0\left[\frac{c-v_1}{c-v_2}\right]=n^{\prime }$

Apparent frequency of reflected sound $=n_0\left[\frac{c+v_1}{C-V}\right]=n^{\prime \prime }$
Beat frequency $=n^{\prime \prime }-n^{\prime }=\frac{2n_0{V}_1}{C-V_2}=\frac{2\times 600\times V_1}{330-30}=4Hz$
$\therefore V_1=1m{s}^{-1}$

When the tuning fork is waxed let its frequency become $k$
Beat frequency $=\frac{2k{v}_1}{C-V_2}=\frac{2\times k\times 1}{330\times 30}=3\text{H}\text{z}$
$\therefore k=450\text{H}\text{z}=$ new frequency of tuning fork.
$k/50=9$