Question

There is a source of sound (a tuning fork) moving towards a reflecting wall will a speed of 30 m/s. The velocity of the sound in air is 330 m/s 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 man becomes 3Hz. If new frequency of tuning fork is { x $\times$ 100 + y $\times$ 10 }, then find (x + y)___

Answer 1

Apparent frequency of source $=n_0\left[\frac{c-v_1}{c-v_s}\right]=n^{\prime }$
Apparent frequency of reflected sound $=n_0\left[\frac{c+v_1}{c-v_s}\right]=m^{\prime \prime }$
Beat frequency $=n^{\prime \prime }-n^{\prime }=\frac{2n_0{v}_1}{c-v_s}=\frac{2\times 600\times v_1}{330-30}=4Hz$
$\therefore v_1=1m/x$
$\therefore$ The source moves with a velocity 1m/s towards the wall.
When the tuning fork is waxed let its frequency become n
Beat frequency $=\frac{2n{v}_1}{c-v_s}=\frac{2\times n\times 1}{330\times 30}=3Hz$
$therefore$ n = 450 Hz = new frequency of tuning fork.