Question

A stationary observer receives a sound from a source of frequency $2000Hz$ moving with constant velocity. The apparent frequency varies with time as shown in fig. The speed of source and maximum value of apparent frequency $\left(fm\right)$ is ( velocity of sound $=300m/sec$ )

Answer 1

From the graph, maximum apparent frequency, $f_m=\frac{(v-v_o)f}{(v-v_s\cos \theta {)}_{min}}$ where $v,v_o,v_s$ velocities of sound, observer and source.
or $f_m=\frac{(v-0)f}{v-v_s\cos 0}=\frac{vf}{(v-v_s)}$
or $v-v_s=\frac{vf}{f_m}.....\left(1\right)$
Minimum apparent frequency , $1800=\frac{(v-0)f}{v-v_s\cos 180}=\frac{vf}{(v+v_s)}$
or $v+v_s=\frac{vf}{1800}...\left(2\right)$
or $\left(1\right)+\left(2\right),2v=vf[\frac{1}{f_m}+\frac{1}{1800}]$
or $\frac{1}{f_m}=[\frac{2}{f}-\frac{1}{1800}]=[\frac{2}{2000}-\frac{1}{1800}]=4.44\times {10}^{-4}$
$\therefore f_m\approx 2250=2300-50=2300-10\left(5\right)$ thus $x=5$