Question

A stationary observer receives a sound from a source of frequency 2000 $Hz$ moving with constant velocity. The apparent frequency varies with time as shown in the figure.

The value of $f_{max}$ is (2300 - 10$x$) $Hz$. Then the value of $x$ is(Take speed of sound = 300 $m/s$ and neglect the time taken by sound to reach the stationary observer)

Answer 1

We know that apparent frequency,
$f_m=\frac{(v-v_o)f}{(v-v_s\cos \theta )}$
where $v,v_o,v_s$ are velocities of sound, observer and source respectively.

Maximum apparent frequency is at $\theta$ minimum i.e., 0,
$f_{max}=\frac{(v-0)f}{v-v_s\cos 0}=\frac{vf}{(v-v_s)}$.

Minimum apparent frequency is at $\theta$ maximum i.e., 180, $f_{min}=\frac{(v-0)f}{v-v_s\cos 180}=\frac{vf}{(v+v_s)}$.

$\frac{1}{f_{max}}=[\frac{2}{f}-\frac{1}{f_{min}}]=[\frac{2}{2000}-\frac{1}{1800}]$

$\therefore f_{max}=2250=2300-50=2300-10\left(5\right)$ thus $x$ = 5.