Answer
$$\boxed{\frac{v_s}{v}=0.236}$$
Work Step by Step
The emitted frequency is $f$
When the sound source moves towards the stationary detector, the detected frequency becomes:
$f^{'}_{app}=\frac{v}{v-v_s}f$
When the sound source moves directly away from the stationary detector, the detected frequency becomes:
$f^{'}_{rec}=\frac{v}{v+v_s}f$
It is given that
$\frac{f^{'}_{app}-f^{'}_{rec}}{f}=0.500$
or, $\frac{\frac{v}{v-v_s}f-\frac{v}{v+v_s}f}{f}=\frac{1}{2}$
or, $\frac{v}{v-v_s}-\frac{v}{v+v_s}=\frac{1}{2}$
or, $\frac{v(v+v_s-v+v_s)}{v^2-v^2_s}=\frac{1}{2}$
or, $\frac{2vv_s}{v^2-v^2_s}=\frac{1}{2}$
or, $\frac{\frac{v_s}{v}}{1-\frac{v^2_s}{v^2}}=\frac{1}{4}$
or, $1-\frac{v^2_s}{v^2}=4\frac{v_s}{v}$
or, $\frac{v^2_s}{v^2}+4\frac{v_s}{v}-1=0$
Let, $\frac{v_s}{v}=x$
Therefore,
$x^2+4x-1=0\;........(1)$
Eq. $1$ is quadratic equation. The solutions are
$x=\frac{-4\pm\sqrt {16+4}}{2}=-2\pm\sqrt {5}$
The speed can not be negative. Therefore, the feasible solution is
$x= -2+\sqrt {5}$
or, $\frac{v_s}{v}=-2+\sqrt {5}$
or, $\boxed{\frac{v_s}{v}=0.236}$
