Answer
$$\boxed{\frac{v_D}{v}=0.25}$$
Work Step by Step
The emitted frequency is $f$
Initially the detector moves at constant velocity directly toward the stationary sound source. During the approach the detected frequency is:
$f^{'}_{app}=\frac{v+v_D}{v}f$
And after passing, the detector moves directly away from the stationary source. During the recession the detected frequency is:
$f^{'}_{rec}=\frac{v-v_D}{v}f$
It is given that
$\frac{f^{'}_{app}-f^{'}_{rec}}{f}=0.500$
or, $\frac{\frac{v+v_D}{v}f-\frac{v-v_D}{v}f}{f}=\frac{1}{2}$
or, $\frac{v_D}{v}=\frac{1}{4}$
or, $\boxed{\frac{v_D}{v}=0.25}$
