A source of sound and detector are moving as shown in the given figure at $t = 0$ . Take velocity of sound wave to be $v_{0}$ .

For this situation mark out the correct statement(s).

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The frequency received by the detector is always greater than $f_{0}$ .

Frequency received by the detector is equal to $f_{0}$ at $t = d \cot θ_{0} / (2 v_{0})$ .

Frequency received by the detector is greater than $f_{0}$ , for $t < t = d \cot θ_{0} / (2 v_{0})$ .

Frequency received by the detector can never be equal to $f_{0}$ .

answer is B, C.

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Detailed Solution

As time increases, the source and detector are relatively approaching each other up to $t = t_{0}$ , where $t_{0}$ is the instant when the source and detector are located perpendicular to direction of motion.
$v_{0} \times t_{0} = \frac{d \cot θ_{0}}{2}$
$t_{0} = \frac{d \cot θ_{0}}{2}$
For $t < t_{0}, f_{a p} > f_{0}$
For $t > t_{0}, f_{a p} < f_{0}$