Question

A source of sound and detector are arranged as shown in Fig. 6.46. The detector is moving along a circle with constant angular speed $\omega$. It starts from the shown location in anti-clockwise at $t=0$ .
(Take velocity of sound in air as $v$ .)
Based on this information answer the question
What is the frequency as received by the detector, when it rotates by an angle $\frac{\pi }{2}$?

• A. $f$
• B. $\frac{\nu -\omega R}{\nu }\times f$
• C. $\frac{\nu -\omega \frac{R}{2}}{\nu }\times f$
• D. $\frac{\nu -\omega R\times \frac{2}{\sqrt{5}}}{\nu }\times f$

Answer 1

Answer: (D)

$\frac{\nu -\omega R\times \frac{2}{\sqrt{5}}}{\nu }\times f$

The location of detector at required instant is shown in Fig. 6.73
We have , ${\nu }_0=R\omega$ [speed of detector]
$f_{ap}=\frac{\nu -{\nu }_{0}cos\theta }{\nu }\times f$
$cos\theta =\frac{2R}{\sqrt{5}R}$
$f_{ap}=\frac{\nu -\omega R\times \frac{2}{\sqrt{5}}}{\nu }\times f$