Question

A fighter plane moving in a vertical loop of radius $R$ with constant speed. The centre of the loop is at a height $h$ directly overhead of an observer standing on the ground. The observer receives maximum frequency of the sound produced by the plane when it is nearest to him. The speed of the plane is $($ Velocity of sound in air is $v.$ Take $R=100$ units, $h=200$ units, $v=340$ units $)$

• A. $100$
• B. $150$
• C. $205$
• D. $225$

Answer 1

Answer: (C)

Let the speed of the plane (source) is $v_{s}$ . Maximum frequency will be observed by the observer when $v_{s}$ is along line $SO$ . The observer receives maximum frequency when the plane is nearest to him.
That is as soon as the wave pulse reaches from $S$ to $O$ with speed $v,$ the plane reaches from $S$ to $S^{'}$ with speed $v_{s}.$
Hence, $t=\frac{S O}{v}=\frac{S S^{'}}{v_{s}}\Rightarrow v_{s}=\left(\frac{S S^{'}}{S O}\right)v=\frac{R \theta }{\sqrt{h^{2} - R^{2}}}v$
Here $cos\theta =\frac{R}{h}$ or $\theta = cos^{- 1}\left(\frac{R}{h}\right)$

Using given values $\theta = cos ^{- 1}\left(\frac{100}{200}\right)=\frac{\pi }{3}$
and $v_{s}=\frac{100 \times \frac{\pi }{3}}{\sqrt{200^{2} - 100^{2}}}\times 340=205$ units