Question

A whistle of frequency 540Hz rotates in a horizontal circle of radius 2m at an angular speed of 15rad/s. the highest frequency heard by the listener at rest with respect to the center of the circle:
(velocity of sound in air = $330m{{s}^{-1}}$ )
a) 590Hz
b) 594Hz
c) 598Hz
d) 602Hz

Answer 1

When the whistle moves in a circular path there is relative motion between the whistle and the listener at the center of the circle. Though the distance between them does not change, at every point on the circle, the velocity of the whistle changes, and it is as if the whistle turns and comes back to the listener. Hence we will use Doppler’s equation to determine the apparent frequency heard by the listener at the center.

Formula used:
${{f}^{'}}=\left( \dfrac{v-{{V}_{O}}}{v-{{V}_{S}}} \right)f$
$v=r\omega$

Complete step-by-step solution:
Let us say there is a relative motion between the observer and the source of sound. Let us say the frequency of the sound is f and moves towards the observer with velocity (${{V}_{O}}$). Similarly let us say the listener moves towards the source of sound with velocity (${{V}_{S}}$). If the speed of sound in air is v, than the apparent frequency (${{f}^{'}}$)of the sound heard by the listener is given by,
${{f}^{'}}=\left( \dfrac{v-{{V}_{O}}}{v-{{V}_{S}}} \right)f$
In the above question the listener is at the centre of the circle i.e. at rest. The whistle moves along the circumference of the circle with angular speed 15rad/s. The relation between the linear velocity (v)and the angular velocity($\omega$) is given in a circle of radius (r) is given by,
$v=r\omega$. hence from this equation the linear velocity of the whistles along the circle is,
$\begin{aligned} & v=r\omega \\\ & \Rightarrow v=2\times 15=30m{{s}^{-1}} \\\ \end{aligned}$
The linear velocity of the body moving along a circular path is given as the tangent to the point. For every point on the circle there exists a different direction of the tangent. Therefore we can imply that velocity of the whistle keeps on changing. If we take the tangent any point we can say that the whistle moves towards the listener. Hence the apparent frequency heard by the listener is given as,
$\begin{aligned} & {{f}^{'}}=\left( \dfrac{v-{{V}_{O}}}{v-{{V}_{S}}} \right)f \\\ & \Rightarrow {{f}^{'}}=\left( \dfrac{330-0}{330-30} \right)540 \\\ & \Rightarrow {{f}^{'}}=\left( \dfrac{330}{300} \right)540 \\\ & \Rightarrow {{f}^{'}}=11\times 54=594Hz \\\ \end{aligned}$
Hence the correct answer of the above question is option b.

Note: If we carefully observe the above answer we can see that the frequency of the sound heard by the observer is greater than the actual frequency of the source. This can be understood if we see the expression of apparent frequency. For the apparent frequency to be greater than the actual frequency, both the observer and the source should move towards each other.