A whistle of frequency 540 Hz rotates in a h | vedclass

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(B) The velocity of the source $v_s$ is given by $v_s = r \omega$. Given $r = 2 \ m$ and $\omega = 15 \ rad/s$,we have $v_s = 2 	imes 15 = 30 \ m/s$.

Vedclass · Accepted Answer

$594$

Vedclass · Answer

(B) The velocity of the source $v_s$ is given by $v_s = r \omega$. Given $r = 2 \ m$ and $\omega = 15 \ rad/s$,we have $v_s = 2 	imes 15 = 30 \ m/s$. The listener is at the center,so the source moves towards the listener when its velocity vector points directly at the center. The highest frequency $f'$ heard by a stationary listener is given by the Doppler effect formula for a source moving towards the listener: $f' = f \left( \frac{v}{v - v_s} ight)$ Substituting the given values: $f' = 540 \left( \frac{330}{330 - 30} ight) = 540 \left( \frac{330}{300} ight) = 540 	imes 1.1 = 594 \ Hz$.

$A$ whistle of frequency $540 \ Hz$ rotates in a horizontal circle of radius $2 \ m$ at an angular speed of $15 \ rad/s$. What is the highest frequency heard by a listener at rest with respect to the centre of the circle (in $Hz$)? (Velocity of sound in air $= 330 \ m/s$)

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