Question

A whistle emitting a sound of frequency 440 Hz is tied to a string of length 1.5 m and rotated with an angular velocity of 20 $rad/s$ in the horizontal plane. Then, the range of frequencies heard by an observer stationed at a large distance from the whistle will be :
$(V_s=330m{s}^{-1})$

• A. 400.0 Hz to 484.0 Hz.
• B. 403.3 Hz to 480.0 Hz.
• C. 400.0 Hz to 480.0 Hz
• D. 403 Hz to 484.0 Hz

Answer 1

The correct option is D 403 Hz to 484.0 Hz

Given : Frequency of the whistle $n=440Hz$

Length of the string $L=1.5m$

Angular velocity of the string $w=20rad/s$

Thus linear velocity of the source (whistle) $v_{source}=v=Lw=1.5\times 20=30m/s$

Let the frequencies heard by the observer from source $S^{\prime }$ and $S$ be $n^{\prime }$ and $n^{\prime \prime }$.

Doppler effect when source $S^{\prime }$ approaches the stationary observer:

$n^{\prime }=n[\frac{v_{sound}}{v_{sound}-v_{source}}]$ $=440[\frac{330}{330-30}]=484Hz$

Doppler effect when source $S$ moves away from the stationary observer:

$n^{\prime \prime }=n[\frac{v_{sound}}{v_{sound}+v_{source}}]$ $=440[\frac{330}{330+30}]=403Hz$

Thus frequency range heard by the observer at large distance is $403Hzto484Hz$