Wave Motion 4 Question 28
26. A whistle emitting a sound of frequency $440 Hz$ is tied to a string of $1.5 m$ length and rotated with an angular velocity of $20 rad / s$ in the horizontal plane. Calculate the range of frequencies heard by an observer stationed at a large distance from the whistle. (Speed of sound $=330 m / s$ ).
(1996, 3M)
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Answer:
Correct Answer: 26. 403.3 Hz, 484 Hz
Solution:
- $v _s=$ Speed of source (whistle) $=R \omega=(1.5)(20) m / s$
$v _s=30 m / s$
Maximum frequency will be heard by the observer in position $P$ and minimum in position $Q$. Now,
$$ f _{\max }=f\left(\frac{v}{v-v _s}\right) $$
where, $v=$ speed of sound in air $=330 m / s$
$$ =(440)\left(\frac{330}{330-30}\right) Hz $$
$$ f _{\max }=484 Hz $$
and
$$ \begin{aligned} f _{\min } & =f\left(\frac{v}{v+v _s}\right)=(440)\left(\frac{330}{330+30}\right) \\ f _{\min } & =403.33 Hz \end{aligned} $$
Therefore, range of frequencies heard by observer is from $484 Hz$ to $403.33 Hz$.
