Question

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

• A. $400.0\text{Hz}to484.0\text{Hz}$
• B. $403.3\text{Hz}to480.0\text{Hz}$
• C. $400.0\text{Hz}to480.0\text{Hz}$
• D. $403.3\text{Hz}to484.0\text{Hz}$

Answer 1

The correct option is D $403.3\text{Hz}to484.0\text{Hz}$

Given that,
Frequency of sound of whistle, $f=440\text{Hz}$
Angular velocity of whistle, $\omega =20\text{rad/s}$
Liner velocity of the whistle, $v_w=1.5\times 20=30\text{m/s}$
Speed of sound in air, $v_s=330\text{m/s}$
Velocity of observer, $v_o=0\text{m/s}$

Let the observer is at $O$ and whistle is rotating in the circle $ABCD$ as shown in the image.
Frequency heard by the observer will be maximum when the source is in position $D$. In this case, The speed of whistle is toward the observer.

Maximum frequency observed,
$f_{max}=\frac{v_s-v_o}{v_s-v_w}\times f$
$f_{max}=\frac{330-0}{330-30}\times 440$
$f_{max}=\frac{330}{300}\times 440$

$f_{max}=484\text{Hz}$

Similarly, frequency heard by the observer will be minimum when the source reaches at the position $B$.
$f_{\text{min}}=\frac{v_s-v_o}{v_s-(-v_w)}\times f$
$f_{\text{min}}=\frac{330}{330+30}\times 440$
$f_{\text{min}}=\frac{330\times 440}{360}=403.3\text{Hz}$