Question

A small source of sound S of frequency 500 Hz is attached to the end of a light string and is whirled in a vertical circle of radius 1.6 m. The string just remains tight when the source is at the highest point. (a) An observer is located in the same vertical plane at a large distance and at the same height as the centre of the circle. The speed of sound in air = 330 m s−1 and g = 10 m s−2. Find the maximum frequency heard by the observer. (b) An observer is situated at a large distance vertically above the centre of the circle. Find the frequency heard by the observer corresponding to the sound emitted by the source when it is at the same height as the centre.

Answer 1

(a) The maximum frequency heard by the observer is $506 \mathrm{Hz}$

(b) The frequency heard by the observer corresponding to the sound emitted by the source when it is at the same height as the centre is $490 \mathrm{Hz}$

Explanation:

(a) The circle radius, $r=1.6 \mathrm{m}$

Where source velocity is u, and its mass = m then the weight mg is just equal to the centrifugal force $\frac{m u^{2}}{r}$ rat the highest point.

$\frac{m u^{2}}{r}=m g$

$u^{2}=g r$

$u=\sqrt{g r}$

The maximum frequency is heard by the listener when the source is at the lowest point of the circle because it is then approaching with the maximum speed u'.  

$\frac{1}{2} m u^{\prime 2}-\frac{1}{2} m u^{2}=m g \times 2 r$

$u^{\prime 2}=4 g r+u^{2}=4 g r+g r=5 g r$

$u^{\prime}=\sqrt{(5 g r)}=\sqrt{(5 \times 10 \times 1.6)}=\sqrt{80}=4 \sqrt{5} \mathrm{m} / \mathrm{s}$

The frequency heard = $\frac{V v}{V-u^{\prime}}=\frac{V v}{V-\sqrt{5 g r}}=\frac{330 * 500}{330-4 \sqrt{5}} H z=514 \mathrm{Hz}$

Note: If we take the source in clockwise direction, it reaches the listener at maximum speed at the top point $u=\sqrt{g r}=\sqrt{(10 * 1.6)}=\sqrt{16}=4 \mathrm{m} / \mathrm{s}$  

Hence the maximum listened frequency

$=\frac{V v}{V-u}=\frac{330 * 500}{(330-4)} H z=330 * \frac{500}{326} H z=506 \mathrm{Hz}$

(b) When the source is at the same height as the centre, it has two locations-one on the left and one on the right.. At one point it goes down while it comes up at the other spot Let there be speed at this stage = U.

Now  the difference of the kinetic energy = change in potential energy .  

$\frac{1}{2} m U^{2}-\frac{1}{2} m u^{2}=m g r$

$U^{2}=u^{2}+2 g r=g r+2 g r=3 g r$

$\mathrm{U}=\sqrt{(3 \mathrm{gr})}=\sqrt{(3 * 10 * 1.6)}=4 \sqrt{3} \mathrm{m} / \mathrm{s}$

Now when the source goes up with U the highest frequency detected.

Maximum frequency heard $=\frac{V_{V}}{V-U}$

$=330 \times \frac{500}{(330-4 \sqrt{3})}$

$=511 \mathrm{Hz}$

The minimum frequency is detected when the source runs down with speed U,

$=\frac{V v}{V-U}$

$=330 * \frac{500}{(330+4 \sqrt{3})}$

$=490 \mathrm{Hz}$