Question

A sound source, fixed at the origin, is continuously emitting sound at a frequency of 660 Hz. The sound travels in air at a speed of 330 m s⁻¹. A listener is moving along the lien x = 336 m at a constant speed of 26 m s⁻¹. Find the frequency of the sound as observed by the listener when he is (a) at y = − 140 m, (b) at y = 0 and (c) at y = 140 m.

Answer 1

Given:
Frequency of sound emitted by the source $n_0$ = 660 Hz
Velocity of sound in air v = 330 ms^$-$1
Velocity of observer $v_0$ = 26 ms⁻¹
Frequency of sound heard by observer n = ?

(a) At y = 140 m:
Frequency of sound heard by the listener, when the source is fixed but the listener is moving towards the source:

$n=\frac{v+v_0}{v}{n}_0$

Here, $v_0=v_0\cos \theta$

On substituting the values, we get:

$$\begin{gathered} n=\frac{v+v_0\cos \theta }{v}{n}_0 \\ =\frac{330+26\times {\displaystyle \frac{140}{364}}}{330}\times 660 \\ =340\times 2=680\mathrm{Hz} \end{gathered}$$

(b) When the observer is at y = 0, the velocity of the observer with respect to the source is zero.
Therefore, he will hear at a frequency of 660 Hz.

(c) When the observer is at y = 140 m:

$n=\frac{v-v_0}{v}\times n_0$
Here, $v_0=v_0\cos \theta$

On substituting the values, we get:

$$\begin{gathered} n=\frac{330-{\displaystyle \frac{26\times 140}{364}}}{330}\times 660 \\ n=\frac{330-10}{330}\times 660=640\mathrm{Hz} \end{gathered}$$