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 `\text { 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:

\[n = \frac{v + v_0 \cos\theta}{v}   n_0 \] 

\[ = \frac{330 + 26 \times \frac{140}{364}}{330} \times 660\] 

\[ = 340 \times 2 = 680 \text{ Hz }\]

(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:

\[n   =   \frac{330 - \frac{26 \times 140}{364}}{330} \times 660\] 

\[n   =   \frac{330 - 10}{330} \times 660 = 640  \text { Hz }\]