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.

#### Solution

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:

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

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