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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.
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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:
\[n = \frac{330 - \frac{26 \times 140}{364}}{330} \times 660\]
\[n = \frac{330 - 10}{330} \times 660 = 640 \text { Hz }\]
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