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प्रश्न
Two identical tuning forks vibrating at the same frequency 256 Hz are kept fixed at some distance apart. A listener runs between the forks at a speed of 3.0m s−1 so that he approaches one tuning fork and recedes from the other figure. Find the beat frequency observed by the listener. Speed of sound in air = 332 m s−1.
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उत्तर
Given:
Speed of sound in air v = 332 ms−1
Velocity of the observer \[v_0\]= 3 `\text {ms}^\(-)`1
Velocity of the source \[v_s\]= 0
Frequency of the tuning forks \[f_0\]= 256 Hz
The apparent frequency\[\left( f_1 \right)\] heard by the man when he is running towards the tuning forks is \[f_1 = \left( \frac{v + v_0}{v} \right) \times f_0\]
On substituting the values in the above equation, we get:
\[f_1 = \left( \frac{332 + 3}{332} \right) \times 256 = 258 . 3 \text { Hz }\]
The apparent frequency \[\left( f_2 \right)\] heard by the man when he is running away from the tuning forks is \[f_2 = \left( \frac{v - v_0}{v} \right) \times f_0\]
On substituting the values in the above equation, we get :
\[f_2 = \left( \frac{332 - 3}{332} \right) \times 256\]
\[ = 253 . 7 \text { Hz }.\]
∴ beats produced by them
=\[f_2 - f_1\]
=258.3 − 253.7 = 4.6 Hz
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