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प्रश्न
Figure shows a person standing somewhere in between two identical tuning forks. each vibrating at 512 Hz. If both the tuning forks move towards right a speed of 5.5 m s−1, find the number of beats heard by the listener. Speed of sound in air = 330 m s−1.
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उत्तर
Given:
Frequency of tuning forks \[f_0\]= 512 Hz
Speed of sound in air v = 330 ms−1
Velocity of tuning forks \[v_s\]= 5.5 ms−1
The apparent frequency \[\left( f_1 \right)\] heard by the person from the tuning fork on the left is given by:
\[f_1 = \left( \frac{v}{v + v_s} \right) \times f_0\]
On substituting the values in the above equation, we get:
\[f_1 = \left( \frac{330}{330 + 55} \right) \times 512\]
\[ = 503 . 60 \text { Hz }\]
Similarly, apparent frequency \[\left( f_2 \right)\] heard by the person from the tuning fork on the right is given by:
\[f_2 = \left( \frac{v}{v - v_s} \right) \times f_0\]
On substituting the values in the above equation, we get:
\[f_2 = \left( \frac{330}{330 - 5 . 5} \right) \times 512\]
\[ = 520 . 68 \text { Hz }\]
∴ beats produced
=\[f_2 - f_1\]
= 520.68 − 503.60 = 17.5 Hz
As the difference is greater than 10 (the persistence of sound for the human ear is 1/10 of a second), the sound gets overlapped, and the observer cannot distinguish between the sounds and the beats.
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