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Karnataka Board PUCPUC Science Class 11

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 - Physics

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Question

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.

Numerical

Solution

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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Chapter 16: Sound Waves - Exercise [Page 356]

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HC Verma Concepts of Physics Vol. 1 [English] Class 11 and 12
Chapter 16 Sound Waves
Exercise | Q 71 | Page 356

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