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Two tuning forks A and B give 4 beats/s ...

Two tuning forks A and B give 4 beats/s when sounded together. If the fork B is loaded with wax 6 beats/s are heard. If the frequency of fork A is 320 Hz, then the natural frequency of the tuning fork B will be

A

316 Hz

B

326 Hz

C

314 Hz

D

320 Hz

Text Solution

Verified by Experts

The correct Answer is:
A
`n_(A) = 320 Hz n_(B) = 324` or 316 Hz. If B is loaded, `n_(B)` decreases. If it is 324 Hz, No. of beats will decrease. But if `n_(B) = 316`, then if it becomes 314 Hz. We can get 6 beats/s
`:. n_(B) = 316 Hz`
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