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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

320

B

316

C

312

D

326

Text Solution

Verified by Experts

The correct Answer is:
B
`n_(B) = n_(A) +- 4`
`= 320 +- 4 = 324 or 315 Hz`
When `n_(B)` is loaded with wax and sounded with the fork `n_(A)` produces 6 b/s
`:. n_(B) = n_(A) +-6`
`= 320 +- 6 = 326 or 314 Hz`
Before loading frequency of tuning fork increases by 2
`:. n_(B) = 316 Hz`
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