Two tuning forks A and B of frequency 512 Hz are sounded together produce 5 beats/s. If the fork A is thin loaded with a piece of wax and found that b

Question

Two tuning forks A and B of frequency 512 Hz are sounded together produce 5 beats/s. If the fork A is thin loaded with a piece of wax and found that b

Two tuning forks A and B of frequency 512 Hz are sounded together produce 5 beats/s. If the fork A is thin loaded with a piece of wax and found that beats occur at shorter intervals. Then the natural frequency of A will be
A. 517
B. 507
C. 512
D. 510

Monjur Alahi

4 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Salim Ahmed Kawsar

Correct Answer - B
`n_(A) = n_(B) +- 5`
`= 512 +- 5 = 517 or 507 Hz`
When `n_(A)` is loaded with little wax and sounded with `n_(B)` it produce beats of shorter time interval, that means beat frequency increases
`n_(A) = 512 +- 6`
`= 518 or 506`
`n_(A) = 507 Hz`

4 views Answered 2023-02-05 15:29