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Question
A tuning fork A, marked 512 Hz, produces 5 beats per second, where sounded with another unmarked tuning fork B. If B is loaded with wax the number of beats is again 5 per second. What is the frequency of the tuning fork B when not loaded?
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Solution
Given that, vA = 512
On loading, frequencies of B decreased
So, in the first case v1 = 5
⇒ vB = vA ± 5
When B is loaded, the frequency of B becomes,
⇒ vB = 512 ± 5
We can write it as,
⇒ vB = 507 or vB = 517
On load in the frequency of B decreased by 507 to lower value of the number of beats will increase so vB ≠ 507
If vB = 517 then its frequency decreases by 10 Hz by then the number of beats will also be the same as 512 – 507 = 5.
Thus, the frequency of the tuning fork B when not loaded is 517 Hz.
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