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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. - Physics

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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?

Short Note

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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Chapter 15: Waves - Exercises [Page 109]

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NCERT Exemplar Physics [English] Class 11
Chapter 15 Waves
Exercises | Q 15.20 | Page 109

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