Question

Please answer with method.
Tuning fork A of frequency 258 Hz gives eight beats with a tuning fork B. When prongs of B are cut and again A and B are sounded, the number of beats heard remains the same. The frequency of B is?

Answer 1

Explanation:

F

A

−F

B

∣=8

∣258−F

B

∣=8

F

B

=250/F

B

=266

When B prongs cut fπ equation increases so that means initially 2+ was = 250

So later become 266

So initially ∣258−250∣=8

so later ∣258−266∣=8

so fπ eq of B is = 250

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