Question

Two sitar strings A and B playing the note ‘Ga’ are slightly out of tune and produce beats of frequency 6 Hz. The tension in the string A is slightly reduced and the beat frequency is found to reduce to 3 Hz. If the original frequency of A is 324 Hz, what is the frequency of B?

Answer 1

Concept :- the difference in frequency is known as the number of beats.

Here,frequency of A ( fA) = 324 Hz
We know,
Frequency of B ( fB) = fA ± beat frequency
= 324 ± 6
= 318Hz or 330Hz
Now, if tension in string slightly reduced then its frequency also reduce from 324 Hz { frequency = 1/wavelength × √{T/u} }
Hence, fB = 318 Hz fB ≠ 330 Hz

Answer 2

Answer:

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