Two sitar strings, A and B, playing the note 'Dha' are slightly out of tune and produce beats at a frequency of 5 Hz. The tension of the string B is slightly increased and the beat frequency is found to decrease by 3 Hz. If the frequency of A is 425 Hz, what is the original frequency of B?

The difference in frequency is known as the number of beats.
Here, frequency of A (fA) = 425 Hz
We know,
Frequency of B (fB) = fA ± beat frequency
= 425 ± 5
= 420 Hz or 430 Hz
Frequency decreases when the tension of string B is increased. This means that the original frequency of B was higher than that of A. Therefore, the original frequency of B was 430 Hz.