Two sitar strings A and B playing a note are slightly out of tune and produces beats of frequency 5 Hz. The tension in string B is slightly increased and beat frequency become 3 Hz. What is the original frequency of B if frequency of A is 427 Hz.
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 422 Hz
The beat frequency is the difference of the two frequencies, \( |f_A - f_B| \). The frequency of a stretched string is \( f \propto \sqrt{T} \), so increasing the tension raises the frequency of B.
\( \displaystyle f_B = f_A \pm 5 = 432\,\mathrm{Hz} \;\text{or}\; 422\,\mathrm{Hz} \)
If \( f_B = 432\,\mathrm{Hz} \), raising it moves it further from 427 Hz, so the beat frequency would rise above 5 Hz.
If \( f_B = 422\,\mathrm{Hz} \), raising it to 424 Hz brings it closer to 427 Hz, and the beat frequency falls to \( 427 - 424 = 3\,\mathrm{Hz} \), as observed.
So the original frequency of B is 422 Hz.