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:
Frequency of string A, fA = 324 Hz
Frequency of string B = fB
Beats frequency, n = 6 Hz
Beats Frequency is given as:
Frequency decreases with a decrease in the tension in a string. This is because frequency is directly proportional to the square root of tension. It is given as:
v ∝ √T
Hence, the beat frequency cannot be 330 Hz
Thus, fB = 318 Hz.
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