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