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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?
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Solution 1
Frequency of string A, fA = 324 Hz
Frequency of string B = fB
Beat’s frequency, n = 6 Hz
Beat's Frequency is given as:
`n = |f_A +- f_B|`
`6 = 324 +- f_B`
`F_B = 330 "Hz" or 318 "Hz"`
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 prop sqrtT`
Hence the beat frequency cannot be 330 Hz
`:.f_n = 318Hz`
Solution 2
Let υ1 and υ2 be the frequencies of strings A and B respectively.
Then, υ1 = 324 Hz, υ2 = ?
Number of beats, b = 6
υ2 = υ1 ± b = 324 ± 6 !.e., υ2 = 330 Hz or 318 Hz
Since the frequency is directly proportional to square root of tension, on decreasing the tension in the string A, its frequency υ1 will be reduced i.e., number of beats will increase if υ2 = 330 Hz. This is not so because number of beats become 3.
Therefore, it is concluded that the frequency υ2 = 318 Hz. because on reducing the tension in the string A, its frequency may be reduced to 321 Hz, thereby giving 3 beats with υ2 = 318 Hz.
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