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

Frequency of string A, $f_{A} = 324 H z$
Frequency of string $B = f_{B}$
Beat’s frequency, $n = 6 H z$
Beat's Frequency is given as:
$n = | f_{A} - f_{B} |$
$6 = | 324 - f_{B} |$
$f_{B} = 330 H z$ or $318 H z$
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:
$f \propto \sqrt{T}$
Hence, the beat frequency cannot be 330 Hz
$∴ f_{B} = 318 H z$

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