Physics Question on simple harmonic motion

Question

Two sitar strings, $A$ and $B$ , playing the note ??ha??are slightly out of tune and produce beats of frequency $5\, Hz$ . The tension of the string $B$ is slightly increased and the beat frequency is found to decrease by $3\, Hz$ . If the frequency of $A$ is $425\, Hz$ , the original frequency of $B$ is :

Answer 1

Solution

Frequency of sitar string A, $f_{1}=425\, Hz$ . Frequency of sitar string $B , f_{2}=(425 \pm 5) Hz$ , that is, either $420\, Hz$ or $430\, Hz$ . On increasing tension in string $B$ , its frequency will increase as $n \propto \sqrt{T}$ , where $T$ is tension. If $430\, Hz$ is correct, then on increasing tension number of beats should have increased. But, the number of beats has decreased. It means $420\, Hz$ is correct. On increasing tension, frequency increases from $420\, Hz$ to $422\, Hz$ and number of beats is 3.