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Q. Two identical piano wires, kept under the same tension $T$ have a fundamental frequency of $600 \,Hz$. The fractional increase in the tension of one of the wires which will lead to occurrence of $6 $ beat's when both the wires oscillate together would be

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Solution:

As $v=\frac{1}{2 L} \sqrt{\frac{T}{\mu}}$
$ \therefore \frac{\Delta v}{v}=\frac{1}{2} \frac{\Delta T}{T}$
$\frac{\Delta T}{T}=2 \frac{\Delta v}{v}=2 \times \frac{6}{600}=0.02$