Two identical piano wires,kept under the same | vedclass

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(B) The fundamental frequency $n$ of a stretched wire is given by $n = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$.Since the wires are identical,$n \propto \sq

Vedclass · Accepted Answer

$0.02$

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(B) The fundamental frequency $n$ of a stretched wire is given by $n = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$.Since the wires are identical,$n \propto \sqrt{T}$.Differentiating this relation,we get $\frac{dn}{n} = \frac{1}{2} \frac{dT}{T}$.For small changes,we can write $\frac{\Delta n}{n} = \frac{1}{2} \frac{\Delta T}{T}$.Here,the beat frequency is $\Delta n = 6\, Hz$ and the original frequency is $n = 600\, Hz$.Substituting these values,we get $\frac{\Delta T}{T} = 2 \left( \frac{\Delta n}{n} ight)$.$\frac{\Delta T}{T} = 2 \left( \frac{6}{600} ight) = 2 	imes 0.01 = 0.02$.

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 the occurrence of $6\, beats/s$ when both the wires oscillate together is:

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