Two identical piano wires,kept under the same | vedclass

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(B) The fundamental frequency of a stretched wire is given by $v = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$.Since $L$ and $\mu$ are constant for identical w

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$0.02$

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(B) The fundamental frequency of a stretched wire is given by $v = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$.Since $L$ and $\mu$ are constant for identical wires,$v \propto \sqrt{T}$.Differentiating both sides,we get $\frac{dv}{v} = \frac{1}{2} \frac{dT}{T}$.Given the initial frequency $v = 600\, Hz$ and the beat frequency $\Delta v = 6\, Hz$,the new frequency $v'$ will be $606\, Hz$ (or $594\, Hz$).Thus,the change in frequency is $\Delta v = 6\, Hz$.The fractional change in tension is $\frac{\Delta T}{T} = 2 \frac{\Delta v}{v}$.Substituting the values: $\frac{\Delta T}{T} = 2 	imes \frac{6}{600} = 2 	imes 0.01 = 0.02$.

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 would be:

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