Two wires are in unison. If the tension in on | vedclass

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(C) The frequency of a stretched wire is given by $n = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$,which implies $n \propto \sqrt{T}$.Taking the derivative,we

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$500$

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(C) The frequency of a stretched wire is given by $n = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$,which implies $n \propto \sqrt{T}$.Taking the derivative,we get $\frac{\Delta n}{n} = \frac{1}{2} \frac{\Delta T}{T}$.Given that the tension increases by $2\%$,i.e.,$\frac{\Delta T}{T} = 0.02$,the change in frequency is $\frac{\Delta n}{n} = \frac{1}{2} 	imes 0.02 = 0.01$ or $1\%$.Let the initial frequency be $n$. The new frequency is $n' = n + 0.01n = 1.01n$.The number of beats produced per second is the difference in frequencies: $n' - n = 5$.$1.01n - n = 5 \implies 0.01n = 5$.$n = \frac{5}{0.01} = 500 \ Hz$.

Two wires are in unison. If the tension in one of the wires is increased by $2\%$,$5$ beats are produced per second. The initial frequency of each wire is .... $Hz$.

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