The fundamental frequency of a sonometer wire | vedclass

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(B) The fundamental frequency of a sonometer wire is given by $n = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$.Since tension $T$ and mass per unit length $\mu$

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decreased by $20 \%$

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(B) The fundamental frequency of a sonometer wire is given by $n = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$.Since tension $T$ and mass per unit length $\mu$ are constant,$n \propto \frac{1}{L}$.Let the initial length be $L_1 = L$ and the new length be $L_2 = L + 0.25L = 1.25L$.The initial frequency $n_1 = 50 \ Hz$.The new fundamental frequency $n_2$ is given by $\frac{n_2}{n_1} = \frac{L_1}{L_2} = \frac{L}{1.25L} = \frac{1}{1.25} = 0.8$.So,$n_2 = 0.8 	imes 50 \ Hz = 40 \ Hz$.The frequency of the second harmonic is $2n$. Initially,$2n_1 = 2 	imes 50 = 100 \ Hz$. Finally,$2n_2 = 2 	imes 40 = 80 \ Hz$.The change in frequency is $100 \ Hz - 80 \ Hz = 20 \ Hz$.The percentage decrease is $\frac{20}{100} 	imes 100 \% = 20 \%$.

The fundamental frequency of a sonometer wire is $50 \ Hz$ for a given length and tension. If the length is increased by $25 \%$ while keeping the tension constant,what is the percentage change in the frequency of the second harmonic?

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