The fundamental frequency of a sonometer wire is 50 Hz for some length and tension. If the length is increased by 25% by keeping tension same then frequency change of second harmonic is

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The fundamental frequency of a sonometer wire is given by $n = \frac{1}{2 L} \frac{T}{m}$ .

The frequency of the second harmonic is $f = 2 n = \frac{1}{L} \frac{T}{m}$ .

Since tension $T$ and linear mass density $m$ are constant, $f \propto \frac{1}{L}$ .

Let the initial length be $L_{1}$ and the initial frequency of the second harmonic be $f_{1}$ .

The new length is $L_{2} = L_{1} + 0.25 L_{1} = 1.25 L_{1} = \frac{5}{4} L_{1}$ .

The new frequency of the second harmonic is $f_{2} = f_{1} \times \frac{L _{1}}{L _{2}} = f_{1} \times \frac{4}{5} = 0.8 f_{1}$ .

Percentage change in frequency is $\frac{f _{2} - f _{1}}{f _{1}} \times 100% = \frac{0.8 f _{1} - f _{1}}{f _{1}} \times 100% = - 20%$ .

Therefore, the frequency of the second harmonic is decreased by $20%$ .

Answer: decreased by $20%$

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