A sonometer wire vibrates in its fundamental mode with a frequency f . If the length of the wire is increased by 25% and the tension remains constant, what is the new fundamental frequency?

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The fundamental frequency of a sonometer wire is $f = \frac{1}{2 L} \frac{T}{μ}$ . Let the initial length be $L_{1} = L$ and initial tension $T_{1} = T$ . The initial frequency is $f_{1} = f$ . So, $f = \frac{1}{2 L} \frac{T}{μ}$ .

The length is increased by $25%$ , so $L_{2} = L + 0.25 L = 1.25 L$ . The tension remains constant, $T_{2} = T$ . The linear mass density $μ$ remains constant. The new fundamental frequency $f_{2}$ is: $f_{2} = \frac{1}{2 ( 1.25 L )} \frac{T}{μ} = \frac{1}{1.25} (\frac{1}{2 L} \frac{T}{μ})$ $f_{2} = \frac{1}{1.25} f = \frac{1}{5/4} f = \frac{4}{5} f = 0.8 f$

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