The fundamental frequency of a sonometer wire | vedclass

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(C) The fundamental frequency of a sonometer wire is given by the formula: $n = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$,where $L$ is the length,$T$ is the

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$\frac{n}{2 \sqrt{3}}$

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(C) The fundamental frequency of a sonometer wire is given by the formula: $n = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$,where $L$ is the length,$T$ is the tension,and $\mu$ is the linear mass density.Since $\mu = 	ext{Area} 	imes 	ext{Density} = (\pi r^2) ho = \pi (\frac{d}{2})^2 ho = \frac{\pi d^2 ho}{4}$,we can write $n \propto \frac{1}{L d} \sqrt{\frac{T}{ho}}$.Given the initial conditions: $n_1 = n$,$L_1 = L$,$T_1 = T$,$d_1 = d$.New conditions: $L_2 = 3L$,$T_2 = 3T$,$d_2 = 2d$.The new frequency $n_2$ is given by: $n_2 = n_1 	imes (\frac{L_1}{L_2}) 	imes (\frac{d_1}{d_2}) 	imes \sqrt{\frac{T_2}{T_1}}$.Substituting the values: $n_2 = n 	imes (\frac{L}{3L}) 	imes (\frac{d}{2d}) 	imes \sqrt{\frac{3T}{T}}$.$n_2 = n 	imes \frac{1}{3} 	imes \frac{1}{2} 	imes \sqrt{3} = \frac{\sqrt{3}}{6} n = \frac{\sqrt{3}}{2 	imes 3} n = \frac{n}{2 \sqrt{3}}$.

The fundamental frequency of a sonometer wire is $n$. If the tension is increased $3$ times,the length is increased $3$ times,and the diameter is increased $2$ times,what will be the new frequency?

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