The fundamental frequency of a sonometer wire | vedclass

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Question

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

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) The fundamental frequency $n$ 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 $\mu = \pi r^2 ho$ (where $r$ is the radius and $ho$ is the density),the frequency can be written as: $n = \frac{1}{2L} \sqrt{\frac{T}{\pi r^2 ho}} = \frac{1}{2Lr} \sqrt{\frac{T}{\pi ho}}$.Given that the diameter $D = 2r$,we have $n \propto \frac{1}{LD} \sqrt{T}$.Let the initial frequency be $n_1 = n$. The new frequency $n_2$ is given by the ratios of the changes:$n_2 = n_1 	imes \left( \frac{L_1}{L_2} ight) 	imes \left( \frac{D_1}{D_2} ight) 	imes \sqrt{\frac{T_2}{T_1}}$.Given $L_2 = 3L_1$,$D_2 = 2D_1$,and $T_2 = 3T_1$:$n_2 = n 	imes \left( \frac{1}{3} ight) 	imes \left( \frac{1}{2} ight) 	imes \sqrt{3} = n 	imes \frac{\sqrt{3}}{6} = \frac{\sqrt{3}}{2 	imes 3} n = \frac{\sqrt{3}}{2 	imes \sqrt{3} 	imes \sqrt{3}} n = \frac{n}{2 \sqrt{3}}$.Thus,the new frequency is $\frac{n}{2 \sqrt{3}}$.

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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