The fundamental frequency of a sonometer wire | vedclass

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(D) The fundamental frequency $n$ of a sonometer wire is given by the formula: $n = \frac{1}{2l} \sqrt{\frac{T}{\mu}}$,where $\mu = \pi r^2 \rho$ is t

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

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(D) The fundamental frequency $n$ of a sonometer wire is given by the formula: $n = \frac{1}{2l} \sqrt{\frac{T}{\mu}}$,where $\mu = \pi r^2 ho$ is the mass per unit length.Thus,$n = \frac{1}{2l} \sqrt{\frac{T}{\pi r^2 ho}} = \frac{1}{2lr} \sqrt{\frac{T}{\pi ho}}$.From this,we see that $n \propto \frac{\sqrt{T}}{lr}$.Let the initial values be $l_1 = l$,$T_1 = T$,and $r_1 = r$. The new values are $l_2 = 3l$,$T_2 = 3T$,and $r_2 = 3r$.The ratio of the frequencies is given by:$\frac{n_2}{n_1} = \sqrt{\frac{T_2}{T_1}} 	imes \frac{l_1}{l_2} 	imes \frac{r_1}{r_2}$$\frac{n_2}{n} = \sqrt{\frac{3T}{T}} 	imes \frac{l}{3l} 	imes \frac{r}{3r}$$\frac{n_2}{n} = \sqrt{3} 	imes \frac{1}{3} 	imes \frac{1}{3} = \frac{\sqrt{3}}{9} = \frac{1}{3\sqrt{3}}$Therefore,$n_2 = \frac{n}{3\sqrt{3}}$.

The fundamental frequency of a sonometer wire is $n$. If the length,tension,and diameter of the wire are all tripled,what is the new fundamental frequency?

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