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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$\frac{n}{4}$

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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 the mass per unit length.Substituting $\mu$,we get $n = \frac{1}{2l} \sqrt{\frac{T}{\pi r^2 \rho}} = \frac{1}{2lr} \sqrt{\frac{T}{\pi \rho}}$.Since $T$ and $\rho$ are constant,$n \propto \frac{1}{lr}$.Let the initial length be $l_1 = l$ and initial radius be $r_1 = r$. The new length $l_2 = 2l$ and new radius $r_2 = 2r$ (since diameter is doubled,radius is also doubled).The new frequency $n'$ is given by: $\frac{n'}{n} = \frac{l_1 r_1}{l_2 r_2} = \frac{l \cdot r}{(2l) \cdot (2r)} = \frac{1}{4}$.Therefore,$n' = \frac{n}{4}$.

The fundamental frequency of a sonometer wire is $n$. If the length and diameter of the wire are doubled while keeping the tension constant,what is the new fundamental frequency?

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