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A uniform wire of length $\mathrm{L}$, diameter $\mathrm{D}$ and density $\rho$ is stretched under a tension T. The correct relation between its fundamental frequency $f$, the length $L$ and the diameter $D$ is
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Verified Answer
The correct answer is:
$f \propto \frac{1}{L D}$
The fundamental frequency is $f=\frac{1}{2 L} \sqrt{\frac{T}{\mu}}$
$$
f=\frac{1}{2 L} \sqrt{\frac{T}{\rho \pi \frac{D^{2}}{4}}}=\frac{1}{L D} \sqrt{\frac{T}{\pi \rho}}
$$
$\therefore$
$$
f \propto \frac{1}{L D}
$$
$$
f=\frac{1}{2 L} \sqrt{\frac{T}{\rho \pi \frac{D^{2}}{4}}}=\frac{1}{L D} \sqrt{\frac{T}{\pi \rho}}
$$
$\therefore$
$$
f \propto \frac{1}{L D}
$$
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