Question

A metal wire of diameter $1mm$ is held on two knife edges separated by a distance of $50cm$. The tension in the wire is $100N$. The wire, vibrating with its fundamental frequency, and a vibrating tuning fork together produce $5$ beats/sec. The tension in the wire is taken reduced to $81N$. When the two are excited, beats are heard again at the same rate.
Calculate
$\left(i\right)$ the frequency of the fork.
$\left(ii\right)$ the density of the material of the wire.

Answer 1

Given rode freq $=5$

$\frac{1}{2l}\sqrt{\frac{100}{14}-w}=\mathrm{5....}\left(1\right)$

Also

$w-\frac{1}{2l}\sqrt{\frac{81}{14}}=\mathrm{5....}\left(2\right)$

$\left(1\right)+\left(2\right)$

$\frac{1}{2L}\left[\frac{1}{\sqrt{14}}\right]=10$

$\Rightarrow \sqrt{14}=\frac{1}{2L\times 10}=\frac{1}{10}$

$\therefore 14=\rho .\pi r^2=0.01$

$\Rightarrow \rho =\frac{0.1}{(0.5\times {10}^{-3}{)}^2\pi }=\frac{40}{\pi }\times {10}^{3}kg/m^3$

from $\left(1\right)$

$\frac{1}{2l}\sqrt{\frac{100}{14}}-v=5$

$\Rightarrow \frac{1}{2l}\sqrt{\frac{100}{0.1}}-v=5$

$\Rightarrow v=100-5=95Hz$