Question

A 40 cm long wire having a mass 3.2 gm and area of cross section $1m{m}^2$ is stretched between the support 40.05 cm apart. In its fundamental mode, it vibrates with a frequency of $\frac{1000}{64}\text{Hz}$. The young's modulus of the wire is ${10}^{x}N/m^2$, then $x$ is.

Answer 1

$\mu =\frac{3.2gm}{40cm}=\frac{3.2\times {10}^{-3}}{40\times {10}^{-2}}=\frac{3.2}{400}=\frac{32}{4000}kg/m$
$l=\frac{\lambda }{2}$
$\Rightarrow \lambda =2l.......\left(1\right)$

$f=\frac{v}{\lambda }=\frac{1}{2l}\sqrt{\frac{T}{\mu }}$
$\Rightarrow \frac{1000}{64}=\frac{1}{2\times 40\times {10}^{-2}}\sqrt{\frac{T}{32/4000}}$
$\Rightarrow {[\frac{1000}{64}\times 2\times 40\times {10}^{-2}]}^2\frac{32}{4000}=T$
$\frac{10000}{64}\times \frac{32}{4000}=T\Rightarrow T=\frac{10}{8}N$
$y=\frac{\frac{10/8}{{10}^{-6}}}{\frac{.05\times {10}^{-2}}{40\times {10}^{-2}}}=\frac{{10}^7}{8}\frac{40}{\left(.05\right)}={10}^{9}N/m^2$.