Question

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

Answer 1

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


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