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 two supports $40.05\text{cm}$ apart. In its fundamental mode, it vibrates with a frequency $\frac{1000}{64}\text{Hz}$. Find the Young's modulus of the wire.

• A. $1\times {10}^9{\text{N-m}}^2$
• B. $2\times {10}^9{\text{N-m}}^2$
• C. $3\times {10}^9{\text{N-m}}^2$
• D. $4\times {10}^9{\text{N-m}}^2$

Answer 1

The correct option is A $1\times {10}^9{\text{N-m}}^2$

As linear mass density is
$\mu =\frac{3.2\text{gm}}{40.05\text{cm}}=\frac{3.2\times {10}^{-3}}{40.05\times {10}^{-2}}\approx \frac{32}{4000}\text{kg/m}$
For fundamental mode, $l=\frac{\lambda }{2}$
$\Rightarrow \lambda =2l$


We use, fundamental frequency
$n_0=\frac{V}{\lambda }=\frac{1}{2l}\sqrt{\frac{T}{\mu }}$
$\Rightarrow \frac{1000}{64}=\frac{1}{2\times 40.05\times {10}^{-2}}\sqrt{\frac{T}{\frac{32}{4000}}}$
$\Rightarrow {[\frac{1000}{64}\times 2\times 40.05\times {10}^{-2}]}^2\frac{32}{4000}=T$
$\Rightarrow {\left(\frac{1000}{64}\right)}^2\times 4\times 1600\times {10}^{-4}\times \frac{32}{4000}\approx T$
$\Rightarrow T\approx \frac{10}{8}\text{N}$

From Hooke's law, Young's modulus can be calculated as
$Y=\frac{T/A}{\Delta L/L}=\frac{10\times 0.4}{8\times {10}^{-6}\times 0.05\times {10}^{-2}}={10}^9{\text{N/m}}^2$