Question

The speed of a transverse wave travelling through a wire having a length $50\text{cm}$and mass $5.0\text{g}$ is $80\text{m/s}$. The area of cross-section of the wire is $1.0{\text{mm}}^2$ and its Young's modulus is $16\times {10}^{11}{\text{N/m}}^2$. Find the extension in the wire with respect to its natural length.

• A. $0.01\text{mm}$
• B. $0.3\text{mm}$
• C. $0.02\text{mm}$
• D. $0.04\text{mm}$

Answer 1

The correct option is C $0.02\text{mm}$

Given: The wave speed is $80\text{m/s}$
The linear mass density of wire is,
$\mu =\frac{m}{l}=\frac{5\times {10}^{-3}\text{kg}}{50\times {10}^{-2}\text{m}}=1.0\times {10}^{-2}\text{kg/m}$
The wave speed is,
$v=\sqrt{\frac{T}{\mu }}$
Thus, the tension in the wire is,
$T=\mu v^2$
$T=\left({10}^{-2}\text{kg/m}\right)\times 6400{\text{m}}^2/{\text{s}}^2$
$\therefore T=64\text{N}$
The Young's modulus is given by
$Y=\frac{\frac{F}{A}}{\frac{\Delta L}{L}}$
The extension in wire is therefore,
$\Delta L=\frac{FL}{AY}$
where applied external force on wire is,
$F=T=64\text{N}$
$\Rightarrow \Delta L=\frac{\left(64\text{N}\right)\left(0.50\text{m}\right)}{(1.0\times {10}^{-6}{\text{m}}^2)\times (16\times {10}^{11}{\text{N/m}}^2)}$
$\therefore \Delta L=2\times {10}^{-5}\text{m}=0.02\text{mm}$