Question

A steel wire of length $1\text{m}$ and mass $0.1\text{kg}$, having a uniform cross-sectional area of ${10}^{-6}{\text{m}}^2$ is rigidly fixed at both ends. The temperature of the wire is lowered by ${20}^{\circ }\text{C}$. If the wire is vibrating in harmonic motion, pick from the following possible frequencies of motion (in $\text{Hz}$)
Given Young's modulus of steel $Y=2\times {10}^{11}{\text{N/m}}^2,\alpha =1.21\times {10}^{-5}/{}^{\circ }\text{C}$

• A. 11
• B. 22
• C. 10
• D. 20

Answer 1

Answer: (A), (B)

The correct options are
A 11
B 22

Given that,
Length of the wire, $l=1\text{m}$
Mass of the wire, $m=0.1\text{kg}$
Cross-sectional area of the wire, $A={10}^{-6}{\text{m}}^2$
Change in temperature, $\Delta T={20}^{\circ }\text{C}$
Change in length of the wire due to change in temperature is
$\Delta l=\alpha l\Delta T$

From $Y=\frac{\frac{T}{A}}{\frac{\Delta l}{l}}$
$\Rightarrow T=YA\frac{\Delta l}{l}$
$\Rightarrow T=\alpha YA\Delta T$
$\Rightarrow T=1.21\times {10}^{-5}\times 2\times {10}^{11}\times {10}^{-6}\times 20=48.4\text{N}$
Speed of wave is
$v=\sqrt{\frac{T}{\mu }}=\sqrt{\frac{48.4}{\left(\frac{0.1}{1}\right)}}=22\text{m/s}$
For harmonic motion,
$l=\frac{n\lambda }{2}$
$\Rightarrow \lambda =\frac{2l}{n}=\frac{2}{n}$
$f=\frac{v}{\lambda }=\frac{22}{2}\times n=11n$
Substituting $n=1,2..$, we get $f=11\text{Hz},22\text{Hz}$ as possible frequencies