Question

A steel wire of length $1$m, mass $0.1$ kg and uniform cross sectional area ${10}^{-6}{m}^2$ is rigidity fixed at both ends. The temperature of the wire is lowered by ${20}^o$C. If transverse waves are set up by plucking the string in the middle, calculate the frequency of fundamental mode of vibration. Young's modulus of steel $=2\times {10}^{11}N/m^2$ and coefficient of linear expansion of steel $=1.21\times {10}^{-5}{C}^{-1}$.

Answer 1

Coefficient of linear expansion is given by
$\alpha =\frac{\Delta l}{l\times \Delta \theta }$ where $\Delta \theta =$ change in temperature
Hence internal strain $=(\alpha .\Delta \theta )$
$\therefore$ Stress $=Y\times$ strain $=Y\alpha \Delta \theta$
Tension $T=\pi r^2\times$ stress $=\pi r^{2}Y\alpha \Delta \theta$
$\therefore v=\sqrt{\frac{T}{m}}$
$\therefore n=\frac{1}{2l}\sqrt{\frac{T}{m}}=\frac{1}{2l}\sqrt{\frac{\pi r^{2}Y\alpha \Delta \theta }{m}}=\frac{1}{2\times 1}\times \sqrt{\frac{{10}^{-6}\times 2\times {10}^{11}\times 1.21\times {10}^{-5}\times 20}{0.1}}=11$Hz.