Question

A steel wire 1m long, mass=0.1kg and uniform cross-sectional area 10^-6 m² is rigidly fixed at both ends. The temperature of the wire is lowered by 20 ^oC. If transverse waves are set up by plucking the string in the middle, calculate the frequency of the fundamental mode of vibration.
{for steel wire y=2 * 10¹¹ N/m² α=1.21 * 10^-5 /^oC}

Answer 1

Dear Student,

$$\begin{gathered} \mathrm{Using}\mathrm{the}\mathrm{formula}\mathrm{of}\mathrm{coefficient}\mathrm{of}\mathrm{linear}\mathrm{expansion}, \\ ∆\mathrm{l}=\mathrm{l\alpha }\times ∆\mathrm{\theta } \\ \mathrm{Also},\mathrm{Y}=\frac{\mathrm{stress}}{\mathrm{strain}} \\ \mathrm{Y}=\frac{{\displaystyle \raisebox{1ex}{$\mathrm{T}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{A}$}\right.}}{{\displaystyle \raisebox{1ex}{$∆\mathrm{l}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{l}$}\right.}} \\ \mathrm{Y}=\frac{{\displaystyle \raisebox{1ex}{$\mathrm{T}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{A}$}\right.}}{{\displaystyle \mathrm{\alpha }\times ∆\mathrm{\theta }}} \\ \mathrm{T}=\mathrm{YA\alpha }∆\mathrm{\theta } \\ \mathrm{Frequency}\mathrm{of}\mathrm{fundamental}\mathrm{mode}\mathrm{of}\mathrm{vibration}, \\ \mathrm{\nu }=\frac{1}{2\mathrm{l}}\sqrt{\frac{\mathrm{T}}{\mathrm{m}}} \\ \mathrm{\nu }=\frac{1}{2\mathrm{l}}\sqrt{\frac{\mathrm{YA\alpha }∆\mathrm{\theta }}{\mathrm{m}}} \\ \mathrm{\nu }=\frac{1}{2\times 1}\sqrt{\frac{2\times {10}^{11}\times {10}^{-6}\times 1.21\times {10}^{-5}\times 20}{0.1}}=11\mathrm{Hz} \end{gathered}$$