Question

A steel wire of length $1m$, mass $0.1kg$ and uniform cross sectional area ${10}^6{m}^2$ is rigidly fixed at both ends. The temp difference of wire is ${20}^{o}C$. If the transverse waves are set up by plucking the string in the middle calculate freq. of fundamental mode of vibration:
($y=2\times {10}^{11}N/m^2,\alpha =1.21\times {10}^5{/}^{o}C$)

Answer 1

given

$\begin{array}{c}l=1m\\ m=0.1kg\\ \gamma =2\times {10}^{11}N/m^2\\ A={10}^6{m}^2\\ \alpha =1.21\times {10}^5{/}^{\circ }C\\ f=?\end{array}$

we know that

$\begin{array}{c}v=\sqrt{\frac{f}{m}}\\ =\sqrt{\frac{\gamma A\Delta \theta }{\delta A}}\\ v=\sqrt{\frac{2\times {10}^{11}\times 1.21\times {10}^5\times 20}{{10}^{-7}\times {10}^2}}\\ v=22\times {10}^{11}m/s\end{array}$