A sonometer wire of length "1.5" m is made of steel. The tension in it produces an elastic strain of 1%. What is the fundamental frequency of steel if density and elasticity of steel are $7.7\times {10}^{3}kg{m}^{-3}$ and $2.2\times {10}^{11}N{m}^{-2}$ respectively?

Fundamental frequency of vibration of wire is $v=\frac{1}{2L}\sqrt{\frac{T}{\mu }}$

where L is the length of the wire, T is the tension in the wire and μ is the mass per length of the wire As μ=ρA where ρ is the density of the material of the wire and A is the area of cross-section of the wire.

$\therefore v=\frac{1}{2L}\sqrt{\frac{T}{\rho A}}$

Here tension is due to elasticity of wire $\therefore T=YA\left[\frac{\Delta L}{L}\right][AsY=\frac{Stress}{Strain}=\frac{TL}{A\Delta L}]$

Hence, $v=\frac{1}{2L}\sqrt{\frac{Y\Delta L}{\rho L}}$

Here, $T=2.2\times {10}^{-11}N{m}^{-2},\rho =7.7\times {10}^{3}kg{m}^{-3}\frac{\Delta L}{L}=0.01,L=1.5m$

Substituting the given values, we get 

$v=\frac{1}{2\times 1.5}\sqrt{\frac{2.2\times {10}^{11}\times 0.01}{7.7\times {10}^3}}=\frac{{10}^3}{3}\sqrt{\frac{2}{7}}Hz=178.2Hz$