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?

Solution :
Fundamental frequency,
$$\eqalign{ & f = \frac{v}{{2\,\ell }} \cr & = \frac{1}{{2\,\ell }}\sqrt {\frac{T}{\mu }} \cr & = \frac{1}{{2\,\ell }}\sqrt {\frac{T}{{A\,\rho }}} \left[ {\because \,\,v = \sqrt {\frac{T}{\mu }} \,{\text{and }}\mu = \frac{m}{\ell }} \right] \cr & {\text{Also, }}Y = \frac{{T\,\ell }}{{A\,\Delta \ell }} \cr & \Rightarrow \,\,\frac{T}{A} = \frac{{Y\,\Delta \ell }}{\ell } \cr & \Rightarrow \,\,f = \frac{1}{{2\,\ell }}\sqrt {\frac{{\gamma \,\Delta \ell }}{{\ell \rho }}} \,\,\,.....\left( {\text{i}} \right) \cr} $$
Putting the value of $$\ell ,\frac{{\Delta \ell }}{\ell },\rho $$   and $$\gamma $$ in eqn. (i) we get,
$$\eqalign{ & f = \sqrt {\frac{2}{7}} \times \frac{{{{10}^3}}}{3} \cr & {\text{or, }}f \approx 178.2\,Hz \cr} $$