Question

A mild-steel wire of length $2L$ and cross-sectional area $A$ is stretched well within elastic limit horizontally between two pillars and a mass $m$ is suspended from the midpoint of the wire as shown in figure. If strain in the wire is equal to $n\times \frac{x^2}{L^2}$. The value of $n$ is .(Assume $x<<L$)

Answer 1

Change in length, $\Delta L=AC-AO$
$={[L^2+x^2]}^{\frac{1}{2}}-L$
For small values of $x$ applying Binomial theorem
$\Rightarrow L[1+\frac{1}{2}\frac{x^2}{L^2}]-L=\frac{x^2}{2L}$
Longitudinal strain $\frac{\Delta L}{L}=\frac{x^2/2L}{L}=\frac{x^2}{2L^2}$
$\Rightarrow n=0.5$