Question

A wire of cross section $A$ is stretched horizontally between two clamps located $2lm$ apart. A weight $Wkg$ is suspended from the mid-point of the wire. If the mid-point sags vertically through a distance $x<1$ the strain produced is

• A. $\frac{2x^2}{l^2}$
• B. $\frac{x^2}{l^2}$
• C. $\frac{x^2}{2l^2}$
• D. None of these

Answer 1

The correct option is C $\frac{x^2}{2l^2}$

Let the angle made with horizontal after sagging be $\theta$. Thus, change in length, $\Delta l=2lsec\theta -2l$
Also, given, $x=ltan\theta$
Thus, we have $\frac{x^2}{l^2}={tan}^2\theta =1-{sec}^2\theta$, or, ${sec}^2\theta =1-\frac{x^2}{l^2}$
We get, $sec\theta =\sqrt{1-\frac{x^2}{l^2}}$
Since x/l is very small, we can se binomial expansion to get, $sec\theta =1-\frac{x^2}{2l^2}$
Thus, we have strain=$\frac{\Delta l}{2l}=\frac{2lsec\theta -2l}{2l}=sec\theta -1=\frac{x^2}{2l^2}$