A mild steel wire of length \(2L\) and cross-sectional area \(A\) is stretched, well within the elastic limit, horizontally between two pillars (figure). A mass \(m\) is suspended from the mid-point of the wire. Strain in the wire is:

    

1. \( \dfrac{x^2}{2 L^2} \) 2. \(\dfrac{x}{\mathrm{~L}} \)
3. \(\dfrac{x^2}{L}\) 4. \(\dfrac{x^2}{2L}\)
(1) Hint: Use Pythagoras theorem to find the change in length.
Step 1: Find the change in length.  
Consider the diagram below
Hence, change in length
L=BO+OC-(BD+DC)
=2BO-2BD                                  ( BO=OC, BD=DC)
=2[BO-BD]
=2x2+L212-L
=2L1+x2L212-1
=2L1+x22L2-1=x2L
Step 2: Find the strain.
Strain=LL=x22L2