Question

A mild steel wire of length 1.0 m and cross-sectional area 0.50 × 10^–2 cm² is stretched, well within its elastic limit, horizontally between two pillars. A mass of 100 g is suspended from the mid-point of the wire. Calculate the depression at the midpoint.

Answer 1

Original length = XZ
Depression = l
The length after mass m, is attached to the wire = XO + OZ
Increase in the length of the wire:
Δl = (XO + OZ) – XZ
Where,

Expanding the expression and eliminating the higher terms:

Hence, the depression at the midpoint is 0.0106 m.