A wire of cross-section A is stretched horizontally between two clamps located 2ℓ meters apart. A weight W kg is suspended from the mid point of the wire. If the mid point sags vertically through a distance x

Engineering

Physics

Elasticity

Question

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

$\frac{x^{2}}{2 ℓ^{2}}$

none of these

$\frac{2 x^{2}}{ℓ^{2}}$

$\frac{x^{2}}{ℓ^{2}}$

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New length = $2 \sqrt{(ℓ^{2} + x^{2})}$
Old length = 2ℓ
strain = $\frac{2 \sqrt{ℓ^{2} + x^{2}} - 2 ℓ}{2 ℓ}$ = $\sqrt{1 + {(\frac{x}{ℓ})}^{2}} - 1$

= 1 + $\frac{1}{2} {(\frac{x}{ℓ})}^{2}$ + higher terms of x²/ℓ² ..... – 1
Neglecting higher terms
Strain = $\frac{1}{2} {(\frac{x}{ℓ})}^{2}$

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