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Question: A wire of length L is clamped at two ends so that it lies horizontally and without tension. A weight...

A wire of length L is clamped at two ends so that it lies horizontally and without tension. A weight W is suspended from the middle point of the wire. The vertical depression is ______ Youngs modulus is Y.

A

2Tl24AY+T2l24A2Y2\sqrt{\frac{2Tl^{2}}{4AY} + \frac{T^{2}l^{2}}{4A^{2}Y^{2}}}

B

2Tl24AYT2l24A2Y2\sqrt{\frac{2Tl^{2}}{4AY} - \frac{T^{2}l^{2}}{4A^{2}Y^{2}}}

C

2Tl24AY\sqrt{\frac{2Tl^{2}}{4AY}}

D

Tl2AY\frac{Tl}{2AY}

Answer

2Tl24AY+T2l24A2Y2\sqrt{\frac{2Tl^{2}}{4AY} + \frac{T^{2}l^{2}}{4A^{2}Y^{2}}}

Explanation

Solution

2T cos θ = W

or T = W2cosθ\frac{W}{2\cos\theta}

∆l = Tl2AY\frac{Tl}{2AY}δ = (12+Δl)2l24\sqrt{\left( \frac{1}{2} + \Delta l \right)^{2} - \frac{l^{2}}{4}}

or δ = (12+Tl2AY)2l24=2Tl24AY+T2l24A2Y2\sqrt{\left( \frac{1}{2} + \frac{Tl}{2AY} \right)^{2} - \frac{l^{2}}{4}} = \sqrt{\frac{2Tl^{2}}{4AY} + \frac{T^{2}l^{2}}{4A^{2}Y^{2}}}