First slide

Elastic behaviour of solids

Question

A metal wire of length L₁ and area of cross-section A is attached to a rigid support. Another metal wire of length L₂ and of the same cross-sectional area is attached to the free end of the first wire. A body of mass M is then suspended from the free end of the second wire. If Y₁ and Y₂ are the Young's moduli of the wires respectively, the effective force constant of the system of two wires is

Moderate

A

$\frac{[(Y_{1} Y_{2}) A]}{[2 (Y_{1} L_{2} + Y_{2} L_{1})]}$
B

$\frac{[(Y_{1} Y_{2}) A]}{{(L_{1} L_{2})}^{1 / 2}}$
C

$\frac{[(Y_{1} Y_{2}) A]}{(Y_{1} L_{2} + Y_{2} L_{1})}$
D

$\frac{[{(Y_{1} Y_{2})}^{1 / 2} A]}{{(L_{1} L_{2})}^{1 / 2}}$

Solution

Using the usual expression for the Young's modulus, the force constant for the wire can be written as
$k = \frac{F}{ΔL} = \frac{YA}{L}$
where the symbols have their usual meanings.
When the two wires are connected together in series, the effective force constant is given by
$k_{eq} = \frac{k_{1} k_{2}}{k_{1} + k_{2}}$
Substituting the corresponding lengths, area of cross sections and the Young's moduli, we get
$k_{eq} = \frac{(\frac{Y_{1} A}{L_{1}}) (\frac{Y_{2} A}{L_{2}})}{\frac{Y_{1} A}{L_{1}} + \frac{Y_{2} A}{L_{2}}} = \frac{(Y_{1} Y_{2}) A}{Y_{1} L_{2} + Y_{2} L_{1}}$

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