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Q. A metal wire of length $L_{1}$ and area of cross-section $A$ is attached to a rigid support. Another metal wire of length $L_{2}$ 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 $Y_{1}$ and $Y_{2}$ are the Young’s moduli of the wires respectively, the effective force constant of the system of two wires is

Mechanical Properties of Solids

Solution:

Using the usual expression for the Young’s modulus, the force constant for the wire can be written as
$k=\frac{F}{\Delta L}$ $=\frac{YA}{L}$
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 Youngs moduli, we get
$k_{eq}$ $=\frac{\left(\frac{Y_{1}A}{L_{1}}\right)\left(\frac{Y_{2}A}{L_{2}}\right)}{\frac{Y _{1}A}{L_{1}}+\frac{Y _{2}A}{L_{2}}}$ $=\frac{\left(Y_{1}Y_{2}\right)A}{Y_{1}L_{2}+Y_{2} L_{1}}$