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

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 Young's moduli, we get
$${k_{eq}} = \frac{{\left( {\frac{{{Y_1}A}}{{{L_1}}}} \right)\left( {\frac{{{Y_2}A}}{{{L_2}}}} \right)}}{{\frac{{{Y_1}A}}{{{L_1}}} + \frac{{{Y_2}A}}{{{L_2}}}}} = \frac{{\left( {{Y_1}{Y_2}} \right)A}}{{{Y_1}{L_2} + {Y_2}{L_1}}}$$