Question

A metal wire of length L1 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 Y1 and $Y_{2}$are the Young’s moduli of the wires respectively, the effective force constant of the system of two wires is

• A. $\frac{\lbrack(Y_{1}Y_{2})A\rbrack}{\lbrack 2(Y_{1}L_{2} + Y_{2}L_{1})\rbrack}$
• B. $\frac{\lbrack(Y_{1}Y_{2})A\rbrack}{(L_{1}L_{2})^{1/2}}$
• C. $\frac{\lbrack(Y_{1}Y_{2})A\rbrack}{(Y_{1}L_{2} + Y_{2}L_{1})}$
• D. $\frac{\lbrack(Y_{1}Y_{2})^{1/2}A\rbrack}{(L_{1}L_{2})^{1/2}}$

Answer 1

Answer: (C)

$\frac{\lbrack(Y_{1}Y_{2})A\rbrack}{(Y_{1}L_{2} + Y_{2}L_{1})}$

Answer 2

: 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 that 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

$K_{eq} = \frac{\left( \frac{Y_{1}A}{L_{1}} \right)\left( \frac{Y_{2}A}{L_{2}} \right)}{\frac{Y_{1}A}{L_{1}} + \frac{Y_{2}A}{L_{2}}} = \frac{(Y_{1}Y_{2})A}{Y_{1}L_{2} + Y_{2}L_{1}}$