Question

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 Young's modules of the wires respectively, the effective force constant of the system of two wires is:

• A. $\frac{\left[\left(Y_1{Y}_2\right)A\right]}{\left[2(Y_1{L}_2+Y_1{L}_1)\right]}$
• B. $\frac{\left[\left(Y_1{Y}_2\right)A\right]}{{\left(L_1{L}_2\right)}^{1/2}}$
• C. $\frac{\left[\left(Y_1{Y}_2\right)A\right]}{\left[(Y_1{L}_2+Y_2{L}_1)\right]}$
• D. $\frac{\left[{\left(Y_1{Y}_2\right)}^{1/2}A\right]}{{\left(L_1{L}_2\right)}^{1/2}}$

Answer 1

Answer: (C)

$\frac{\left[\left(Y_1{Y}_2\right)A\right]}{\left[(Y_1{L}_2+Y_2{L}_1)\right]}$

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
$k_{eq}=\frac{\left(\frac{Y_{1}A}{L_1}\right)\left(\frac{Y_{2}A}{L_2}\right)}{\frac{Y_{1}A}{L_1}+\frac{Y_{2}A}{L_2}}=\frac{\left(Y_1{Y}_2\right)A}{Y_1{L}_2+Y_2{L}_1}$