Question

A mass $M$ is suspended by two springs $A$ and $B$ of force constants $k_{1}$ and $k_{2}$ respectively as shown in the diagram. The total stretch of springs in equilibrium is:

• A. $\frac{M g}{k_{1} + k_{2}}$
• B. $\frac{2 Mg}{\left(k_{1} + k_{2}\right)}$
• C. $\frac{M g}{2 \left(k_{1} + k_{2}\right)}$
• D. $\frac{M g \left(k_{1} + k_{2}\right)}{k_{1} k_{2}}$

Answer 1

Answer: (D)

In series combination of spring, equivalent spring constant,
$\frac{1}{k_{eq}}=\frac{1}{k_{1}}+\frac{1}{k_{2}}$
$\Rightarrow k_{eq}=\frac{k_{1} k_{2}}{k_{1} + k_{2}}$
At equilibrium,
$mg=k_{eq}x$
$\Rightarrow x=\frac{m g \left(k_{1} + k_{2}\right)}{k_{1} k_{2}}$