Question

A spiral spring of sufficient stiffness is hanging vertically from the ceiling. Initially, it is in its natural length. A body of mass $m$ is attached to its lower end and is gradually lowered from this position, till it attains equilibrium. This stretches the spring by a length $d$. If the same body, attached to the same spring in the same initial position, is allowed to fall suddenly, what would be the maximum elongation of the spring in this case ?
(Neglect mass of the spring.)

• A. $d$
• B. $2d$
• C. $3d$
• D. $\frac{1}{2}d$

Answer 1

The correct option is B $2d$

In the first case, when spring attains equilibrium.

$F_{spring}=mg$

$kd=mg\Rightarrow d=\frac{mg}{k}$

When the body is allowed to fall suddenly, it does not stop in its previous equilibrium position because of kinetic energy, and that extension can be calculated by balancing of energy, the decrease in gravitational potential energy will be equal to the energy stored in the spring.

$\therefore$ decrease in gravitational PE = increase in elastic PE

Let $d^{\prime }$ be the maximum compression of the spring. So

$mg{d}^{\prime }=\frac{1}{2}k{d}^{\prime 2}\Rightarrow d^{\prime }=\frac{2mg}{k}$

$\therefore d^{\prime }=2d$

Therefore, option $\left(B\right)$ is the right choice.