Question

A body is attached to the lower end of a vertical spiral spring and it is gradually lowered to its equilibrium position.This stretches the spring by a length d.If the same body attached to the same spring is allowed to fall suddenly, what would be the maximum stretching in this case?

• A. d
• B. 2d
• C. 3d
• D. 1/2d

Answer 1

The correct option is B 2d

As is known, in equilibrium position,

$F=mg=Kd$ or $K=\frac{mg}{d}$

When the object falls freely , its gravitation $P.E$. is converted into elastic $P.E$. of the spring

i.e.,$mgx=\frac{1}{2}K{x}^2=\frac{1}{2}\times \frac{mg}{d}\times x^2$

$\Rightarrow$ $x=2d$