Question

A mass $M$ is allowed to fall on a pedestal fixed on the top of a vertical spring. If the height of the mass was $H$ from the pedestal and the compression of the spring is $d$ then the force constant of the spring is given by:

• A. $Mg\frac{(H+d)}{d^2}$
• B. $2Mg\frac{(H-d)}{d^2}$
• C. $\frac{Mg}{2}\frac{H}{d^2}$
• D. $2Mg\frac{(H+d)}{d^2}$

Answer 1

The correct option is D $2Mg\frac{(H+d)}{d^2}$

Given that the height of the mass was $H$ from the pedestal and the compression of the spring is $d$ hence, the loss in potential energy of mass $M$ is $Mg(H+d)$ and gain in potential energy of spring is $\frac{1}{2}k{d}^2$ (where, $k$ is the force constant of spring) are equal.

$\therefore \frac{1}{2}k{d}^2=Mg(H+d)$

$\therefore k=\frac{2Mg(H+d)}{d^2}$