A body of mass $m$ falls from a height $h$ onto the pan of a spring balance. The masses of the pan and spring are negligible. The force constant of the spring is $k$ . The body sticks to the pan and oscillates simple harmonically. Find the time period of oscillations and amplitude of motion.

Let the spring is compressed by $y$ , then by conservation of mechanical energy, we have

             $mg(h+y) = \frac{1}{2}k{y}^2$

or         $\frac{k{y}^2}{2}-mgy-mgh = 0$

or          $y=\frac{mg\pm \sqrt{{\left(mg\right)}^2+4\times {\displaystyle \frac{k}{2}}\times mgh}}{a\left(\frac{k}{2}\right)}$

or          $y = \frac{mg}{k}\pm \frac{\sqrt{{\left(mg\right)}^2+2kmgh}}{k}$

where $\frac{mg}{k}$ is known as static deflection and $\frac{\sqrt{{\left(mg\right)}^2+2kmgh}}{k}$ is known as amplitude of motion.

The time period of motion is given by $T = 2\mathrm{\pi }\sqrt{\frac{\mathrm{m}}{\mathrm{k}}}$ .