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.

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$2 \sqrt{\frac{m}{k}}, \frac{\sqrt{{(mg)}^{2} + kmgh}}{k}$

$2 π \sqrt{\frac{m}{k}}, \frac{\sqrt{{(mg)}^{2} + 2 kmgh}}{k}$

$2 \sqrt{\frac{m}{k}}, \frac{\sqrt{(mg) + 2 kmgh}}{k}$

$2 π \sqrt{\frac{m}{k}}, \frac{\sqrt{{(mg)}^{2} - 2 kmgh}}{k}$

answer is C.

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Detailed Solution

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

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

or $\frac{k y^{2}}{2} - m g y - m g h = 0$

or $y = \frac{m g \pm \sqrt{{(m g)}^{2} + 4 \times \frac{k}{2} \times m g h}}{a (\frac{k}{2})}$

or $y = \frac{m g}{k} \pm \frac{\sqrt{{(m g)}^{2} + 2 k m g h}}{k}$

where $\frac{m g}{k}$ is known as static deflection and $\frac{\sqrt{{(m g)}^{2} + 2 k m g h}}{k}$ is known as amplitude of motion.

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

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