Question

A body of mass $m$ is released from a height $h$ to a scale pan hung from a spring. The spring constant of the spring is $k$, the mass of the scale pan is negligible and the body does not bounce relative to the pan; then the amplitude of vibration is

• A. $\frac{mg}{k}\sqrt{1-\frac{2hk}{mg}}$
• B. $\frac{mg}{k}$
• C. $\frac{mg}{k}+\frac{mg}{k}\sqrt{1+\frac{2hk}{mg}}$
• D. $\frac{mg}{k}-\frac{mg}{k}\sqrt{1-\frac{2hk}{mg}}$

Answer 1

Answer: (C)

$\frac{mg}{k}+\frac{mg}{k}\sqrt{1+\frac{2hk}{mg}}$

Decrease in potential energy of the mass when the pan gets lowered by distance $y$ (due to mass hitting on the pan) $=mg(h+y)$, where $h$ is the height through which the mass falls on the pan. Increases in elestic potential of the spring $=\frac{1}{2}k{y}^2$ (according to law of conservation of energy)
or $mg(h+y)=\frac{1}{2}k{y}^2$
or $k{y}^2-2mgy-2mgh=0$
$\therefore y=\frac{2mgh\pm \sqrt{4m^2{g}^2+8mghk}}{2k}$
$=\frac{mg}{k}\pm \frac{mg}{k}\sqrt{(1+\frac{2hk}{mg})}$
The velocity of the pan will be maximum at the time of the collision and will be zero at the lowest position.
Hence $y$ should be the amplitude of oscillation.
So, amplitude of vibration $=[\frac{mg}{k}+\frac{mg}{k}\sqrt{(1+\frac{2hk}{mg})}]$