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

$\frac{m g}{k} \sqrt{1 - \frac{2 h k}{m g}}$

B

$\frac{m g}{k}$

C

$\frac{m g}{k} + \frac{m g}{k} \sqrt{1 + \frac{2 h k}{m g}}$

D

$\frac{m g}{k} - \frac{m g}{k} \sqrt{1 - \frac{2 h k}{m g}}$

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The correct Answer is:C

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

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Updated on:21/07/2023

Class 11 PHYSICS SIMPLE HARMONIC MOTION

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