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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

mgk12hkmg

B

mgk

C

mgk+mgk1+2hkmg

D

mgkmgk12hkmg

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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) =mg(h+y), where h is the height through which the mass falls on the pan. Increases in elastic potential of the spring =12ky2 (according to low of conservation of energy)
or mg(h+y)=12ky2
or ky22mgy2mgh=0
y=2mg±4m2g2+8mghk2k
=mgk±mgk(1+2hkmg)
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=[mgk+mgk(1+2hkmg)]

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Updated on:21/07/2023

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