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

Options

A. `(mg)/(k)sqrt(1-(2hk)/(mg))`
B. `(mg)/(k)`
C. `(mg)/(k)+(mg)/(k)sqrt(1+(2hk)/(mg))`
D. `(mg)/(k)-(mg)/(k)sqrt(1-(2hk)/(mg))`

Answer

C

Explanation

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 `=(1)/(2)ky^2` (according to low of conservation of energy)
or `mg(h+y)=(1)/(2)ky^2`
or `ky^2-2mgy-2mgh=0`
`y=(2mg+-sqrt(4m^2g^2+8mghk))/(2k)`
`=(mg)/(k)+-(mg)/(k)sqrt((1+(2hk)/(mg))`
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`=[(mg)/(k)+(mg)/(k)sqrt((1+(2hk)/(mg)))]`