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

Explanation

As the pan is of negligible mass, there is no loss of kinetic energy even though the collision is inelastic. The meachanical energy of the body `m` in the fiels generated by the joing action of both the gravity force and the elastic force is conserved `i.e.,DeltaE=0`. During the motion of the body `m` from the initial to the final ( positon of maxiumu compression of the spring) position `DeltaT=0`, and therefore` DeltaU=DeltaU_(gr)+DeltaU_(sp)=0`
or `-mg(h+x)+(1)/(2)kx^(2)=0`
On solving the quadratic equation `:`
`x=(mg)/(k)+-sqrt((m^(2)g^(2))/(k^(2))+(2mgh)/(k))`
As minus sign is not acceptable
`x=(mg)/(k)+sqrt((m^(2)g^(2))/(k^(2))+(2mgh)/(k))`
If the body `m` were at rest on the spring, the corresponding position of `m` will be its equilibrium position and at this position the resultant force on the body `m` will be zero. Therefore the equilibrium compression `Deltax` (say ) due to the body `m` will be given by
`k Deltax= mg ` or `Deltax=mg//k`
Therefore separation betweent the equilibrium position and one of the extremen position i.e. the sought amplitude lt brgt `a=x-Deltax+sqrt((m^(2)g^(2))/(k^(2))+(2mgh)/(k))`
The mechanical energy of oscillation which is conserved equals `E=U_(extreme)`, because at the extreme position kinetic energy becomes zero .
Althought the weight of body `m` is a conservative force, it is not restoring in this problem, hence `U_(extreme)` is only concerned with the spring force. Therefore
`E=U_(extreme)=(1)/(2)ka^(2)=mgh+(m^(2)g^(2))/(2k)`