A body of mass 2m is connected to another body of mass m as shown in figure. The mass 2m performs vertical S.H.M. Then find out the maximum amplitude of 2m such that mass m doesn't lift up from the ground.

Explanation

In the given situation 2m mass is in equilibrium condition. Let assume spring is compressed `x_(0)` distance from its natural length.
`implies kx_(0) = mg`
`implies x_(0) = (2mg)/(k)`
The lower block will be lift up, only in the case when the spring force on it will be greater than equal to mg and in upward direction
`implies` kx' = mg
`implies x' = (mg)/(k)`
Above situation arises when 2m block moves upward mg/k from natural length as shown in figure Block m doesn't lift up if the maximum amplitude of the 2m block is
`=(2mg)/(k) + (mg)/(k) = (3mg)/(k)`