Two discs, each having mass `m`, are attached rigidly to the ends of a vertical spring. One of the discs rests on a horizontal surface and the other produces a compression `x_0` on the spring when it is in equilibrium. How much further must the spring be compressed so that when the force causing compression is removed, the extension of the spring will be able to lift the lower disc off the table?

Options

A. `0.5 y`
B. `3y`
C. `2y`
D. `y`

Answer

C

Explanation

Let upper block is pushed down by `x`, at equilibrium `mg = ky`, i.e, weight of upper of upper block is balanced by spring When it is deformed bu `y`, upper block will perform `SHM` with amplitude `x` equilibrium position, lower block will leave surface when spring is extended by `y`, means upper block is at distance `2y` from its mean position. That should be upper extreme position of upper block. So amplitude `x = 2y`
Alternate
at equilibrium of upper block `mg = ky`
lower plate will leave the surface if the extension in spring is `y`
Let upper plate is displaaced by `x` downward and left
so by energy conservation between compressed to extended positions
`0 + (1)/(2) k (x + y)^(2) = mg(x + 2y) = (1)/(2) ky^(2)`
`rArr (1)/(2) kx^(2) + (1)/(2) ky^(2) + kxy = mgx + mg2y + (1)/(2) ky^(2) rArr x = 2y`