Consider the system shown in figure. Pulley, string and spring are ideal and `Deltam lt lt M`. Initially spring is in it’s natural length and both the blocks are at rest. (Assume that initially `Deltam` was situated at origin). Maximum y coordinate of `Deltam` in subsequent motion is xmg/k then value of x is .

Options

A. 1
B. 2
C. 3
D. 4

Explanation

As `Deltam lt lt m`
So, we can assume that motion of mass M will not be influenced by `Deltam`. Now, when total force on mass M by energy conservation
`rArr kx=mg rArr x=(mg)/k`
Now, maximum downwards displacement of M
`2x=2(Mg)/k`
As block `Deltam` is connected to mass M so its maximum upwards displacement `=(2 Mg)/k`