Figure shows two discs of same mass m. They are rigidy attached to a spring of stiffness k. The system is in equlibrium. From this equilibrium position, the upper disc is pressed down slowly by a distance x and released. Find the minimum vlue of x, if the lower disc is just lifted off the ground.

Explanation

Let us first find out the minimum elongation (x') needed to lift the lower discc from the ground . THe free body digram of the lower disct at this instant will be

The system given in the question is in equlibrium thus for the equlibrium of upper disc initial compression `(x_(0))` in spring is

now from this equlibrium state spring is further compressed by an amount x and released. Now let us apply energy conservation from this compressed state and final state. Let us take the reference for gravitation potential energy at unstreatched position then
`1/2k(x+x_(0))^(2)=-mg(x+x_(0))=1/2kx^(2)+mgx'`
Now, `x_(0)=(mg)/(k)`
`x=(mg)/(k)`
on solving, we get
`x=(2mg)/(k)`