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 other produces a compression \(x_o\) 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 ?

\(2x_o\)

A is the position of the spring when it is in its normal uncompressed length. The upper disc compresses the spring by \(x_o\) when spring is in equilibrium.

So, \(k x_o = mg\). Hence, B is the equilibrium position of the spring. Let its be further compressed by y and released. After releasing it can be proved that the spring will go up to the position D so that BC = BD.

Extension in the spring at this position = \(y - x_o\)

Now, for lifting up of the lower disc : \(k (y - x_o) = mg\)

\(\Rightarrow y = \frac{mg}{k} + x_o\)

\(\Rightarrow y = x_o + x_o = 2 x_o\)