A disc of mass 3 m and a dise of mass m are connected by a massless sping of stiffness k. The heavier is disc placed on the ground with the spring vertical and lighter disc on top from its equilibrium position the upper disc is pushed down by a distance `delta` and released. Then.

Options

A. if `deltagt(3mg)/k` the lower desc will bounce up.
B. if `delta=(2 mg)/k` maximum normal reaction fron on lower disc = 6mg .
C. if `delta=(2 mg)/k` maximum normal reaction from ground on lower disc = 4 mg.
D. if `delta gt (4 mg)/k`, the lower disc will bounce up

Answer

B, D

Explanation

At equilirium
.
`kdelta_(0) =mg`
or `delta_(0) =(mg)/k`
where, `delta_(0) ="compression"`
(b) `delta_("Total") =delta + delta_(0) =(3mg)/k`
`F_(max) =kdelta_(max) =k((3mg)/k) =3mg ("downward")`

`:. N_(max) =3mg + F_(max) =6mg`
(d) If `deltagt(4mg)/k`, the upper block will move a distance `xgt(4mg)/k-delta_(0)` or `xgt(3mg)/k` from natural length.
Hence in this case, extension
`xgt(3mg)/k`
or `F =kxgt3mg` (upward on lower block)
So lower block will bounce up.