A disc of mass 3 m and a disc of mass m are connected by a massless spring 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 `delta gt 3mg//k`, the lower disc will bounce up
B. if `delta=2 mg//k`, maximum normal reaction from ground on lower disc `=6 mg`
C. if `delta = 2 mg//k`, maximum normal reaction from ground on lower disc `= 4 mg`
D. if `delta gt 4mg//k`, the lower disc will bounce up

Answer

B, D

Explanation

From energy conservation

`1/2kdelta_(1^(2)) = mg(delta_(1) + x_(1)) + 1/2kx_(1^(2))`
`rArr delta_(1^(2)) - (2mg)/(k)delta_(1) = (15m^(2)g^(2))/(k^(2)) = 0`
`rArr delta_(1) = (5mg)/(k) = delta = (4mg)/(k)`
`rArr` If `delta ge (4mg)/(k)` the lower disk will bounce up.
Now if `delta = (2mg)/(k)` then maximum normal reaction
from ground on lower disk
`N = 3mg + k(x_(0) + delta) = 6mg`