SHMHard
Question
A disc of mass 3 m and a disc of mass m are connected by massless spring of stiffness k. The heavier disc is placed in on the ground with spring vertical and lighter disc on top. From its equilibrum position, the upper disc is pushed down by a distance δ and released. Then
Options
A.if δ > 3mg /k, the lower disc will bounce up
B.if δ = 2 mg /k, maximum normal reaction from ground on lower disc = 6 mg
C.if δ = 2 mg /k, maximum normal reaction from ground on lower disc = 4 mg
D.if δ = 2 mg /k, the lower disc will bounce up
Solution
From energy conservation
At equilibrium position To leave surface
mg = kx0 kx1 = 3mg
kδ12 = mg (δ1 + x1) + 
kx12
⇒δ12 -
= 0
⇒ δ1
⇒ If δ >
the lower disk will bounce up.
Now If δ =
then maximum normal reaction from ground on lower disk
N = 3mg + k(x0 + δ) = 6mg
At equilibrium position To leave surface
mg = kx0 kx1 = 3mg
kδ12 = mg (δ1 + x1) + kx12⇒δ12 -
= 0⇒ δ1
⇒ If δ >
the lower disk will bounce up.Now If δ =
then maximum normal reaction from ground on lower disk N = 3mg + k(x0 + δ) = 6mg
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