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

Create a free account to view solution

View Solution Free
Topic: SHM·Practice all SHM questions

More SHM Questions

For a particle executing simple harmonic motion, the acceleration is proportional to...Which vector in the figures best represents the acceleration of a pendulum mass at the intermediate point in its swing?...One end of a spring of force constant k is fixed to a fixed vertical wall and the other to a block of mass m resting on ...The magnitude of average acceleration in half time period from equilibrium position in a simple harmonic motion is...The function sin2(ωt) represents...