A mass of 2kg resting upon a horizontal table is connected t

A mass of 2kg resting upon a horizontal table is connected to a spring whose stiffness is 4N/m. If the damping constant is 4 N-sec/m and the mass is displaced 1 m to the right of the equilibrium position and released from rest, find the motion of the mass and its furthest displacement from the equilibrium position. (Equation m + b + lambda x = f(t) becomes 2 + 4 + 4x = 0. The direction is to the right in the positive x-direction, and the initial data is: x(0) = 1, (0) = 0.

Solution

Given differential equation

2X\"+4X\'+4X=0

The Auxilary equation is

2m²+4m+4=0

m²+2m+2=0

m₁=-1+i and m₂=-1-i

The General solution of the differential equation is

X(t)=C₁e^-tCos(t)+C₂e^-tSIn(t)

Given at t=0=>X(0)=1

C₁+0=1=>C₁=1

X\'(t)=C₁[-e^-tSIn(t)-e^-tCos(t)] +C₂[e^-tCos(t)-e^-tsin(t)]

Given at t=0=>X\'(t)=0

0=C₁[-0-1]+C₂[1-0]

C₂=C₁ =1

X(t) =e^-t[Cost+Sint]