Laplace Transforms: Vertical Motion of a Coupled Spring System

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Problem Statement

Figure 1. Coupled Spring System.

Derive the system of differential equations describing the straight-line vertical motion of the coupled spring shown in Figure 1. Use Laplace transform to solve the system when

k1=k2=k3=1

,

m1=m2=1

, and

x1(0)=0

,

x′1(0)=−1

,

x2(0)=0

, and

x'2(0)=1

.

Solution

At positions x1 and x2, the masses m1 and m2 are in equilibrium. Thus, the motion equations for m1 and m2 are,

m1x¨1=−k1x1+k2(x2−x1)
∴ m1x¨1+k1x1−k2(x2−x1)=0
m2x¨2=−k2(x2−x1)−k3x2
∴ m2x¨2+k2(x2−x1)−k3x2=0


where m1x′'1 and m2x′'2 represent the Newton's Second Law of Motion and −k1x1+k2(x2−x1) and −k2(x2−x1)−k3x2 represent the net forces acting in the masses.

Laplace Transform

Applying the Laplace Transform to the motion equations for this systems, we obtain,

ℒ[m1x¨1+k1x1−k2(x2−x1)]=0
m1[s2X1(s)−sx1(0)−x˙1(0)]+k1X1(s)−k2(X2(s)−X1(s))=0
X1(s)(m1s2+k1+k2)=m1(sx1(0)−x˙1(0))+k2X2(s)
X1(s)=m1(sx1(0)+x˙1(0))+k2X2(s)(m1s2+k1+k2)
ℒ[m2x¨2+k2(x2−x1)+k3x2]=0
m2[s2X2(s)−sx2(0)−x˙2(0)]+k2(X2(s)−X1(s))+k3X2(s)=0
X2(s)(m2s2+k2+k3)=m2(sx2(0)−x˙2(0))+k2X1(s)
X2(s)=m2(sx2(0)+x˙2(0))+k2X1(s)(m2s2+k2+k3)