Laplace Transforms: Vertical Motion of a Coupled Spring System

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

$k_1=k_2=k_3=1$

,

$m_1=m_2=1$

, and

$x_1(0)=0$

,

$x^{\prime }1(0)=-1$

,

$x_2(0)=0$

, and

$x{\text{'}}_2(0)=1$

.

Solution

At positions $x_1$ and $x_2$, the masses $m_1$ and $m_2$ are in equilibrium. Thus, the motion equations for $m_1$ and $m_2$ are,

$m_1{\ddot{x}}_1=-k_1{x}_1+k_2(x_2-x_1)$
∴ $m_1{\ddot{x}}_1+k_1{x}_1-k_2(x_2-x_1)=0$

$m_2{\ddot{x}}_2=-k_2(x_2-x_1)-k_3{x}_2$
∴ $m_2{\ddot{x}}_2+k_2(x_2-x_1)-k_3{x}_2=0$

where $m_1{x}^{\prime }{\text{'}}_1$ and $m_2{x}^{\prime }{\text{'}}_2$ represent the Newton's Second Law of Motion and $-k_1{x}_1+k_2(x_2-x_1)$ and $-k_2(x_2-x_1)-k_3{x}_2$ represent the net forces acting in the masses.

Laplace Transform

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

${ℒ}[m_1{\ddot{x}}_1+k_1{x}_1-k_2(x_2-x_1)]=0$
${ℒ}[m_2{\ddot{x}}_2+k_2(x_2-x_1)+k_3{x}_2]=0$