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 $x_{1}(0)=0$, $x'1(0)=-1$, $x_{2}(0)=0$, and $x'_{2}(0)=1$.

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 $m_{1}x''_{1}$ and $m_{2}x''_{2}$ 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,

${\mathcal {L}}[m_{1}{\ddot {x}}_{1}+k_{1}x_{1}-k_{2}(x_{2}-x_{1})]=0$
$m_{1}[s^{2}X_{1}(s)-sx_{1}(0)-{\dot {x}}_{1}(0)]+k_{1}X_{1}(s)-k_{2}(X_{2}(s)-X_{1}(s))=0$
$X_{1}(s)(m_{1}s^{2}+k_{1}+k_{2})=m_{1}(sx_{1}(0)-{\dot {x}}_{1}(0))+k_{2}X_{2}(s)$
$X_{1}(s)={\dfrac {m_{1}(sx_{1}(0)+{\dot {x}}_{1}(0))+k_{2}X_{2}(s)}{(m_{1}s^{2}+k_{1}+k_{2})}}$

${\mathcal {L}}[m_{2}{\ddot {x}}_{2}+k_{2}(x_{2}-x_{1})+k_{3}x_{2}]=0$
$m_{2}[s^{2}X_{2}(s)-sx_{2}(0)-{\dot {x}}_{2}(0)]+k_{2}(X_{2}(s)-X_{1}(s))+k_{3}X_{2}(s)=0$
$X_{2}(s)(m_{2}s^{2}+k_{2}+k_{3})=m_{2}(sx_{2}(0)-{\dot {x}}_{2}(0))+k_{2}X_{1}(s)$
$X_{2}(s)={\dfrac {m_{2}(sx_{2}(0)+{\dot {x}}_{2}(0))+k_{2}X_{1}(s)}{(m_{2}s^{2}+k_{2}+k_{3})}}$