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 and plugging the values of $k_1$, $k_2$, $k_3$, $m_1$, $m_2$, $x_1(0)$, $x^{\prime }1(0)$, $x_2(0)$, and $x{\text{'}}_2(0)$ for this systems, we obtain,

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

$m_1[s^2{X}_1(s)-s{x}_1(0)-{\dot{x}}_1(0)]+k_1{X}_1(s)-k_2(X_2(s)-X_1(s))=0$
 
$X_1(s)(m_1{s}^2+k_1+k_2)=m_1(s{x}_1(0)-{\dot{x}}_1(0))+k_2{X}_2(s)$

$X_1(s)=\frac{m_1(s{x}_1(0)+{\dot{x}}_1(0))+k_2{X}_2(s)}{(m_1{s}^2+k_1+k_2)}$

$X_1(s)=\frac{1(s0+(-1)]+1X_2(s)}{(1s^2+1+1)}$

$X_1(s)=\frac{X_2(s)-1}{(s^2+2)}$


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

$m_2[s^2{X}_2(s)-s{x}_2(0)-{\dot{x}}_2(0)]+k_2(X_2(s)-X_1(s))+k_3{X}_2(s)=0$

$X_2(s)(m_2{s}^2+k_2+k_3)=m_2(s{x}_2(0)-{\dot{x}}_2(0))+k_2{X}_1(s)$

$X_2(s)=\frac{m_2(s{x}_2(0)+{\dot{x}}_2(0))+k_2{X}_1(s)}{(m_2{s}^2+k_2+k_3)}$

$X_2(s)=\frac{1(s0+1)+1X_1(s)}{(1s^2+1+1)}$

$X_2(s)=\frac{X_1(s)+1}{(s^2+2)}$

Finally, solving for $X_1(s)$ and $X_2(s)$ yields,

$X_1(s)=\frac{-1}{s^2+3}$

$X_2(s)=\frac{1}{s^2+3}$

Inverse Laplace Transform

First, we recognize that

$sin(kt)={{ℒ}}^{-1}\left\{\frac{k}{s^2+k^2}\right\}$

On the other hand, we identify that $k^2=3$, and so $k=\sqrt{3}$. Hence, we fix the expression by multiplying and dividing by $\sqrt{3}$,

${{ℒ}}^{-1}[X_1(s)]={{ℒ}}^{-1}\left[\frac{-1}{s^2+3}\right]=-\frac{1}{\sqrt{3}}{{ℒ}}^{-1}\left[\frac{\sqrt{3}}{s^2+3}\right]=-\frac{1}{\sqrt{3}}sin(\sqrt{3}t)$

${{ℒ}}^{-1}[X_2(s)]={{ℒ}}^{-1}\left[\frac{1}{s^2+3}\right]=\frac{1}{\sqrt{3}}{{ℒ}}^{-1}\left[\frac{\sqrt{3}}{s^2+3}\right]=\frac{1}{\sqrt{3}}sin(\sqrt{3}t)$

A plot of the system displacement is shown on Figure 2.

Initial-Value & Final-Value Theorem

The initial-value and final-value theorem can be useful the finding the behavior of a functionat small and large times respectively. By definition, the Initial-Value Theorem is,

$\underset{s\to \mathrm{\infty }}{\lim }sX(s)=x(0)$

and the Final-Value Theorem is,

$\underset{s\to 0}{\lim }sX(s)=x(\mathrm{\infty })$

Thus, applying both theorems to our the Laplace Transforms,

$\underset{s\to \mathrm{\infty }}{\lim }s{X}_1(s)=\underset{s\to \mathrm{\infty }}{\lim }s\frac{-1}{s^2+3}=\underset{s\to \mathrm{\infty }}{\lim }\frac{-s}{s^2+3}=\frac{-\mathrm{\infty }}{{\mathrm{\infty }}^2+3}=0=x_1(0)$

$\underset{s\to \mathrm{\infty }}{\lim }s{X}_2(s)=\underset{s\to \mathrm{\infty }}{\lim }s\frac{1}{s^2+3}=\underset{s\to \mathrm{\infty }}{\lim }\frac{s}{s^2+3}=\frac{\mathrm{\infty }}{{\mathrm{\infty }}^2+3}=0=x_2(0)$

$\underset{s\to 0}{\lim }s{X}_1(s)==\underset{s\to 0}{\lim }s\frac{1}{s^2+3}=\underset{s\to 0}{\lim }\frac{s}{s^2+3}=\frac{0}{0^2+3}=0=x_1(\mathrm{\infty })$

$\underset{s\to 0}{\lim }s{X}_2(s)==\underset{s\to 0}{\lim }s\frac{1}{s^2+3}=\underset{s\to 0}{\lim }\frac{s}{s^2+3}=\frac{0}{0^2+3}=0=x_2(\mathrm{\infty })$

Bode Plot

The Bode plot can be easily done using a program like Octave or MATLAB.