Laplace Transforms: Vertical Motion of a Coupled Spring System

Figure 1. Coupled Spring System.

Problem Statement

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 and plugging the values of $k_{1}$, $k_{2}$, $k_{3}$, $m_{1}$, $m_{2}$, $x_{1}(0)$, $x'1(0)$, $x_{2}(0)$, and $x'_{2}(0)$ for this systems, we obtain,

${\mathcal {L}}[m_{1}{\ddot {x}}_{1}+k_{1}x_{1}-k_{2}(x_{2}-x_{1})]=0$

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

$X_{1}(s)(m_{1}s^{2}+k_{1}+k_{2})=m_{1}(sx_{1}(0)-{\dot {x}}_{1}(0))+k_{2}X_{2}(s)$

$X_{1}(s)={\dfrac {m_{1}(sx_{1}(0)+{\dot {x}}_{1}(0))+k_{2}X_{2}(s)}{(m_{1}s^{2}+k_{1}+k_{2})}}$

$X_{1}(s)={\dfrac {1(s0+(-1)]+1X_{2}(s)}{(1s^{2}+1+1)}}$

$X_{1}(s)={\dfrac {X_{2}(s)-1}{(s^{2}+2)}}$

${\mathcal {L}}[m_{2}{\ddot {x}}_{2}+k_{2}(x_{2}-x_{1})+k_{3}x_{2}]=0$

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

$X_{2}(s)(m_{2}s^{2}+k_{2}+k_{3})=m_{2}(sx_{2}(0)-{\dot {x}}_{2}(0))+k_{2}X_{1}(s)$

$X_{2}(s)={\dfrac {m_{2}(sx_{2}(0)+{\dot {x}}_{2}(0))+k_{2}X_{1}(s)}{(m_{2}s^{2}+k_{2}+k_{3})}}$

$X_{2}(s)={\dfrac {1(s0+1)+1X_{1}(s)}{(1s^{2}+1+1)}}$

$X_{2}(s)={\dfrac {X_{1}(s)+1}{(s^{2}+2)}}$

Finally, solving for $X_{1}(s)$ and $X_{2}(s)$ yields,

$X_{1}(s)={\dfrac {-1}{s^{2}+3}}$

$X_{2}(s)={\dfrac {1}{s^{2}+3}}$

Inverse Laplace Transform

Figure 2. Coupled Spring System Displacement.

First, we recognize that

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

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 ${\sqrt {3}}$,

$x_{1}(t)={\mathcal {L}}^{-1}[X_{1}(s)]={\mathcal {L}}^{-1}\left[{\dfrac {-1}{s^{2}+3}}\right]=-{\dfrac {1}{\sqrt {3}}}{\mathcal {L}}^{-1}\left[{\dfrac {\sqrt {3}}{s^{2}+3}}\right]=-{\dfrac {1}{\sqrt {3}}}sin({\sqrt {3}}t)$

$x_{2}=(t){\mathcal {L}}^{-1}[X_{2}(s)]={\mathcal {L}}^{-1}\left[{\dfrac {1}{s^{2}+3}}\right]={\dfrac {1}{\sqrt {3}}}{\mathcal {L}}^{-1}\left[{\dfrac {\sqrt {3}}{s^{2}+3}}\right]={\dfrac {1}{\sqrt {3}}}sin({\sqrt {3}}t)$

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,

$\lim _{s\rightarrow \infty }sX(s)=x(0)\,$

and the Final-Value Theorem is,

$\lim _{s\rightarrow 0}sX(s)=x(\infty )\,$

Thus, applying both theorems to our the Laplace Transforms,

$\lim _{s\rightarrow \infty }sX_{1}(s)=\lim _{s\rightarrow \infty }s{\dfrac {-1}{s^{2}+3}}=\lim _{s\rightarrow \infty }{\dfrac {-s}{s^{2}+3}}={\dfrac {-\infty }{\infty ^{2}+3}}=0=x_{1}(0)\,$

$\lim _{s\rightarrow 0}sX_{1}(s)==\lim _{s\rightarrow 0}s{\dfrac {1}{s^{2}+3}}=\lim _{s\rightarrow 0}{\dfrac {s}{s^{2}+3}}={\dfrac {0}{0^{2}+3}}=0=x_{1}(\infty )\,$

$\lim _{s\rightarrow \infty }sX_{2}(s)=\lim _{s\rightarrow \infty }s{\dfrac {1}{s^{2}+3}}=\lim _{s\rightarrow \infty }{\dfrac {s}{s^{2}+3}}={\dfrac {\infty }{\infty ^{2}+3}}=0=x_{2}(0)\,$

$\lim _{s\rightarrow 0}sX_{2}(s)==\lim _{s\rightarrow 0}s{\dfrac {1}{s^{2}+3}}=\lim _{s\rightarrow 0}{\dfrac {s}{s^{2}+3}}={\dfrac {0}{0^{2}+3}}=0=x_{2}(\infty )\,$

Bode Plot

Figure 3. Bode Diagram.

The Bode plot can be easily done using a program like Octave or MATLAB. The code is displayed below. From Figure 3 we may notice that the Amplitude vs. Frequency plot for both functions overlaps. On the other hand, the Phase vs. Frequency plot for both functions have opossite magnitudesNotice that both functions overlaps.

h1=tf([-1],[1 0 3]);
h2=tf([1],[1 0 3]);
bode(h1,'b',h2,'r');