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,

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

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

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

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

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

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

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

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

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

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

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 Motion.

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 oscillatory motion 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 for the Transfer Functions

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

and

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

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. The peak amplitude occurs at about 2 seconds, as well as the phase switching as shown in the Phase vs. Frequency plot.

h1=tf([-1],[1 0 3]); 
h2=tf([1],[1 0 3]); 
bode(h1,'b',h2,'-.r');
legend('H_1(s)','H_2(s)')
grid on;

Magnitude Frequency Response

Considering the Transfer Functions $H_{1}(s)$ and $H_{2}(s)$ described in the Bode Plot section, we notice that there are no values for the s-variable that make $H_{1}(s)$ and $H_{2}(s)$ equal zero.

Convolution

$y(t)=f(t)*h(t)=\int _{-\infty }^{\infty }{f(\tau )h(t-\tau )d\tau }=\int _{\infty }^{-\infty }{f(t-\tau )h(\tau )d\tau }$