# Laplace Transforms: Vertical Motion of a Coupled Spring System

By Jimmy Apablaza

This problem is described in Exercise 14, Section 7.6 (page 323) of the A first Course in Differential Equations textbook, 8ED (ISBN 0-534-41878-3).

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 ${\displaystyle k_{1}=k_{2}=k_{3}=1{\frac {}{}}}$, ${\displaystyle m_{1}=m_{2}=1{\frac {}{}}}$, and ${\displaystyle x_{1}(0)=0{\frac {}{}}}$, ${\displaystyle x'1(0)=-1{\frac {}{}}}$, ${\displaystyle x_{2}(0)=0{\frac {}{}}}$, and ${\displaystyle x'_{2}(0)=1{\frac {}{}}}$.

# Solution

At positions ${\displaystyle x_{1}{\frac {}{}}}$ and ${\displaystyle x_{2}{\frac {}{}}}$, the masses ${\displaystyle m_{1}{\frac {}{}}}$ and ${\displaystyle m_{2}{\frac {}{}}}$ are in equilibrium. Thus, the motion equations for ${\displaystyle m_{1}{\frac {}{}}}$ and ${\displaystyle m_{2}{\frac {}{}}}$ are,

${\displaystyle m_{1}{\ddot {x}}_{1}=-k_{1}x_{1}+k_{2}(x_{2}-x_{1})}$
${\displaystyle m_{1}{\ddot {x}}_{1}+k_{1}x_{1}-k_{2}(x_{2}-x_{1})=0}$
${\displaystyle m_{2}{\ddot {x}}_{2}=-k_{2}(x_{2}-x_{1})-k_{3}x_{2}}$
${\displaystyle m_{2}{\ddot {x}}_{2}+k_{2}(x_{2}-x_{1})-k_{3}x_{2}=0}$

where ${\displaystyle m_{1}x''_{1}{\frac {}{}}}$ and ${\displaystyle m_{2}x''_{2}{\frac {}{}}}$ represent the Newton's Second Law of Motion and ${\displaystyle -k_{1}x_{1}+k_{2}(x_{2}-x_{1}){\frac {}{}}}$ and ${\displaystyle -k_{2}(x_{2}-x_{1})-k_{3}x_{2}{\frac {}{}}}$ represent the net forces acting in the masses.

## Laplace Transform

Applying the Laplace Transform to the motion equations and plugging the values of ${\displaystyle k_{1}{\frac {}{}}}$, ${\displaystyle k_{2}{\frac {}{}}}$, ${\displaystyle k_{3}{\frac {}{}}}$, ${\displaystyle m_{1}{\frac {}{}}}$, ${\displaystyle m_{2}{\frac {}{}}}$, ${\displaystyle x_{1}(0){\frac {}{}}}$, ${\displaystyle x'1(0){\frac {}{}}}$, ${\displaystyle x_{2}(0){\frac {}{}}}$, and ${\displaystyle x'_{2}(0){\frac {}{}}}$ for this systems, we obtain,

${\displaystyle {\mathcal {L}}[m_{1}{\ddot {x}}_{1}+k_{1}x_{1}-k_{2}(x_{2}-x_{1})]=m_{1}[s^{2}X_{1}(s)-sx_{1}(0)-{\dot {x}}_{1}(0)]+k_{1}X_{1}(s)-k_{2}(X_{2}(s)-X_{1}(s))=0}$
${\displaystyle X_{1}(s)(m_{1}s^{2}+k_{1}+k_{2})=m_{1}(sx_{1}(0)-{\dot {x}}_{1}(0))+k_{2}X_{2}(s)}$
${\displaystyle X_{1}(s)={\dfrac {m_{1}(sx_{1}(0)+{\dot {x}}_{1}(0))+k_{2}X_{2}(s)}{(m_{1}s^{2}+k_{1}+k_{2})}}}$
${\displaystyle X_{1}(s)={\dfrac {1(s0+(-1)]+1X_{2}(s)}{(1s^{2}+1+1)}}}$
${\displaystyle X_{1}(s)={\dfrac {X_{2}(s)-1}{(s^{2}+2)}}}$
${\displaystyle {\mathcal {L}}[m_{2}{\ddot {x}}_{2}+k_{2}(x_{2}-x_{1})+k_{3}x_{2}]=m_{2}[s^{2}X_{2}(s)-sx_{2}(0)-{\dot {x}}_{2}(0)]+k_{2}(X_{2}(s)-X_{1}(s))+k_{3}X_{2}(s)=0}$
${\displaystyle X_{2}(s)(m_{2}s^{2}+k_{2}+k_{3})=m_{2}(sx_{2}(0)-{\dot {x}}_{2}(0))+k_{2}X_{1}(s)}$
${\displaystyle X_{2}(s)={\dfrac {m_{2}(sx_{2}(0)+{\dot {x}}_{2}(0))+k_{2}X_{1}(s)}{(m_{2}s^{2}+k_{2}+k_{3})}}}$
${\displaystyle X_{2}(s)={\dfrac {1(s0+1)+1X_{1}(s)}{(1s^{2}+1+1)}}}$
${\displaystyle X_{2}(s)={\dfrac {X_{1}(s)+1}{(s^{2}+2)}}}$

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

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

## Inverse Laplace Transform

Figure 2. Coupled Spring System Motion.

First, we recognize that

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

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

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

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 function at small and large times respectively. By definition, the Initial-Value Theorem is,

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

and the Final-Value Theorem is,

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

Thus, applying both theorems to our the Laplace Transforms,

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

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

## Bode Plot

Figure 3. Bode Diagram.

The Bode plot for the Transfer Functions

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

and

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

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 ${\displaystyle {\sqrt {3}}\approx 1.73{\frac {}{}}}$ 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 ${\displaystyle H_{1}(s){\frac {}{}}}$ and ${\displaystyle H_{2}(s){\frac {}{}}}$ described in the Bode Plot section, we notice that there are no values for the s-variable that make ${\displaystyle H_{1}(s){\frac {}{}}}$ and ${\displaystyle H_{2}(s){\frac {}{}}}$ equal zero.

From the Bode Plot, we notice that the break point occurs at ≈ 1.73 seconds, so the

## Convolution

By definition, the convolution of two functions is,

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

where ${\displaystyle h_{1}{\frac {}{}}}$ refers to the inverse Laplace Transform. Assuming that ${\displaystyle \tau {\frac {}{}}}$ is a dummy variable of integration, and ${\displaystyle f(t){\frac {}{}}}$ is the impulse function ${\displaystyle (f_{1}(t)=f_{2}(t)=\delta (t){\frac {}{}})}$, the convolution for our system is,

${\displaystyle y_{1}(t)=\int _{0}^{t}{\delta (t-\tau )h_{1}(\tau )d\tau }=-{\dfrac {1}{\sqrt {3}}}\int _{0}^{t}{\delta (t-\tau )sin({\sqrt {3}}\tau )d\tau }=\left[-{\dfrac {1}{\sqrt {3}}}sin({\sqrt {3}}\tau )\right]_{\tau =t}=-{\dfrac {1}{\sqrt {3}}}sin({\sqrt {3}}t)}$
${\displaystyle y_{2}(t)=\int _{0}^{t}{\delta (t-\tau )h_{2}(\tau )d\tau }={\dfrac {1}{\sqrt {3}}}\int _{0}^{t}{\delta (t-\tau )sin({\sqrt {3}}\tau )d\tau }=\left[{\dfrac {1}{\sqrt {3}}}sin({\sqrt {3}}\tau )\right]_{\tau =t}={\dfrac {1}{\sqrt {3}}}sin({\sqrt {3}}t)}$

which is the same solution yields by the inverse Laplace Transform

## State Equation Model

By definition, the the state equation is stated as

${\displaystyle {\underline {\dot {x}}}={\widehat {A}}\,{\underline {x}}+{\widehat {B}}\,{\underline {u}}}$

Now, consider the motion equations described in the Solution section,

${\displaystyle m_{1}{\ddot {x}}_{1}+k_{1}x_{1}-k_{2}(x_{2}-x_{1})=m_{1}{\ddot {x}}_{1}+k_{1}x_{1}-k_{2}x_{2}+k_{2}x_{1}=0}$

${\displaystyle m_{2}{\ddot {x}}_{2}+k_{2}(x_{2}-x_{1})-k_{3}x_{2}=m_{2}{\ddot {x}}_{2}+k_{2}x_{2}-k_{2}x_{1}-k_{3}x_{2}=0}$

Solving for ${\displaystyle {\ddot {x}}_{1}}$ and ${\displaystyle {\ddot {x}}_{2}}$ yields,

${\displaystyle {\ddot {x}}_{1}=-{\dfrac {k_{1}}{m_{1}}}x_{1}+{\dfrac {k_{2}}{m_{1}}}x_{2}-{\dfrac {k_{2}}{m_{1}}}x_{1}}$
${\displaystyle {\ddot {x}}_{2}={\dfrac {-k_{2}}{m_{2}}}x_{2}+{\dfrac {k_{2}}{m_{2}}}x_{1}+{\dfrac {k_{3}}{m_{2}}}x_{2}}$

Finally, we let ${\displaystyle x_{1}{\frac {}{}}}$, ${\displaystyle {\dot {x}}_{1}{\frac {}{}}}$, ${\displaystyle x_{2}{\frac {}{}}}$, and ${\displaystyle {\dot {x_{2}}}{\frac {}{}}}$ be the state variables. Thus,

${\displaystyle {\begin{bmatrix}{\dot {x}}_{1}\\{\ddot {x}}_{1}\\{\dot {x}}_{2}\\{\ddot {x}}_{2}\end{bmatrix}}={\begin{bmatrix}0&1&0&0\\-{\frac {1}{m_{1}}}(k_{1}+k_{2})&0&{\frac {k_{2}}{m_{1}}}&0\\0&0&0&1\\{\frac {k_{2}}{m_{2}}}&0&{\frac {1}{m_{2}}}(k_{3}-k_{2})&0\end{bmatrix}}{\begin{bmatrix}x_{1}\\{\dot {x}}_{1}\\x_{2}\\{\dot {x}}_{2}\end{bmatrix}}}$