Laplace Transforms: Vertical Motion of a Coupled Spring System

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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 , , and , , , and .


At positions and , the masses and are in equilibrium. Thus, the motion equations for and are,

where and represent the Newton's Second Law of Motion and and represent the net forces acting in the masses.

Laplace Transform

Applying the Laplace Transform to the motion equations and plugging the values of , , , , , , , , and for this systems, we obtain,

Finally, solving for and yields,

Inverse Laplace Transform

Figure 2. Coupled Spring System Motion.

First, we recognize that

On the other hand, we identify that , and so . Hence, we fix the expression by multiplying and dividing by ,

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,

and the Final-Value Theorem is,

Thus, applying both theorems to our the Laplace Transforms,

Bode Plot

Figure 3. Bode Diagram.

The Bode plot for the Transfer Functions


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 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]); 
grid on;

Magnitude Frequency Response

Considering the Transfer Functions and described in the Bode Plot section, we notice that there are no values for the s-variable that make and equal zero.

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


By definition, the convolution of two functions is,

where refers to the inverse Laplace Transform. Assuming that is a dummy variable of integration, and is the impulse function , the convolution for our system is,

which is the same solution yields by the inverse Laplace Transform

State Equation Model

By definition, the the state equation is stated as

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

Solving for and yields,

Finally, we let , , , and be the state variables. Thus,