Coupled Oscillator: Double Pendulum: Difference between revisions

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'''Assumptions:'''
'''Assumptions:'''
* the system oscillates vertically under the influence of gravity.
* the system oscillates vertically under the influence of gravity.
* the mass of both rod are neligible
* the mass of both rod are negligible
* no dumpung forces act on the system
* no damping forces act on the system
* positive direction to the right.
* positive direction to the right.


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</math>
</math>


Let's plug some numbers. Knowing <math>g=32</math>, and assuming that <math>m_1=3</math>, <math>m_2=1</math>, and <math>l_1=l_2=16</math>, the constants defined previously become,
Let's plug some numbers. It's known that <math>g=32</math>. In addition, we assume that <math>m_1=3</math>, <math>m_2=1</math>, and <math>l_1=l_2=16</math>, so the constants defined previously become,


: <math>A=1024 \quad B=256 \quad C=2048 \quad D=256 \quad E=512</math>
: <math>A=1024 \quad B=256 \quad C=2048 \quad D=256 \quad E=512</math>
Line 106: Line 106:
</math>
</math>


=== Eigenvalues ===
=== Eigenvalues & Eigenvectors ===
The eigenvalues are obtained from <math>\widehat{A}</math>'s identity matrix,
The eigenvalues and eigenvectors are easily obtained with the help of a TI-89 calculator. First, we consider the <math>\widehat{A}</math>'s identity matrix,


: <math>
: <math>
Line 123: Line 123:
</math>
</math>


According to my TI-89, the eigenvalues are,
Once we define the <math>\widehat{A}</math> matrix, the eigenvalues are determined by using the '''''eigVi()''''' function,


: <math>\lambda_1= 2 \mathbf{i}</math>
: <math>\lambda_1= 2 \mathbf{i}</math>
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: <math>\lambda_4= -1.1547 \mathbf{i}</math>
: <math>\lambda_4= -1.1547 \mathbf{i}</math>


and the eigenvectors,
On the other hand, we use the '''''eigVc()''''' function to find the eigenvectors,


: <math>
: <math>
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Now, we plug the eigenvalues and eigenvectors to produce the standar equation,
Now, we plug the eigenvalues and eigenvectors to produce the standar equation,


: <math>x = c_1 k_1 e^{\lambda_1 t} + c_2 k_2 e^{\lambda_2 t} + c_3 k_3 e^{\lambda_3 t} + c_4 k_4 e^{\lambda_4 t}</math>
: <math>\underline{x} = c_1 \underline{k_1} e^{\lambda_1 t} + c_2 \underline{k}_2 e^{\lambda_2 t} + c_3 \underline{k_3} e^{\lambda_3 t} + c_4 \underline{k_4} e^{\lambda_4 t}</math>


: <math>
: <math>


\underline{x}
x
=
=
c_1
c_1
Line 230: Line 230:
e^{-1.1547 \mathbf{i}}
e^{-1.1547 \mathbf{i}}
</math>
</math>

=== Matrix Exponential ===
The matrix exponential is,

: <math>\dot{\bar{z}}=\hat{A}\bar{z}=TAT^{-1}\bar{z}</math>

where

: <math>
A
=
\begin{bmatrix}
0 & 1 & 0 & 0 \\
-\dfrac{8}{3} & 0 & \dfrac{2}{3} & 0 \\
0 & 0 & 0 & 1 \\
\dfrac{8}{3} & 0 & -\dfrac{8}{3} & 0 \\
\end{bmatrix}
</math>,

and

: <math>
T^{-1}
=
[k_1|k_2|k_3|k_4]
=
\begin{bmatrix}
-0.2 \mathbf{i} & 0.2 \mathbf{i} & -0.29277 \mathbf{i} & 0.29277 \mathbf{i} \\
0.4 & 0.4 & 0.33806 & 0.33806 \\
0.4 \mathbf{i} & -0.4 \mathbf{i} & 0.58554 \mathbf{i} & 0.58554 \mathbf{i} \\
-0.8 & -0.8 & -0.67621 & 0.67621 \\
\end{bmatrix}
</math>,

so

: <math>
T
=
(T^{-1})^{-1}
=
[k_1|k_2|k_3|k_4]^{-1}
=
\begin{bmatrix}
1.25 \mathbf{i} & 0.625 & -0.625 \mathbf{i} & -0.3125 \\
-1.25 \mathbf{i} & 0.625 & 0.625 \mathbf{i} & -0.3125 \\
0.853913 \mathbf{i} & 0.73951 & 0.426956 \mathbf{i} & 0.369755 \\
-0.853913 \mathbf{i} & 0.73951 & -0.426956 \mathbf{i} & 0.369755 \\
\end{bmatrix}
</math>

Again, we can resort to the TI-89 calculator. As it is mentioned above, the matrix exponential is obtained by typing '''''eigVc(a)^-1*a*eigVc(a)''''', where '''''a''''' is the <math>\widehat{A}</math> matrix. Thus,

: <math>
\dot{\bar{z}}
=
\hat{A}\bar{z}
=
TAT^{-1}\bar{z}
=
\begin{bmatrix}
2 \mathbf{i} & 0 & 0 & 0 \\
0 & -2 \mathbf{i} & 0 & 0 \\
0 & 0 & 1.1547 \mathbf{i} & 0 \\
0 & 0 & 0 & -1.1547 \mathbf{i} \\
\end{bmatrix}
\bar{z}
</math>

So, the exponential matrix becomes,

:<math>
\dot{\underline{x}}=\widehat{A}\underline{x}=T^{-1}e^{\hat{A}t}T\underline{x}=e^{At}\underline{x}
</math>

where

:<math>
e^{\hat{A}t}
=
\begin{bmatrix}
e^{2 \mathbf{i} t} & 0 & 0 & 0 \\
0 & e^{-2 \mathbf{i} t} & 0 & 0 \\
0 & 0 & e^{1.1547 \mathbf{i} t} & 0 \\
0 & 0 & 0 & e^{-1.1547 \mathbf{i} t} \\
\end{bmatrix}
</math>

Hence,

: <math>
\dot{\underline{x}}
=
\begin{bmatrix}
-0.2 \mathbf{i} & 0.2 \mathbf{i} & -0.29277 \mathbf{i} & 0.29277 \mathbf{i} \\
0.4 & 0.4 & 0.33806 & 0.33806 \\
0.4 \mathbf{i} & -0.4 \mathbf{i} & 0.58554 \mathbf{i} & 0.58554 \mathbf{i} \\
-0.8 & -0.8 & -0.67621 & 0.67621 \\
\end{bmatrix}
\begin{bmatrix}
e^{2 \mathbf{i} t} & 0 & 0 & 0 \\
0 & e^{-2 \mathbf{i} t} & 0 & 0 \\
0 & 0 & e^{1.1547 \mathbf{i} t} & 0 \\
0 & 0 & 0 & e^{-1.1547 \mathbf{i} t} \\
\end{bmatrix}
\begin{bmatrix}
1.25 \mathbf{i} & 0.625 & -0.625 \mathbf{i} & -0.3125 \\
-1.25 \mathbf{i} & 0.625 & 0.625 \mathbf{i} & -0.3125 \\
0.853913 \mathbf{i} & 0.73951 & 0.426956 \mathbf{i} & 0.369755 \\
-0.853913 \mathbf{i} & 0.73951 & -0.426956 \mathbf{i} & 0.369755 \\
\end{bmatrix}
\underline{x}
</math>

so, the matrix exponential can be solved using matrix multiplication.

Latest revision as of 12:46, 13 December 2009

By Jimmy Apablaza

This problem is described in Page 321-322, Section 7.6 of the A first Course in Differential Equations textbook, 8ED (ISBN 0-534-41878-3).

Figure 1. Coupled Pendulum.‎

Problem Statement

Consider the double-pendulum system consisting of a pendulum attached to another pendulum shown in Figure 1.

Assumptions:

  • the system oscillates vertically under the influence of gravity.
  • the mass of both rod are negligible
  • no damping forces act on the system
  • positive direction to the right.

The system of differential equations describing the motion is nonlinear


In order to linearize these equations, we assume that the displacements and are small enough so that and . Thus,

Solution

Since our concern is about the motion functions, we will assign the masses and , the rod lenghts and , and gravitational force constants to different variables as follows,

Hence,

Solving for and we obtain,

Therefore,

State Space

Let's plug some numbers. It's known that . In addition, we assume that , , and , so the constants defined previously become,

Hence, the state space matrix is,

Eigenvalues & Eigenvectors

The eigenvalues and eigenvectors are easily obtained with the help of a TI-89 calculator. First, we consider the 's identity matrix,

Once we define the matrix, the eigenvalues are determined by using the eigVi() function,

On the other hand, we use the eigVc() function to find the eigenvectors,

Standard Equation

Now, we plug the eigenvalues and eigenvectors to produce the standar equation,

Matrix Exponential

The matrix exponential is,

where

,

and

,

so

Again, we can resort to the TI-89 calculator. As it is mentioned above, the matrix exponential is obtained by typing eigVc(a)^-1*a*eigVc(a), where a is the matrix. Thus,

So, the exponential matrix becomes,

where

Hence,

so, the matrix exponential can be solved using matrix multiplication.