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Coupled Oscillator: Double Pendulum - Class Wiki

Coupled Oscillator: Double Pendulum

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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).

Contents


(thumbnail)
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

(m_1+m_2)l_1^2\theta_1^{\prime\prime} + m_2l_1l_2\theta_2^{\prime\prime}cos(\theta_1-\theta_2) + m_2l_1l_2(\theta_2^{\prime})^2sin(\theta_1-\theta_2) + (m_1+m_2)l_1gsin\theta_1 = 0
m_2l_1^2\theta_2^{\prime\prime} + m_2l_1l_2\theta_1^{\prime\prime}cos(\theta_1-\theta_2) - m_2l_1l_2(\theta_1^{\prime})^2sin(\theta_1-\theta_2) + m_2l_2gsin\theta_2 = 0


In order to linearize these equations, we assume that the displacements \theta_1 and \theta_2 are small enough so that cos(\theta_1-\theta_2)\approx1 and sin(\theta_1-\theta_2)\approx0. Thus,

(m_1+m_2)l_1^2\theta_1^{\prime\prime} + m_2l_1l_2\theta_2^{\prime\prime} + (m_1+m_2)l_1g\theta_1 = 0
m_2l_1^2\theta_2^{\prime\prime} + m_2l_1l_2\theta_1^{\prime\prime} + m_2l_2g\theta_2 = 0

Solution

Since our concern is about the motion functions, we will assign the masses m_1 and m_2, the rod lenghts l_1 and l_1, and gravitational force g constants to different variables as follows,

A=(m_1+m_2)l_1^2 \quad B=m_2l_1l_2 \quad C=(m_1+m_2)l_1g \quad D=m_2l_1^2 \quad E=m_2l_2g

Hence,

A\theta_1^{''} + B\theta_2^{''} + C\theta_1 = 0
D\theta_2^{''} + B\theta_1^{''} + E\theta_2 = 0

Solving for \theta_1^{''} and \theta_2^{''} we obtain,

\theta_1^{''} = - \left ( \dfrac{B}{A} \right ) \theta_2^{''} - \left ( \dfrac{C}{A} \right ) \theta_1
\theta_2^{''} = - \left ( \dfrac{B}{D} \right ) \theta_1^{''} - \left ( \dfrac{E}{D} \right ) \theta_2

Therefore,

\theta_1^{''} = - \left ( \dfrac{CD}{AD+B^2} \right ) \theta_1 - \left ( \dfrac{BE}{AD+B^2} \right ) \theta_2
\theta_2^{''} = \left ( \dfrac{BC}{AD+B^2} \right ) \theta_1 - \left ( \dfrac{AE}{AD+B^2} \right ) \theta_2

State Space


\begin{bmatrix}
\theta_1^{'} \\ \theta_1^{''} \\ \theta_2^{'} \\ \theta_2^{''}
\end{bmatrix}

=

\widehat{A} \, \underline{x}(t) + \widehat{B} \, \underline{u}(t)

=

\begin{bmatrix}
0                   & 1 & 0                   & 0 \\
 & & & \\
\dfrac{-CD}{AD-B^2} & 0 & \dfrac{BE}{AD-B^2}  & 0 \\
 & & & \\
0                   & 0 & 0                   & 1 \\
 & & & \\
\dfrac{BC}{AD-B^2}  & 0 & \dfrac{-AE}{AD-B^2} & 0 \\
\end{bmatrix}

\begin{Bmatrix}
\theta_1 \\ \theta_1^{'} \\ \theta_2 \\ \theta_2^{'}
\end{Bmatrix}

+

\widehat{0}

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

A=1024 \quad B=256 \quad C=2048 \quad D=256 \quad E=512

Hence, the state space matrix is,


\begin{bmatrix}
\theta_1^{'} \\ \theta_1^{''} \\ \theta_2^{'} \\ \theta_2^{''}
\end{bmatrix}

=

\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}

\begin{Bmatrix}
\theta_1 \\ \theta_1^{'} \\ \theta_2 \\ \theta_2^{'}
\end{Bmatrix}

Eigenvalues & Eigenvectors

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


[\lambda I-A]

=

\begin{bmatrix}
\lambda       & 1       & 0             & 0       \\
-\dfrac{8}{3} & \lambda & \dfrac{2}{3}  & 0       \\
0             & 0       & \lambda       & 1       \\
\dfrac{8}{3}  & 0       & -\dfrac{8}{3} & \lambda \\
\end{bmatrix}

Once we define the \widehat{A} matrix, the eigenvalues are determined by using the eigVi() function,

\lambda_1= 2 \mathbf{i}
\lambda_2= -2 \mathbf{i}
\lambda_3= 1.1547 \mathbf{i}
\lambda_4= -1.1547 \mathbf{i}

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


k_1

=

\begin{bmatrix}
-0.2 \mathbf{i} \\
0.4             \\
0.4 \mathbf{i}  \\
-0.8            \\
\end{bmatrix}

\quad

k_2

=

\begin{bmatrix}
0.2 \mathbf{i} \\
0.4             \\
-0.4 \mathbf{i}  \\
-0.8            \\
\end{bmatrix}

\quad

k_3

=

\begin{bmatrix}
-0.29277 \mathbf{i} \\
0.33806             \\
0.58554 \mathbf{i}  \\
-0.67621            \\
\end{bmatrix}

\quad

k_4

=

\begin{bmatrix}
0.29277 \mathbf{i} \\
0.33806             \\
0.58554 \mathbf{i}  \\
0.67621            \\
\end{bmatrix}

Standard Equation

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

\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}


\underline{x}
=
c_1
\begin{bmatrix}
-0.2 \mathbf{i} \\
0.4             \\
0.4 \mathbf{i}  \\
-0.8            \\
\end{bmatrix}
e^{2 \mathbf{i}}
+
c_2
\begin{bmatrix}
0.2 \mathbf{i} \\
0.4             \\
-0.4 \mathbf{i}  \\
-0.8            \\
\end{bmatrix}
e^{-2 \mathbf{i}}
+
c_3
\begin{bmatrix}
-0.29277 \mathbf{i} \\
0.33806             \\
0.58554 \mathbf{i}  \\
-0.67621            \\
\end{bmatrix}
e^{1.1547 \mathbf{i}}
+
c_4
\begin{bmatrix}
0.29277 \mathbf{i} \\
0.33806             \\
0.58554 \mathbf{i}  \\
0.67621            \\
\end{bmatrix}
e^{-1.1547 \mathbf{i}}

Matrix Exponential

The matrix exponential is,

\dot{\bar{z}}=\hat{A}\bar{z}=TAT^{-1}\bar{z}

where


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}
,

and


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}
,

so


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}

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 \widehat{A} matrix. Thus,


\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}

So, the exponential matrix becomes,


\dot{\underline{x}}=\widehat{A}\underline{x}=T^{-1}e^{\hat{A}t}T\underline{x}=e^{At}\underline{x}

where


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}

Hence,


\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}

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

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