Coupled Oscillator: Double Pendulum

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 neligible
• no dumpung forces act on the system
• positive direction to the right.

The system of differential equations describing the motion is nonlinear

$(m_{1}+m_{2})l_{1}^{2}\theta _{1}^{\prime \prime }+m_{2}l_{1}l_{2}\theta _{2}^{\prime \prime }cos(\theta _{1}-\theta _{2})+m_{2}l_{1}l_{2}(\theta _{2}^{\prime })^{2}sin(\theta _{1}-\theta _{2})+(m_{1}+m_{2})l_{1}gsin\theta _{1}=0$

$m_{2}l_{1}^{2}\theta _{2}^{\prime \prime }+m_{2}l_{1}l_{2}\theta _{1}^{\prime \prime }cos(\theta _{1}-\theta _{2})-m_{2}l_{1}l_{2}(\theta _{1}^{\prime })^{2}sin(\theta _{1}-\theta _{2})+m_{2}l_{2}gsin\theta _{2}=0$

In order to linearize these equations, we assume that the displacements $\theta _{1}$ and $\theta _{2}$ are small enough so that $cos(\theta _{1}-\theta _{2})\approx 1$ and $sin(\theta _{1}-\theta _{2})\approx 0$. Thus,

$(m_{1}+m_{2})l_{1}^{2}\theta _{1}^{\prime \prime }+m_{2}l_{1}l_{2}\theta _{2}^{\prime \prime }+(m_{1}+m_{2})l_{1}g\theta _{1}=0$

$m_{2}l_{1}^{2}\theta _{2}^{\prime \prime }+m_{2}l_{1}l_{2}\theta _{1}^{\prime \prime }+m_{2}l_{2}g\theta _{2}=0$

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,

$A=(m_{1}+m_{2})l_{1}^{2}\quad B=m_{2}l_{1}l_{2}\quad C=(m_{1}+m_{2})l_{1}g\quad D=m_{2}l_{1}^{2}\quad E=m_{2}l_{2}g$

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

${\begin{bmatrix}\theta _{1}^{'}\\\theta _{1}^{''}\\\theta _{2}^{'}\\\theta _{2}^{''}\end{bmatrix}}={\widehat {A}}\,{\underline {x}}(t)+{\widehat {B}}\,{\underline {u}}(t)={\begin{bmatrix}0&1&0&0\\&&&\\{\dfrac {-CD}{AD+B^{2}}}&0&{\dfrac {BE}{AD+B^{2}}}&0\\&&&\\0&0&0&1\\&&&\\{\dfrac {BC}{AD+B^{2}}}&0&{\dfrac {-AE}{AD+B^{2}}}&0\\\end{bmatrix}}{\begin{Bmatrix}\theta _{1}\\\theta _{1}^{'}\\\theta _{2}\\\theta _{2}^{'}\end{Bmatrix}}+{\widehat {0}}$

Plugging the constants yields,

${\begin{bmatrix}\theta _{1}^{'}\\\theta _{1}^{''}\\\theta _{2}^{'}\\\theta _{2}^{''}\end{bmatrix}}={\begin{bmatrix}0&1&0&0\\&&&\\{\dfrac {-l_{1}(m_{1}+m_{2})g}{l_{1}^{2}(m_{1}+m_{2})+l_{2}^{2}m_{2}}}&0&{\dfrac {l_{2}^{2}m_{2}g}{l_{1}(l_{1}^{2}(m_{1}+m_{2})+l_{2}^{2}m_{2})}}&0\\&&&\\0&0&0&1\\&&&\\{\dfrac {l_{2}(m_{1}+m_{2})g}{l_{1}^{2}(m_{1}+m_{2})+l_{2}^{2}m_{2}}}&0&{\dfrac {-l_{2}(m_{1}+m_{2})g}{l_{1}^{2}(m_{1}+m_{2})+l_{2}^{2}m_{2}}}&0\\\end{bmatrix}}{\begin{Bmatrix}\theta _{1}\\\theta _{1}^{'}\\\theta _{2}\\\theta _{2}^{'}\end{Bmatrix}}$

Let's plug some numbers. Knowing $g=32$, and assuming that $m_{1}=3$, $m_{2}=1$, and $l_{1}=l_{2}=16$, the state space matrix becomes,

${\begin{bmatrix}\theta _{1}^{'}\\\theta _{1}^{''}\\\theta _{2}^{'}\\\theta _{2}^{''}\end{bmatrix}}={\begin{bmatrix}0&1&0&0\\-{\dfrac {8}{5}}&0&-{\dfrac {8}{5}}&0\\0&0&0&1\\{\dfrac {8}{5}}&0&-{\dfrac {8}{5}}&0\\\end{bmatrix}}{\begin{Bmatrix}\theta _{1}\\\theta _{1}^{'}\\\theta _{2}\\\theta _{2}^{'}\end{Bmatrix}}$

Laplace Transform

First, we determine the eigenvalues of the ${\widehat {A}}$ matrix,

$[sI-A]={\begin{bmatrix}s&1&0&0\\-{\dfrac {8}{5}}&s&-{\dfrac {8}{5}}&0\\0&0&s&1\\{\dfrac {8}{5}}&0&-{\dfrac {8}{5}}&s\\\end{bmatrix}}$

Hence, the inverse matrix (thanks TI-89!) is,

$[sI-A]^{-1}={\begin{bmatrix}{\dfrac {5s(5s^{2}+8)}{25s^{4}+80s^{2}+128}}&-{\dfrac {5(5s^{2}+8)}{25s^{4}+80s^{2}+128}}&-{\dfrac {40s}{25s^{4}+80s^{2}+128}}&{\dfrac {40}{25s^{4}+80s^{2}+128}}\\{\dfrac {8(5s^{2}+16)}{25s^{4}+80s^{2}+128}}&{\dfrac {5s(5s^{2}+8)}{25s^{4}+80s^{2}+128}}&-{\dfrac {40s^{2}}{25s^{4}+80s^{2}+128}}&-{\dfrac {40s}{25s^{4}+80s^{2}+128}}\\{\dfrac {40s}{25s^{4}+80s^{2}+128}}&-{\dfrac {40}{25s^{4}+80s^{2}+128}}&{\dfrac {5s(5s^{2}+8)}{25s^{4}+80s^{2}+128}}&-{\dfrac {5(5s^{2}+8)}{25s^{4}+80s^{2}+128}}\\-{\dfrac {40s^{2}}{25s^{4}+80s^{2}+128}}&{\dfrac {40s}{25s^{4}+80s^{2}+128}}&{\dfrac {8(5s^{2}+16)}{25s^{4}+80s^{2}+128}}&{\dfrac {5s(5s^{2}+8)}{25s^{4}+80s^{2}+128}}\\\end{bmatrix}}$