The Double Pendulum

A double pendulum consists of two balls hanging from strings: a string of length $L_1$ connects a fixed support to a ball of mass $m_1$, and a string of length $L_2$ connects this ball to another ball of mass $m_2$. We will assume that the objects are hanging almost vertically and swinging slowly, so that we can use a linear approximation to their motion.

Let $\theta_1$ and $\theta_2$ be the angles of the two strings from the vertical. If $x_1$ and $x_2$ are the $x$ coordinates of the two objects, where $x=0$ is directly below the support, we have $x_1 = L_1 \sin \theta_1$ and $x_2 = x_1 + L_2 \sin \theta_2$. Since the angles are small, we can approximate $x_1 \approx L_1 \theta_1$ and $x_2 \approx L_1 \theta_1 + L_2 \theta_2$. The tensions in the strings are $T_1 \approx (m_1 + m_2) g$ and $T_2 \approx m_2 g$ respectively (the second string supports the weight of one ball, while the first string supports two; the weight of the strings is neglected). The net force in the $x$ direction on the first ball is then $-T_1 \sin \theta_1 + T_2 \sin \theta_2 \approx -T_1 \theta_1 + T_2 \theta_2$, while the net force in the $x$ direction on the second ball is $-T_2 \sin\theta_2 \approx - T_2 \theta_2$. According to Newton's Laws, these forces are equal to $m_1 x_1''$ and $m_2 x_2''$ respectively. Writing these in terms of $\theta_1$ and $\theta_2$, and using the approximations again, we have

$$m_1 L_1 \theta_1'' \approx -(m_1 + m_2) g \theta_1 + m_2 g \theta_2$$

$$m_2 L_1 \theta_1'' + m_2 L_2 \theta_2'' \approx -m_2 g \theta_2$$

Thus we obtain a second-order linear system to approximate the motion:

$$\theta_1'' = -\frac{(m_1 + m_2) g}{m_1 L_1} \theta_1 + \frac{m_2 g}{m_1 L_1} \theta_2$$

$$\theta_2'' = \frac{(m_1 + m_2) g}{m_1 L_2} \theta_1 - \frac{(m_1+m_2) g}{m_1 L_2} \theta_2$$

For simplicity, we will assume the masses are equal: $m_1 = m_2$. With $\alpha = g/L_1$ and $\beta = g/L_2$, then, we can write our system as $\Theta'' = A \Theta$ where

$$A = \pmatrix{-2\alpha & \alpha \cr 2\beta & -2\beta\cr}$$

Now this is a second order system, and we have only studied first order systems. But we can make it into a 4 by 4 first order system by introducing additional variables $v_1 = \theta_1'$ and $v_2 = \theta_2'$. Then we have

$$\pmatrix{ \theta_1\cr \theta_2\cr v_1 \cr v_2\cr }' = \pmatr... ...& 0 & 0\cr} \pmatrix{ \theta_1\cr \theta_2\cr v_1 \cr v_2\cr }$$

Thus we need four linearly independent solutions to make a fundamental set. But it's easier to work with our 2 by 2 matrix $A$ instead of a 4 by 4 matrix. If we take $\Theta = \exp(c t) {\bf u}$ where ${\bf u}$ is a constant vector, then $\Theta'' = c^2 \exp(c t) {\bf u}$, so this is a solution if $A {\bf u}= c^2 {\bf u}$, i.e. ${\bf u}$ is an eigenvector of $A$ for eigenvalue $c^2$. Assuming our matrix $A$ has two distinct eigenvalues, neither of them 0, each eigenvalue has two square roots, and this will give us our four linearly independent solutions.

For ease of calculation, let's try $\alpha = 3$ and $\beta = 4$. The characteristic equation of $$A = \pmatrix{-6 & 3\cr 8 & -8\cr}$$ is $r^2 + 14 r + 24$. The eigenvalues are thus $-2$ and $-12$. The corresponding eigenvectors are $$\pmatrix{3\cr 4\cr}$$ and $$\pmatrix{1\cr -2\cr}$$. So our fundamental set of solutions is

$${\rm e}^{\sqrt{2}it} \pmatrix{3\cr 4\cr},\qquad {\rm e}^{-\s... ...{1\cr -2\cr},\qquad {\rm e}^{-\sqrt{12}it} \pmatrix{1\cr -2\cr}$$

Or, taking real and imaginary parts, we could use

$$\cos(\sqrt{2} t) \pmatrix{3\cr 4\cr},\quad \sin(\sqrt{2} t)... ...trix{1\cr -2\cr},\quad \sin(\sqrt{12} t) \pmatrix{1\cr -2\cr}$$

Each pair of solutions represents a ``normal mode'' of the system, in which all parts of the system oscillate sinusoidally with one frequency. There is a mode with angular frequency $\sqrt{2}$ where the two angles are in phase, and one with angular frequency $\sqrt{12}$ where $\theta_1$ and $\theta_2$ are 180 degrees out of phase.

In the physical model I made, the string lengths were $L_1 = 60$ cm and $L_2 = 43$ cm, so $\alpha = 16.33$ and $\beta = 22.79$. The resulting solutions are

$$\cos \mbox{(or sin)}(3.33 t) \pmatrix{16.33\cr 21.59\cr},\quad \cos \mbox{(or sin)}(8.2 t) \pmatrix{16.33\cr -34.49\cr}$$

Again we have a ``slow'' mode (period $2\pi/3.33 = 1.89$ seconds) with $\theta_1$ and $\theta_2$ in phase, and a ``fast'' mode (period $2 \pi/8.2 = 0.77$ seconds) with $\theta_1$ and $\theta_2$ 180 degrees out of phase. Here are animations of these modes (the timing of the animations is not likely to be correct, and will vary depending on your computer and browser). The angles are exaggerated for ease of viewing: for swings this large, the linear approximations would not be very good.