1.3 Two Coupled Oscillators

1.3.1 Double Pendulum

**Figure 1.2:** A double pendulum.
$\scalebox{0.7}{ \includegraphics{doublePendulum.eps} }$

Equations of Motion

The positions of the masses, relative to their equilibrium positions are given by

$\displaystyle x_1$ $\textstyle =$ $\displaystyle \ell \sin \theta_1$

$\displaystyle y_1$ $\textstyle =$ $\displaystyle \ell (1-\cos \theta_1)$

$\displaystyle x_2$ $\textstyle =$ $\displaystyle \ell (\sin \theta_1 + \sin \theta_2)$

$\displaystyle y_2$ $\textstyle =$ $\displaystyle \ell (2 - \cos \theta_1 - \cos \theta_2)$ (1.14)

where $\ell$ is the length of the strings. To first order, $\sin \theta \approx \theta$ and $\cos \theta \approx 1$ . Hence, for small angles, and Eq.'s 1.14 become

$\displaystyle x_1$ $\textstyle \approx$ $\displaystyle \ell \theta_1$

$\displaystyle y_1$ $\textstyle \approx$ $\displaystyle 0$

$\displaystyle x_2$ $\textstyle \approx$ $\displaystyle \ell (\theta_1 + \theta_2)$

$\displaystyle y_2$ $\textstyle \approx$ $\displaystyle 0$ (1.15)
By Eq.'s 1.15, the motions of the masses are purely horizontal to first order in the angles. We need only consider the horizontal forces on the masses,

$\displaystyle F_{1 net x} =$ $\textstyle -T_1 \sin \theta_1 + T_2 \sin \theta_2$ $\displaystyle \approx -T_1 \theta_1 + T_2 \theta_2$

$\displaystyle F_{2 net x} =$ $\textstyle -T_2 \sin \theta_2$ $\displaystyle \approx -T_2 \theta_2$ (1.16)
For small angles, the tensions are $T_1 \approx 2mg$ and $T_2 \approx mg$ . Newton's second law yields

$\displaystyle m\ddot{x_1} \approx$ $\textstyle m \ell \ddot{\theta_1}$ $\displaystyle \approx -mg (2 \theta_1 - \theta_2)$

$\displaystyle m\ddot{x_2} \approx$ $\textstyle m \ell (\ddot{\theta_1} + \ddot{\theta_2})$ $\displaystyle \approx -mg \theta_2$ (1.17)
Eq.'s 1.17 can be rearranged to find the equations of motion of the system, a system of two coupled second order differential equations,

$\displaystyle \ddot{\theta_1}$ $\textstyle =$ $\displaystyle -\frac{g}{\ell} (2 \theta_1 - \theta_2)$

$\displaystyle \ddot{\theta_2}$ $\textstyle =$ $\displaystyle 2 \frac{g}{\ell} (\theta_1 - \theta_2).$ (1.18)

Normal Modes

A system with two degrees of freedom has two independent normal modes of oscillation in which both pendulums oscillate with the same frequency

$\displaystyle \theta_1$ $\textstyle =$ $\displaystyle C_1 \cos(\omega t + \phi)$

$\displaystyle \theta_2$ $\textstyle =$ $\displaystyle C_2 \cos(\omega t + \phi)$ (1.19)

but possibly with different amplitudes. With these solutions, Eq.'s 1.18 become

$\displaystyle (2\frac{g}{\ell}-\omega^2) \theta_1 - \frac{g}{\ell} \theta_2$ $\textstyle =$ $\displaystyle 0$

$\displaystyle 2 \frac{g}{\ell} \theta_1 - (2 \frac{g}{\ell}-\omega^2) \theta_2$ $\textstyle =$ $\displaystyle 0$ (1.20)
Note that Eq.'s 1.19 indicate that both pendulums oscillate with the same phase $\phi$ . However, the constants and can have opposite sign, which corresponds to a phase difference of $\pi$ . That is, in a normal mode, the pendulums oscillate either in phase or with opposing phase.
The angular displacements $\theta_1$ and $\theta_2$ can be eliminated from Eq.'s 1.20 to find

$\begin{displaymath} -(2\frac{g}{\ell}-\omega^2)(2\frac{g}{\ell}-\omega^2) + 2 \frac{g^2}{\ell^2} = 0 \end{displaymath}$

which is a quadratic in $\omega^2$ with solutions

$\begin{displaymath} \omega_{\pm} = \sqrt{\frac{g}{\ell} (2 \pm \sqrt{2})} \end{displaymath}$ (1.21)
We can now relate the amplitudes of the two pendulums by plugging the solutions of Eq.'s 1.19 and the normal mode frequencies of Eq.'s 1.21 into Eq.'s 1.18,

$\begin{displaymath} \frac{C_{1+}}{C_{2+}} = -\frac{1}{\sqrt{2}}, \frac{C_{1-}}{C_{2-}} = \frac{1}{\sqrt{2}} \end{displaymath}$ (1.22)
The normal mode solutions are given by Eq.'s 1.19 with the frequencies of Eq.'s 1.21 and amplitude ratios of Eq.'s 1.22,

$\displaystyle \theta_{1+}$ $\textstyle =$ $\displaystyle C_+ \cos(\omega_+ t + \phi_+)$

$\displaystyle \theta_{2+}$ $\textstyle =$ $\displaystyle -\sqrt{2} C_+ \cos(\omega_+ t + \phi_+)$

$\displaystyle \theta_{1-}$ $\textstyle =$ $\displaystyle C_- \cos(\omega_- t + \phi_-)$

$\displaystyle \theta_{2-}$ $\textstyle =$ $\displaystyle \sqrt{2} C_- \cos(\omega_- t + \phi_-)$ (1.23)

which we may wish to write in the alternative form

$\displaystyle \theta_{1+}$ $\textstyle =$ $\displaystyle A_+ \cos(\omega_+ t) + B_+ \sin(\omega_+ t)$

$\displaystyle \theta_{2+}$ $\textstyle =$ $\displaystyle -\sqrt{2} A_+ \cos(\omega_+ t) - \sqrt{2} B_+ \sin(\omega_+ t)$

$\displaystyle \theta_{1-}$ $\textstyle =$ $\displaystyle A_- \cos(\omega_- t) + B_- \sin(\omega_- t)$

$\displaystyle \theta_{2-}$ $\textstyle =$ $\displaystyle -\sqrt{2} A_- \cos(\omega_- t) - \sqrt{2} B_- \sin(\omega_- t)$ (1.24)
We only need the explicit solutions (Eq.'s 1.24) if we need to predict the positions and velocities of the pendulums at particular times based on initial conditions. The amplitude ratios $\frac{C_{1\pm}}{C_{2\pm}}$ and frequencies $\omega_\pm$ give a general picture of the motion of the system.
Animation : doubleModes.py (see Appendix A on running animations)

General Solutions

The normal mode solutions are solutions to linear differential equations ( $\theta_1$ , $\theta_2$ and their derivatives only appear to the first power), so they obey the principle of superposition. Hence, we need both normal mode solutions to describe the behavior of the system in general.
The general solutions of the equations of motion are linear combinations of the normal mode solutions,

$\displaystyle \theta_1$ $\textstyle =$ $\displaystyle A_+ \cos(\omega_+ t) + B_+ \sin(\omega_+ t)$

$\displaystyle + A_- \cos(\omega_- t) + B_- \sin(\omega_- t)$

$\displaystyle \theta_2$ $\textstyle =$ $\displaystyle - \sqrt{2} A_+ \cos(\omega_+ t) - \sqrt{2} B_+ \sin(\omega_+ t)$

$\displaystyle + \sqrt{2} A_- \cos(\omega_- t) + \sqrt{2} B_- \sin(\omega_- t)$ (1.25)
The constants , , , and are determined by initial angles $\theta_{1o}$ and $\theta_{2o}$ and initial angular velocities $\dot{\theta}_{1o}$ and $\dot{\theta}_{2o}$ .
We write the general solutions in terms of sines and cosines instead of cosines with phase constants, because it is a more convenient form for applying initial conditions.

1.3.2 Beads on a String : Transverse Modes

**Figure 1.3:** Transverse displacements of two beads on a massless string.
$\scalebox{0.7}{ \includegraphics{2Beads.eps} }$

Equations of Motion

Consider the system shown in Fig. 1.3. The beads have equal mass , and the strings are massless and have equal length $\ell$ . We will neglect gravity.
For small oscillations, $\cos \theta \approx 1$ , and we can take the tension in the strings to have constant magnitude . Small displacements give the tension vectors vertical components.
Newton's second law gives

$\displaystyle m \ddot{y_1}$ $\textstyle =$ $\displaystyle -T_o \sin \theta_1 + T_o \sin \theta_2$

$\textstyle =$ $\displaystyle -T_o \left( \frac{y_1}{\ell} + \frac{y_1-y_2}{\ell} \right)$

$\displaystyle m \ddot{y_2}$ $\textstyle =$ $\displaystyle - T_o \sin \theta_2 - T_o \sin \theta_3$

$\textstyle =$ $\displaystyle -T_o \left(\frac{y_2-y_1}{\ell} + \frac{y_2}{\ell} \right)$ (1.26)
Eq.'s 1.26 can be simplified to give equations of motion,

$\displaystyle \ddot{y_1}$ $\textstyle =$ $\displaystyle - \frac{T_o}{m \ell} (2y_1 - y_2)$

$\displaystyle \ddot{y_2}$ $\textstyle =$ $\displaystyle - \frac{T_o}{m \ell} (2y_2 - y_1)$ (1.27)

Normal Modes

For the normal modes, the solutions to the equations of motion (Eq.'s 1.27) take the form

$\displaystyle y_1$ $\textstyle =$ $\displaystyle C_1 \cos(\omega t + \phi)$

$\displaystyle y_2$ $\textstyle =$ $\displaystyle C_2 \cos(\omega t + \phi)$ (1.28)
Applying Eq.'s 1.27 to Eq.'s 1.28 yields

$\displaystyle (2\frac{T_o}{m \ell}-\omega^2) y_1 - \frac{T_o}{m \ell} y_2$ $\textstyle =$ $\displaystyle 0$

$\displaystyle -\frac{T_o}{m \ell} y_1 + (2\frac{T_o}{m \ell}-\omega^2) y_2$ $\textstyle =$ $\displaystyle 0$ (1.29)

which reduce to

$\begin{displaymath} (\omega^2)^2 -4 \frac{T_o}{m \ell} \omega^2 + 3 \left( \frac{T_o}{m \ell} \right)= 0 \end{displaymath}$ (1.30)

with solutions

$\begin{displaymath} \omega_+ = \sqrt{3 \frac{T_o}{m \ell}}, \omega_- = \sqrt{\frac{T_o}{m\ell}}. \end{displaymath}$ (1.31)
The solutions (Eq.'s 1.28) with the normal mode frequencies (Eq.'s 1.31) in the equations of motion (Eq.'s 1.27) give amplitude ratios

$\begin{displaymath} \frac{C_{1+}}{C_{2+}} = -1, \frac{C_{1-}}{C_{2-}} = 1 \end{displaymath}$ (1.32)
Animation : transverseModes.py (see Appendix A on running animations)

General Solutions

Just as for the double pendulum, the general solutions to the equations of motion (Eq.'s 1.27) are linear combinations of the normal mode solutions,

$\displaystyle y_1$	$\textstyle =$	$\displaystyle A_+ \cos(\omega_+ t) + B_+ \sin(\omega_+ t)$
		$\displaystyle + A_- \cos(\omega_- t) + B_- \sin(\omega_- t)$
$\displaystyle y_2$	$\textstyle =$	$\displaystyle - A_+ \cos(\omega_+ t) - B_+ \sin(\omega_+ t)$
		$\displaystyle + A_- \cos(\omega_- t) + B_- \sin(\omega_- t)$	(1.33)

where the amplitudes $A_\pm$ , $B_\pm$ are determined by initial positions $y_{1o}, y_{2o}$ and initial velocities $\dot{y}_{1o}, \dot{y}_{2o}$ .

Updated Wed Jan 18 09:51:28 2006

$\displaystyle x_1$	$\textstyle =$	$\displaystyle \ell \sin \theta_1$
$\displaystyle y_1$	$\textstyle =$	$\displaystyle \ell (1-\cos \theta_1)$
$\displaystyle x_2$	$\textstyle =$	$\displaystyle \ell (\sin \theta_1 + \sin \theta_2)$
$\displaystyle y_2$	$\textstyle =$	$\displaystyle \ell (2 - \cos \theta_1 - \cos \theta_2)$	(1.14)

$\displaystyle x_1$	$\textstyle \approx$	$\displaystyle \ell \theta_1$
$\displaystyle y_1$	$\textstyle \approx$	$\displaystyle 0$
$\displaystyle x_2$	$\textstyle \approx$	$\displaystyle \ell (\theta_1 + \theta_2)$
$\displaystyle y_2$	$\textstyle \approx$	$\displaystyle 0$	(1.15)

$\displaystyle F_{1 net x} =$	$\textstyle -T_1 \sin \theta_1 + T_2 \sin \theta_2$	$\displaystyle \approx -T_1 \theta_1 + T_2 \theta_2$
$\displaystyle F_{2 net x} =$	$\textstyle -T_2 \sin \theta_2$	$\displaystyle \approx -T_2 \theta_2$	(1.16)

$\displaystyle m\ddot{x_1} \approx$	$\textstyle m \ell \ddot{\theta_1}$	$\displaystyle \approx -mg (2 \theta_1 - \theta_2)$
$\displaystyle m\ddot{x_2} \approx$	$\textstyle m \ell (\ddot{\theta_1} + \ddot{\theta_2})$	$\displaystyle \approx -mg \theta_2$	(1.17)

$\displaystyle \ddot{\theta_1}$	$\textstyle =$	$\displaystyle -\frac{g}{\ell} (2 \theta_1 - \theta_2)$
$\displaystyle \ddot{\theta_2}$	$\textstyle =$	$\displaystyle 2 \frac{g}{\ell} (\theta_1 - \theta_2).$	(1.18)

$\displaystyle \theta_1$	$\textstyle =$	$\displaystyle C_1 \cos(\omega t + \phi)$
$\displaystyle \theta_2$	$\textstyle =$	$\displaystyle C_2 \cos(\omega t + \phi)$	(1.19)

$\displaystyle (2\frac{g}{\ell}-\omega^2) \theta_1 - \frac{g}{\ell} \theta_2$	$\textstyle =$	$\displaystyle 0$
$\displaystyle 2 \frac{g}{\ell} \theta_1 - (2 \frac{g}{\ell}-\omega^2) \theta_2$	$\textstyle =$	$\displaystyle 0$	(1.20)

$\displaystyle \theta_{1+}$	$\textstyle =$	$\displaystyle C_+ \cos(\omega_+ t + \phi_+)$
$\displaystyle \theta_{2+}$	$\textstyle =$	$\displaystyle -\sqrt{2} C_+ \cos(\omega_+ t + \phi_+)$
$\displaystyle \theta_{1-}$	$\textstyle =$	$\displaystyle C_- \cos(\omega_- t + \phi_-)$
$\displaystyle \theta_{2-}$	$\textstyle =$	$\displaystyle \sqrt{2} C_- \cos(\omega_- t + \phi_-)$	(1.23)

$\displaystyle m \ddot{y_1}$	$\textstyle =$	$\displaystyle -T_o \sin \theta_1 + T_o \sin \theta_2$
	$\textstyle =$	$\displaystyle -T_o \left( \frac{y_1}{\ell} + \frac{y_1-y_2}{\ell} \right)$
$\displaystyle m \ddot{y_2}$	$\textstyle =$	$\displaystyle - T_o \sin \theta_2 - T_o \sin \theta_3$
	$\textstyle =$	$\displaystyle -T_o \left(\frac{y_2-y_1}{\ell} + \frac{y_2}{\ell} \right)$	(1.26)

$\displaystyle \ddot{y_1}$	$\textstyle =$	$\displaystyle - \frac{T_o}{m \ell} (2y_1 - y_2)$
$\displaystyle \ddot{y_2}$	$\textstyle =$	$\displaystyle - \frac{T_o}{m \ell} (2y_2 - y_1)$	(1.27)

$\displaystyle (2\frac{T_o}{m \ell}-\omega^2) y_1 - \frac{T_o}{m \ell} y_2$	$\textstyle =$	$\displaystyle 0$
$\displaystyle -\frac{T_o}{m \ell} y_1 + (2\frac{T_o}{m \ell}-\omega^2) y_2$	$\textstyle =$	$\displaystyle 0$	(1.29)