Coupled oscillators

Consider a system like shown, with two masses connected by springs between two stationary walls. $x_1$ and $x_2$ are the masses’ respective displacements from equilibrium.

We first model the equations of motion of each mass.

\begin{align*} m \ddot x_1 &= -kx_1 + k(x_2-x_1) \\ m \ddot x_2 &= -k(x_2-x_1) - k x_2. \end{align*}

We guess the solution to these equations will be $x=\mathrm{Re}[z]$ , $z=Ae^{i \omega t}$ . Now we could substitute the solution into our system of equations and solve for the frequencies and amplitudes, but there’s an easier way.

We first define several vectors.

\ddot{\mathbf x} = \begin{bmatrix} \ddot x_1 \\ \ddot x_2 \end{bmatrix}, \quad \mathbf A = \begin{bmatrix} A_1 \\ A_2 \end{bmatrix}, \quad \mathbf x = \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \mathbf A e^{i \omega t}.

Next we rewrite the equations of motion in normal form, substituting $\omega_0^2 = k/m$ .

\begin{align*} \ddot x_1 &= -2 \omega_0^2 x_1 + \omega_0^2 x_2 \\ \ddot x_2 &= -2 \omega_0^2 x_2 + \omega_0^2 x_1. \end{align*}

Then we write the relation between position and acceleration as a matrix multiplication.

\begin{align*} \ddot{\mathbf x} = \mathbf{Mx} = \begin{bmatrix} -2 \omega_0^2 & \omega_0^2 \\ \omega_0^2 & -2 \omega_0^2 \end{bmatrix} \mathbf x. \end{align*}

Recall that from the solution we guessed, $\ddot{\mathbf x} = -\omega^2 \mathbf x$ . We substitute this back into the equation.

-\omega^2 \mathbf A e^{i \omega t} = \begin{bmatrix} -2 \omega_0^2 & \omega_0^2 \\ \omega_0^2 & -2 \omega_0^2 \end{bmatrix} \mathbf A e^{i \omega t }.

\left( \begin{bmatrix} -2 \omega_0^2 & \omega_0^2 \\ \omega_0^2 & -2 \omega_0^2 \end{bmatrix} + \mathbf I \omega^2 \right) \mathbf A = \mathbf 0.

The solution to $\mathbf{MA}=0$ is only interesting (nonzero) when the determinate of the matrix is $0$ .

\begin{vmatrix} \omega^2 -2 \omega_0^2 & \omega_0^2 \\ \omega_0^2 & \omega^2 -2 \omega_0^2 \end{vmatrix} = \mathbf 0.

To solve this we define $f= \omega^2 / \omega_0^2$ .

\omega_0^2 \begin{vmatrix} f-2 & 1 \\ 1 & f-2 \end{vmatrix} = \mathbf 0.

We find $(f-1)(f-3)=0$ , so $\omega = \{\pm \omega_0, \pm \sqrt{3} \omega_0\}.$

Note

Note the $\pm$ in the solutions for $\omega$ . We can write the solution for some $\pm\omega$ as $x = c_1 e^{i\omega t} + c_2 e^{-i\omega t}$ . We use the property that a superposition of any two sinusoids of the same frequency can be written as a single phase-shifted sinusoid. In its complex form we say $x = c_3 e^{i(\omega t - \phi)}$ , in its real form $x = c_3 \cos(\omega t - \phi) = c_4 \cos(\omega t) + c_5 \sin(\omega t)$ .

Recall that we guessed our solution would look like

\begin{align*} \mathbf x &= \mathbf A \mathrm{Re} \left[ e^{i \omega t } \right ] \\ &= \mathbf A \left( C \cos(\omega t) + D \sin(\omega t) \right). \end{align*}

Since we have two values for $\omega$ , we will have two such solutions. Let $\omega_+ = \sqrt{3} \omega_0$ , $\omega_- = \omega_0$ . We plug these values in to the $(\mathbf M + \mathbf I \omega^2) \mathbf A = 0$ equation to find the ratio of the relative amplitudes $A_1$ and $A_2$ .

\begin{align*} \left( \begin{bmatrix} -2 \omega_0^2 & \omega_0^2 \\ \omega_0^2 & -2 \omega_0^2 \end{bmatrix} + \mathbf I \omega^2 \right) \mathbf A = \mathbf 0 \end{align*}

\begin{align*} (\omega^2 -2 \omega_0^2) A_1 + \omega_0^2 A_2 = 0 \end{align*}

Plugging in $\omega = \omega_-$ we find $A_1 = A_2$ .

Plugging in $\omega = \omega_+$ we find $A_1 = -A_2$ .

We take the superposition of the two particular solutions to find the general solution.

\begin{align*} x_1(t) &= C ( \cos(\omega_+ t) + \sin(\omega_+ t) ) + D (\cos(\omega_- t) + \sin(\omega_- t)) \\ x_2(t) &= C ( \cos(\omega_+ t) + \sin(\omega_+ t) ) - D (\cos(\omega_- t) + \sin(\omega_- t)). \end{align*}

The two particular solutions are called the normal modes of the system. It is interesting to note that the behavior of this system from any initial condition can be represented as a linear combination of the two normal modes.

Note

This is the general solution for this particular system. Different systems will have different relative amplitudes and frequencies.