Coupled Harmonic Oscillator - A. Freddie Page, Imperial College London
In this worksheet, we will go step by step through the coupled harmonic oscillator system, building up our intuition as we go. This worksheet is interactive; at every stage you should feel free to experiment with the elements in the page.
Grab one of the masses with your mouse, try moving it around and throwing it.

Before we dive into the full coupled system, it's worth looking at a single oscillator - e.g. a mass on a spring. Recall that this system has the equation of motion, $ \ddot x_1(t) = -\frac{k_1}{m_1} x_1(t) $ for spring constant $ k_1 $ and mass $ m_1 $. This system oscillates in harmonic motion with a frequency $ ω_1 = \sqrt{\frac{k_1}{m_1}} $. Using this, the equation of motion can be re-written as, $ \ddot x_1(t) = -ω_1^2 x_1(t) $ with a solution $ x_1(t) = A_1 \cos(ω_1 t + φ_1) $.

In this worksheet, we will focus on frequencies rather than spring constants. We'll define our springs from now on in terms of their natural frequency (we can always convert back, with $ k_1 = m_1 ω_1^2 $ ).

Experiment with different natural frequencies and masses.
For a single mass with frequency held constant, altering the mass doesn't affect the motion.

Harmonic motion is also referred to as circular motion, since if we plot the displacement of our oscillator, $ x_1 $, against its velocity (scaled by frequency to get the dimensions right), $ \frac{\dot{x}_1}{ω_1} $, then in phase space, the particle moves in a circle. The interpretation of this is that the system is constantly exchanging potential energy, $ \frac{1}{2} k_1 x_1^2 $, and kinetic energy, $ \frac{1}{2} k_1 \left( \frac{\dot{x}_1}{ω_1} \right )^2 $.

Let's add another mass - We now have two un-coupled oscillators.
The motion of each mass is independent of the other.
Each mass has it's own amplitude, frequency, and phase.
The masses do not exchange any energy between each other.

Let's do a bit of book-keeping here. There are two modes of oscillation of this system.
The first mode is where mass 1 oscillates at a frequency $ ω_1 $, with mass 2 stationary…

… And the second mode is where mass 2 oscillates at a frequency $ω_2$, with mass 1 stationary.

Because these modes are independent of each other, they are in some sense orthogonal to each other; they are normal modes.

The equations of motion for this system are simple. They're just the equation for the single particle duplicated.
$ \ddot x_1(t) = -ω_1^2 x_1(t) $
$ \ddot x_2(t) = -ω_2^2 x_2(t) $
With Solutions, $ x_1(t) = A_1 \cos(ω_1 t + φ_1) $ and $ x_2(t) = A_2 \cos(ω_2 t + φ_2) $.

So far so good, but things get more complicated once we introduce another spring; this couples the two masses together, and allows them to exchange energy.
The modes of the system are now not so simple.
See what happens when you move one of the masses.

Now the first mass does not move independently of the second, and the system as a whole is not moving in harmonic motion with a well defined frequency.
See how the particles in phase space are not moving in circular, or even elliptical orbits any more.

Let's look deeper to see what's going on. We have new equations of motion.
$ {\color{blue}\ddot x_1(t)} = -{\color{blue}ω_1^2 x_1(t)} + \frac{\color{green}j}{\color{blue}m_1} [{\color{red} x_2(t)} - {\color{blue}x_1(t)} ] $
$ {\color{red}\ddot x_2(t)} = +\frac{\color{green}j}{\color{red}m_2} [ {\color{blue} x_1(t)} - {\color{red}x_2(t)} ] - {\color{red}ω_2^2 x_2(t)} $
There are now cross coupling terms for the middle spring, with a spring constant $j$.

Again, we can put this in terms of a natural frequency for the spring. Defining, $ω_j = \sqrt{\frac{j}{μ}}$,
where $μ = \frac{m_1 m_2}{m_1 + m_2}$ is the reduced mass of the system. The equations of motion are then,
$ {\color{blue}\ddot x_1(t)} = -{\color{blue}ω_1^2 x_1(t)} + \frac{μ}{\color{blue}m_1} {\color{green} ω_j^2} [ {\color{red} x_2(t)} - {\color{blue} x_1(t)} ] $
$ {\color{red}\ddot x_2(t)} = + \frac{μ}{\color{red}m_2} {\color{green} ω_j^2} [ {\color{blue} x_1(t)} - {\color{red} x_2(t)} ] - {\color{red}ω_2^2 x_2(t)} $.

A good technique in Physics, when trying to build an intuition of a new system, is to take the parameters to extreme limits.
Let's consider what happens when the coupling frequency is much larger than the natural frequencies of each mass-spring pair. $ω_j \gg ω_1, ω_2$. In this case the masses exchange energy with each other much faster than they exchange their own kinetic and potential energy.

One of the modes of the system will be the combined masses locked together and oscillating under the pull of a combined spring constant, with a frequency, $ω_M = \sqrt{\frac{k_1 + k_2}{m_1 + m_2}}$.
As an exercise, re-write this in terms of the natural frequencies $ω_1$ and $ω_2$. Show that the combined frequency is always inbetween the two.
As such this translation mode has a frequency, $ω_M$ much smaller than the coupling frequency, $ω_j$.

The other mode of the system is the vibration mode. Here the masses oscillate about a stationary centre of mass. We won't go into details here, but if we changed coordinate system to the (non-inertial) rest frame of one of the masses, fictitious forces would make the other mass appear as if it had a reduced mass, $μ$. This mode therefore oscillates at the coupling frequency, $ω_j = \sqrt{\frac{j}{μ}}$.
See what happens if you change the value of the masses.

Of course, in principle, a combination of the translation and vibration mode can be excited at the same time.
This isn't a mode of the system, because it's not oscillating at a single frequency, but it is a general solution to the system.

Let's recap. At zero coupling strength, there are two normal modes at frequencies $ω_1$ and $ω_2$, where one mass oscillates with the other stationary and vice-versa.
Then at high coupling strengths, there is a low frequency translation mode, $ω_M$, and a high frequency vibration mode, $ω_j$, where both particles move, in phase and 180° out of phase, respectively.
For intermediate coupling strengths, the modes are a mixture of the two cases…

Let's start with the high frequency mode. Try adjusting the coupling frequency, $ω_j$.
For low coupling strengths, only mass 1 will be in motion (it has the higher natural frequency). Whereas for high coupling strengths, the system will enter the vibration mode.
For intermediate coupling, the system will oscillate at a new frequency, $ω_{u_1}$, which starts at $ω_1$ and approaches $ω_j$ from above, as $ω_j$ increases.

The case is similar with the low frequency mode, $ω_{u_2}$. This will range from $ω_2$ and approach $ω_M$ from below, as $ω_j$ increases.
The way the masses move with respect to each other will transition between only mass 2 moving and both masses moving in phase in the translation mode.

Another limit we can study is if one of the masses is significantly bigger than the other, i.e. $m_2 \gg m_1$.
Here mass 1 has very little influence on mass 2, but mass 2 can influence mass 1 significantly.
Either mass 2 is stationary, in which case it acts like a fixed wall, then mass 1 will move under the combined strength of springs $ω_1$ and $ω_j$.
This is the upper branch, it's frequency is the hyperbola, $ω_{u_1} = \sqrt{ω_1^2 + ω_j^2}$.

The other mode is the lower branch, where mass 2 is moving. Since it is dominant over mass 1, it will oscillate at it's natural frequency, $ω_2$, and mass 1 must follow.
The frequency of this mode is constant for all coupling strengths, at $ω_{u_2} = ω_M = ω2$.

Something particularly interesting happens if we swap the relative sizes of springs 1 and 2.
We still have the two mode frequencies $ω_{u_1}$ and $ω_{u_2}$ limiting to a hyperbola and a constant, but this time the modes swap character right at the point where the curves would otherwise cross.
This behaviour is called an avoided crossing, and pops up occasionally in physics.
Experiment with swapping the relative sizes of the natural frequencies and coupling frequency.

With all this in mind, finding normal modes is the process of finding what frequencies the system oscillates at, and how each of the components move with respect to each other at those frequencies.

Let's formalise this process. We'll write the equations of motion in matrix form. At this stage, this is just a book-keeping exercise, to keep the equations tidy.
$ \begin{bmatrix} \ddot{x}_1(t) \\ \ddot{x}_2(t) \end{bmatrix} = -\begin{bmatrix} ω_1^2 + \frac{μ}{m_1}ω_j^2 & -\frac{μ}{m_1}ω_j^2 \\ -\frac{μ}{m_2}ω_j^2 & ω_2^2 + \frac{μ}{m_2}ω_j^2 \end{bmatrix} \begin{bmatrix} x_1(t) \\ x_2(t) \end{bmatrix} $.

In principle, $\begin{bmatrix} x_1(t) \\ x_2(t) \end{bmatrix}$ has a complicated general solution, as animated below. In order to find the normal modes, we can use the differential equations trick: assume we already have a normal mode, plug it into the equations of motion, and test the implications.

Let's assume a trial solution,
$ \begin{bmatrix} x_1(t) \\ x_2(t) \end{bmatrix} \rightarrow \mathbf{u}(t) = A \hat{\mathbf{u}}\cos(ω t + φ) $, where $ \hat{\mathbf{u}} $ is a constant 2D vector.
Then the second derivative also has simple form, $ \begin{bmatrix} \ddot{x}_1(t) \\ \ddot{x}_2(t) \end{bmatrix} \rightarrow -ω^2 A \hat{\mathbf{u}}\cos(ω t + φ) = -ω^2\mathbf{u}(t) $.

We can then feed our test solution back into the equations of motion,
$ -ω^2 A\hat{\mathbf{u}} \cos(ωt + φ) = -A\begin{bmatrix} ω_1^2 + \frac{μ}{m_1}ω_j^2 & -\frac{μ}{m_1}ω_j^2 \\ -\frac{μ}{m_2}ω_j^2 & ω_2^2 + \frac{μ}{m_2}ω_j^2 \end{bmatrix} \hat{\mathbf{u}} \cos(ωt + φ) $,
which we can simplify by dividing out common terms…

This gives us,
$ ω^2 \hat{\mathbf{u}} = \begin{bmatrix} ω_1^2 + \frac{μ}{m_1}ω_j^2 & -\frac{μ}{m_1}ω_j^2 \\ -\frac{μ}{m_2}ω_j^2 & ω_2^2 + \frac{μ}{m_2}ω_j^2 \end{bmatrix} \hat{\mathbf{u}} $,
or, $ ω^2 \hat{\mathbf{u}} = \mathbf{M} \hat{\mathbf{u}} $, which is independent of $A$ and $φ$, but gives us a condition for $ω$ and $\hat{\mathbf{u}}$.

That is, our trial solution, $\mathbf{u}(t) = A \hat{\mathbf{u}} \cos(ωt+φ)$, is only a normal mode if $ω$ and $\hat{\mathbf{u}}$ satisfy the matrix eigenvalue equation, $\mathbf{M}\hat{\mathbf{u}} = ω^2\hat{\mathbf{u}}$.
We won't explicitly solve the system in this worksheet, except to say there will be two solutions,
$ω_{u_1}, \hat{\mathbf{u}}_1 = \begin{bmatrix}1 \\ r_1\end{bmatrix}$ and $ω_{u_2}, \hat{\mathbf{u}}_2 = \begin{bmatrix}r_2 \\ 1\end{bmatrix}$, and we've seen these frequencies previously.

The normal modes of the system, defined by their frequency, $ω$, and configuration of components $\hat{\mathbf{u}}$, oscillate independently of each other. By calculating the normal modes, we move from a description of the system with simple parts and complicated motion ($\mathbf{x}(t)$) to a description with complicated parts but simple motion ($\mathbf{u}(t)$). We have solved for two un-coupled oscillators in an abstract space.
Use the new controls below to alter the normal mode amplitudes directly.

Coupled Harmonic Oscillator - A. Freddie Page, Imperial College London
This is the end of the worksheet, please feel free to email me any feedback.
From here you might find it instructive to skim through the worksheet again with all the controls enabled, or go ahead to the last page with extra stats.

Extra Stats

$A_1$		$A_2$		$ω_{u_1}$		$ω_{u_2}$
$φ_1$		$φ_2$		$r_1$		$r_2$
$k_1$		$k_2$		$j$		$η$