**COUPLED HARMONIC OSCILLATORS**

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A coupled oscillator consists of two simple harmonic oscillators which are connected, or “coupled” by an intermediary force or potential. In such a system, the oscillation of one particle depends not just on one potential, but *two*. One of the most basic examples of a coupled oscillator is the following mass — spring system:

As shown above, the two masses are connected not just to their own springs on the far left and right, but also through the means of a “mediator” spring at the center. Therefore, it is easy to see that the state, or position of mass *m1 *affects not just the force on itself, but also the force on mass *m2*. For example, if *m1 *is shifted to the right, then not only is there a leftward force pulling it to, well, the left, but the compression of the central spring contributes as well. Further, because the middle spring is now compressed, *m2* also experiences a *rightward*force. It’s easy to see how the whole system oscillates not just by virtue of the left and right springs, but also the central springs.

Another popular example of a coupled oscillator is the *coupled pendulum*. It is basically a set of pendulums, generally two, connected by a string or a wooden plank (or any material, for that matter).

As seen above, this simple setup is the most common “coupled oscillator” that can be formed — it’s extremely simple to make. In fact, Christiaan Huygens had famously observed the phenomenon of synchronisation in pendulum clocks connected by a wooden beam in the 17th century. Two pendulum clocks, starting at different positions, seem to eerily “synchronise” with one another, being either in phase or out of phase. However, Huygens findings were before Newton published his *Principia*, so we shall be turning to Newton to help us find explicit time — dependent solutions to a coupled harmonic oscillator, such as the spring above.

*TIME — DEPENDENT SOLUTIONS*

To start off, we shall make a few simplifying assumptions. First, that the edge springs in image 1 have the same spring constant *k *and that the blocks have the same mass *m. *The central spring will have a spring constant *K*. *x1(t)* and *x2(t) *are the positions of block 1 and 2 with respect to time.

Therefore, *k = ω^2m*, where *ω *is the natural frequency of the uncoupled oscillator.

*K = Ω^2m*, where *Ω *is the natural frequency of the central spring when connected to only one of the masses. Now, the equations of motion, after all simplifications have been made, are as follows:

These two can be simplified into a second order matrix equation.

It turns out that solutions to this equation are of the form

Here, ** e1 **and

**are the eigenvectors of the above matrix corresponding to eigenvalues λ1 and λ2. Now, the eigenvalues and eigenvectors are as follows:**

*e2*Therefore, the full solution is

Which can be further simplified to

Their graphs are:

As shown in image 4, when one oscillator is at its peak energy, the other is generally at its bottom/trough. This is because the two oscillators transfer energy to each other via the intermediary spring.

One extremely beautiful thing to realize is that, unlike a normal oscillator, this simplified coupled oscillator has not one, but *two* natural frequencies corresponding to the square roots of the eigenvalues, and as such a coupled oscillator can have multiple resonant frequencies. This is especially useful in strings, which can be thought of as *multiple* connected oscillators, and therefore has many resonant frequencies.

## LIMITING CASES

We shall now explore limiting cases of the oscillator, specifically the two cases *ω << Ω* and *ω >> Ω, *which correspond to strong and weak coupling respectively.

*Strong coupling:*

In the case of *ω << Ω*, the solutions simplify in the following ways:

This doesn’t look different from our explicit solution, but a graph makes it much clearer.

This is a graph of the positions of both strongly coupled oscillators. Now, this is nothing like image 3, in that this one is much more sinusoidal. Further, the two sinusoids seem to be out of phase with each other, nearly πradians out of phase. Further, it turns out that the individual phases δ and ε don’t impact the phase difference between the solutions. Therefore, in this limiting case in which the middle spring is much, much stronger than the ones on the side, it will effectively “control” the two masses, pushing them back and forth out of phase. This observation is like the “synchronization” Huygens observed.

*Weak coupling:*

In the case of *ω >> Ω*, the solutions are:

These solutions are pure, uncoupled sinusoids, which is expected in case of this extremely weak coupling between the two masses. The graph is:

As is visible, the two solutions are sinusoids. However, one difference here is that the phase angles α and β *do impact the overall phase difference,* as shown in the image below, where the phases are arbitrarily set so that both waves are in phase.

## GENERIC OSCILLATORS

In fact, an oscillator doesn’t even need a spring to be coupled, just a *coupling potential*. Consider the coupling potential Φ(x1, x2). Because the motion is simple harmonic, it turns out that the trivial solutions *x1 = 0* and *x2 = 0 *are, in fact, stable. Therefore, for small values of* x1* and *x2 *the functions

and

can be expressed as:

Now, because Φ is a potential field, it follows from Newton’s 3rd law that

After all this math, one can now define the energy of a generic coupled oscillator:

Now we introduce a slightly modified Euler — Lagrange equation, using energy instead as we now have an expression for total energy *E*:

After using all equations above, we get the equations of motion as:

We will now define the second partial derivative of Φ at (0,0) as seen above as *K. *Therefore, we get:

These equations of motion describe the generic coupled harmonic oscillator and include the *strength of coupling K *of the oscillator.

As this article concludes, I would like readers to consider this fact: if one looks at the potential energy of only *one* mass in a coupled oscillator, essentially writing it in terms of two variables: *x1* and *t*, where the time dependence comes from the fact that *x2* is written explicitly in terms of time, one realizes that when energy is lost, it is either stored in the spring or in the other mass. But what about momentum? When the first mass loses momentum, it obviously goes to the other mass, but looking *purely* at the first mass, it would seem as though, in a time varying potential, its momentum was “hidden”, in a way. So, my last question: can one “hide” momentum like one can hide energy?