Coupled oscillators: the sympathy of pendulums

The puzzle: the pendulum that stops itself¶

Hang two identical pendulums from the same slightly slack string. Start one swinging and leave the other hanging still. Come back a minute later and you find the opposite: the one you started is hanging dead, and the one you never touched is swinging.

Wait the same time again and it has reversed. Nothing is being damped away — a minute after that, the first pendulum is swinging as widely as when you let it go.

Christiaan Huygens noticed something like this in 1665, from a sickbed, watching two clocks hung on a common beam settle into step with each other. He called it “an odd kind of sympathy”.

Hold on to how strange the first pendulum’s behaviour is. It slows to a complete stop without any friction, and then, unprompted, starts again.

Predict before you calculate¶

Commit to an answer for each before running anything. Write them down.

You start pendulum 1 alone and wait a long time. Where is the energy — mostly still in pendulum 1, mostly in pendulum 2, or split evenly between them?
How many ways are there to start the pair so that no energy is ever exchanged? Zero, one, two, infinitely many?
You stiffen the coupling between them. Does the trading get faster or slower?
Does either of the no-exchange motions, if they exist, happen at exactly the frequency a single uncoupled pendulum would have?

Explore the model¶

Three animations, all generated from the same wavelab.coupled code the laboratory runs.

One mass started alone, with the two displacements in the middle and the energy held by each mass on the right. Watch the bars rather than the masses: they pulse in antiphase beneath the dashed line, which marks the total. The total never moves. Pendulum 1 does not slow down because something is draining it — nothing is — but because its energy is somewhere else, and it comes back.

The two motions that never trade anything. Top: both masses move together, and the green coupling spring keeps exactly its rest length the whole time. Bottom: they move in opposite directions, and the coupling spring stretches and compresses twice a cycle. Each of these repeats for ever at a single frequency, with no sloshing at all.

The same motion as the first animation, in two pictures at once. Left: energy per mass, which trades back and forth. Right: energy per mode, which does not move. Neither picture is more true than the other. The right-hand one is simply the picture in which the answer is already two constant numbers, and finding it is what this module is for.

Model specification

System: two identical point masses, each held to a wall by a spring of stiffness $k$ and joined to each other by a coupling spring of stiffness $k_c$ — equivalently two small-angle pendulums on a shared support; the observables are the two displacements, their velocities, and the energy attributed to each mass or to each mode.
Dynamics: $M\ddot{\mathbf{x}} + K\mathbf{x} = 0$ with $M = m\,\mathbb{1}$ , integrated by velocity Verlet and solved exactly by decomposition into two independent oscillators.
Boundary: both outer springs anchored to rigid supports that never move and never absorb anything.
Ensemble: a single deterministic trajectory per choice of initial conditions; measurement noise, where added, is Gaussian, independent and seeded.
Ignored: damping and driving — the pair here is free and conservative; also spring masses, the pendulum’s large-angle nonlinearity, and the escapement mechanics of the real clocks that gave Huygens his observation.
Valid when: displacements stay in the linear range of the real springs (small angles for pendulums), and the time step stays well below the period of the faster mode.
Failure modes: amplitudes large enough for nonlinearity to couple the modes, at which point superposition itself fails; damping added carelessly, since the modes stay independent only for special damping; and reading the site picture as though it were fundamental.

Run the laboratory now, then come back. Three things are worth doing before you read on:

Start one mass and time how long the energy takes to arrive at the other and return. Then stiffen the coupling and time it again.
Hunt for the no-exchange starts by hand. Set $x_2/x_1$ to different ratios, release from rest, and watch the energy bars. Two ratios make the sloshing stop completely; find them before anyone tells you what they are.
Having found them, check whether either one swings at the frequency a single mass on its anchor spring would have.

Derive the result¶

Two masses, three springs. Let $x_1$ and $x_2$ be the displacements from rest. Mass 1 feels its own anchor spring pulling it back, and the coupling spring, which cares only about how much further apart the two masses are than at rest:

m\ddot{x}_1 = -k x_1 - k_c\,(x_1 - x_2), \qquad m\ddot{x}_2 = -k x_2 - k_c\,(x_2 - x_1).

(1)

Each equation contains the other mass’s coordinate. That is what “coupled” means, and it is why neither can be solved on its own. Collect them:

M\ddot{\mathbf{x}} + K\mathbf{x} = 0, \qquad M = \begin{pmatrix} m & 0 \\ 0 & m \end{pmatrix}, \qquad K = \begin{pmatrix} k + k_c & -k_c \\ -k_c & k + k_c \end{pmatrix}.

(2)

The off-diagonal entries of $K$ are the coupling and nothing else. Set $k_c = 0$ and the matrix is diagonal, the two equations fall apart, and the masses never learn of each other’s existence.

Look for a motion at a single frequency. The prediction section asked whether such motions exist. Assume one does and see what it must satisfy. Try $\mathbf{x} = \mathbf{a}\cos\omega t$ , with a fixed shape $\mathbf{a}$ and one frequency for both masses. Two derivatives bring down $-\omega^2$ , and the cosine divides out:

\left(K - \omega^2 M\right)\mathbf{a} = 0 .

(3)

A matrix times a nonzero vector giving zero means the matrix has no inverse, so its determinant vanishes:

\det\begin{pmatrix} k + k_c - m\omega^2 & -k_c \\ -k_c & k + k_c - m\omega^2 \end{pmatrix} = \left(k + k_c - m\omega^2\right)^2 - k_c^2 = 0 .

(4)

That is a difference of two squares, so it factors immediately: $k + k_c - m\omega^2 = \pm k_c$ . Two roots, and with them two shapes:

\omega_s = \sqrt{\frac{k}{m}} = \wnat, \qquad \mathbf{a}_s \propto \begin{pmatrix} 1 \\ 1 \end{pmatrix}; \qquad\qquad \omega_a = \sqrt{\frac{k + 2k_c}{m}}, \qquad \mathbf{a}_a \propto \begin{pmatrix} 1 \\ -1 \end{pmatrix}.

(5)

There is the answer to prediction 2: exactly two. And to prediction 4, which is the more interesting one.

These two motions are the normal modes. Their existence is not obvious and their number is not accidental: two masses, two modes. And because the equations are linear, any sum of the two is also a solution — which turns out to be every solution there is.

Normal coordinates: the same fact as a change of variables. Add and subtract the two equations of motion and define

q_+ = \frac{x_1 + x_2}{\sqrt2}, \qquad q_- = \frac{x_1 - x_2}{\sqrt2}.

(6)

The addition gives $m\ddot{q}_+ = -k q_+$ , because the coupling terms cancel exactly. The subtraction gives $m\ddot{q}_- = -(k + 2k_c) q_-$ , because they reinforce. So

\ddot{q}_+ + \wnat^2 q_+ = 0, \qquad \ddot{q}_- + \omega_a^2 q_- = 0 .

(7)

Two independent module-01 oscillators, with no trace of each other. Nothing was thrown away — $q_\pm$ carry exactly the information $x_{1,2}$ do — and yet in these coordinates the problem has no coupling in it at all. The coupling was never in the physics; it was in the choice of coordinates, and a different choice removes it. That sentence is the whole of Part II, and the rest of the course keeps finding new systems it applies to.

Starting one mass. Now solve the puzzle. Release from rest with $x_1 = A$ , $x_2 = 0$ . In normal coordinates that is $q_+ = q_- = A/\sqrt2$ : equal amounts of both modes. Each then oscillates at its own frequency, and converting back,

x_1(t) = A\cos\!\left(\frac{\Delta\omega}{2}t\right)\cos\bar\omega t, \qquad x_2(t) = A\sin\!\left(\frac{\Delta\omega}{2}t\right)\sin\bar\omega t,

(8)

with $\Delta\omega = \omega_a - \omega_s$ and $\bar\omega = (\omega_s + \omega_a)/2$ .

This is module 00’s beat identity, unchanged. Two nearby frequencies added give a fast oscillation at their mean inside a slow envelope at half their difference — except that here the two envelopes belong to two different objects, and where one is large the other is small. The energies go as $\cos^2$ and $\sin^2$ of $\Delta\omega t/2$ , so they hand over completely and come back:

T_{\text{ex}} = \frac{2\pi}{\Delta\omega} .

(9)

There is the answer to the puzzle. Nothing keeps score; the two modes simply run at slightly different rates, and the pattern of who-has-the-energy is set by how far out of step they have drifted. Full transfer at $T_{\text{ex}}/2$ , full return at $T_{\text{ex}}$ .

How big is the splitting? For weak coupling, $k_c \ll k$ , expand the square root:

\omega_a = \wnat\sqrt{1 + \frac{2k_c}{k}} \approx \wnat\left(1 + \frac{k_c}{k}\right) \qquad\Longrightarrow\qquad \Delta\omega \approx \wnat\,\frac{k_c}{k}, \qquad T_{\text{ex}} \approx \frac{2\pi}{\wnat}\,\frac{k}{k_c} .

(10)

Stiffer coupling, wider splitting, faster trading — the answer to prediction 3, and it is inverse rather than merely monotonic. Turn it around and it becomes a measuring instrument: a slow, clean exchange is the signature of a small splitting, and timing the sloshing measures a frequency difference far too small to see on any spectrum.

The estimate is a weak-coupling estimate and should be trusted like one — it runs 1% low at $k_c/k = 0.02$ , 5% at 0.1, and 17% at 0.5, where the expansion has no business being used.

Verify computationally¶

Each check below is also a test in the project’s suite, so these claims cannot silently rot.

1. The hand-solved modes. The frequencies and shapes derived above by determinant, against normal_mode_solve working numerically on the same matrices: they agree to 10^-15, at couplings from $k_c/k = 0.006$ to 5. The symmetric frequency is $\wnat$ to the last bit every time.

2. The falsifying experiment. Released with one mass displaced, site_energies is tracked through a full exchange period.

3. The exchange is the beat. The simulated $x_1(t)$ is laid over phasors.beat_signal evaluated at the two mode frequencies. They agree to better than $5\times 10^{-3}$ of the amplitude, and the integrator knows nothing about either mode — so a measurement of module 00’s identity in a new setting, not a restatement of it.

4. The scaling law. $T_{\text{ex}}$ is measured across a decade of coupling strengths and fitted on log-log axes. The exponent comes out -0.991 against the predicted -1 over $k_c/k \in [0.005, 0.05]$ . Extend the sweep into strong coupling and it drifts to -0.830, which is the approximation failing exactly where it was derived to fail.

5. Modes hold still while sites slosh. The same trajectory, read both ways: the site energies traverse essentially the whole total, while the modal energies are constant to a part in 10⁵. The bound does not grow when the run is made four times longer — the integrator is symplectic, so its error oscillates instead of accumulating.

Transfer the idea¶

More of everything. Two masses gave two modes. Twenty give twenty, and the counting is not a coincidence; module 07 does the general case and then lets the number of masses run away to infinity, which is where waves come from.
Two coupled LC circuits. Two identical resonant circuits sharing a capacitor obey these matrices with the module-05 dictionary applied twice. Energy sloshes between the circuits at a rate set by the splitting, and the design of band-pass filters is the art of choosing it.
Molecules. Carbon dioxide is a mass between two others, and its symmetric stretch — both oxygens moving out together — is a genuine normal mode with its own frequency, distinct from the antisymmetric stretch. Which modes absorb infrared light and which do not is the reason the molecule matters to the climate.
Two-level systems. Module 52-quantum-optics describes an atom trading a quantum of excitation with a light field at a rate set by the splitting between two coupled states. The arithmetic is this module’s arithmetic, and the exchange has the same name in reverse: a splitting measured in the frequency domain, a trading period measured in time.

Check your understanding¶

Two identical pendulums hang from a shared support, weakly coupled. You start one swinging and leave the other still, then watch for several minutes. There is no friction. What happens to the energy?
A. It travels completely to the second pendulum, then completely back, over and over. B. Most of it stays in the first pendulum, with a small amount leaking back and forth. C. It settles into an even split, with both pendulums swinging at half amplitude for ever. D. The first pendulum gradually stops and stays stopped, its energy lost to the support.
Check your answer
Correct answer: A — It travels completely to the second pendulum, then completely back, over and over.
- A. Right. Starting one pendulum excites both normal modes equally, and they run at slightly different frequencies, so they drift out of step and back into step. When they are out of step the motion has moved entirely to the other pendulum. At weak coupling the first pendulum’s energy falls below 0.1% of the total before returning.
- B. This is the intuition to give up. Nothing limits the transfer: the two mode amplitudes are equal, so the beat between them goes all the way to zero for one pendulum and all the way to full for the other. Simulate it and watch the energy bars — the first pendulum stops completely, without any friction to stop it.
- C. An even split would need something to dissipate the difference and hold the state there, and the model is perfectly conservative. The motion is periodic: it passes through the even split twice per exchange period and keeps going.
- D. It does stop — that part is right and is the surprise. But nothing is lost: there is no damping term anywhere in the model, and the total energy is constant throughout. Wait one more half exchange period and the first pendulum is swinging as widely as when you released it.
A coupled pair has normal-mode angular frequencies omega_s = 4.000 rad/s and omega_a = 4.195 rad/s. How long does it take, in seconds, for the energy to pass from one oscillator to the other and return?
Check your answer
Correct answer: 32.2 (± 1.0)
T_ex = 2 pi / (omega_a - omega_s) = 2 pi / 0.195 = 32.2 s. Only the splitting matters; the mode frequencies themselves cancel out entirely, which is why timing a slow exchange is a way to measure a frequency difference far too small to resolve on a spectrum. Full transfer happens at half this, about 16 s.
For two identical coupled oscillators, how many ways are there to start them from rest so that no energy is ever exchanged between them, and what are those starts?
A. Two — displace both equally in the same direction, or displace both equally in opposite directions. B. One — displace both equally in the same direction, so the coupling spring does nothing. C. Infinitely many — any start with x1 = x2 in some fixed ratio will do. D. None — any start at all will eventually exchange energy, given enough time.
Check your answer
Correct answer: A — Two — displace both equally in the same direction, or displace both equally in opposite directions.
- A. Right, and the count is not a coincidence: two masses give two normal modes. In the first the coupling spring never changes length, so the pair swings at the uncoupled frequency omega0. In the second it is worked hardest, so the frequency is higher, sqrt((k + 2 k_c)/m). Every other start is a mixture of these two, and mixtures slosh.
- B. That is one of them, and it is the easier one to guess. But the opposite start — equal displacements in opposite directions — also oscillates at a single frequency and never trades. It is the mode in which the coupling spring works hardest, which is why its frequency is the higher of the two.
- C. Only two ratios work, +1 and -1. Try any other in simulation and the energy sloshes. The reason is that the determinant condition has exactly two roots, and each root fixes its shape completely.
- D. The two special starts genuinely never exchange, for ever. That is what makes them worth a name, and finding them is the point of the module: they are the motions in which the coupled system behaves like a single simple oscillator.
Two masses m sit on a frictionless table. Mass 1 is joined to a left wall by a spring of stiffness k1, mass 2 to a right wall by a spring k2, and the two masses to each other by a spring k_c. Write the equations of motion and assemble the matrices M and K. Then say how you would recognise, from K alone, whether the two oscillators are coupled at all.
Check your answer
Newton for each mass. Mass 1 feels its own wall spring, -k1 x1, and the coupling spring, which responds to the change in separation, -k_c (x1 - x2). Mass 2 likewise, with -k2 x2 and -k_c (x2 - x1). So

m x1’’ = -k1 x1 - k_c (x1 - x2), m x2’’ = -k2 x2 - k_c (x2 - x1).

Collecting into M x’’ + K x = 0 gives M = m * identity and

K = [[k1 + k_c, -k_c], [-k_c, k2 + k_c]].

Two things are worth noticing. The coupling appears on the diagonal as well as off it — each mass is held a little more stiffly than its own wall spring alone would hold it, because the coupling spring also resists its motion. And K is symmetric, which is not a coincidence: it follows from the potential energy being a single scalar function, and it is what guarantees real frequencies and non-sloshing modes. Coupling is visible in K as the off-diagonal entries. If K is diagonal the equations separate completely and the masses never influence each other. Note that this is a statement about the coordinates as much as about the physics: the whole point of normal coordinates is that they are the ones in which K is diagonal.


5. Two identical oscillators with natural frequency omega0 = 10.0 rad/s are weakly coupled with k_c/k = 0.040. Using the weak-coupling estimate, what is the mode splitting omega_a - omega_s, in rad/s?

:::{dropdown} Check your answer
**Correct answer:** 0.4 (± 0.04)

Delta omega approx omega0 k_c / k = 10.0 * 0.040 = 0.40 rad/s. The exact value is slightly smaller, 0.398, because sqrt(1 + 2 k_c/k) is a little below 1 + k_c/k; at k_c/k = 0.04 that is a 0.5% correction. The exchange time is 2 pi / 0.40 = 16 s, about 25 oscillation periods, so the pair swings many clean times while its amplitude drifts — the signature of weak coupling.

Show that the normal coordinates q_plus = (x1 + x2)/sqrt(2) and q_minus = (x1 - x2)/sqrt(2) turn the two coupled equations of motion into two independent ones. Then explain what has actually been achieved, given that no information was added or removed.
Check your answer
Start from m x1’’ = -k x1 - k_c (x1 - x2) and m x2’’ = -k x2 - k_c (x2 - x1).

Add them. The coupling terms are -k_c (x1 - x2) and -k_c (x2 - x1), which are equal and opposite, so they cancel exactly: m (x1 + x2)‘’ = -k (x1 + x2), that is q_plus’’ + omega0^2 q_plus = 0. Subtract them. Now the coupling terms reinforce, contributing -2 k_c (x1 - x2), giving m (x1 - x2)‘’ = -(k + 2 k_c)(x1 - x2), that is q_minus’’ + omega_a^2 q_minus = 0 with omega_a^2 = (k + 2 k_c)/m. Each is a module-01 oscillator with no reference to the other. What has been achieved is not new information — q_plus and q_minus carry exactly what x1 and x2 carry, and the transformation is invertible — but a new description. The coupling was never a property of the physics; it was a property of the coordinates chosen to describe it, and a different choice removes it entirely. That is why the modes are worth the trouble. In the site picture you must solve two equations that refer to each other; in the mode picture you solve two you already know the answer to, then transform back. Module 07 does the same thing for N masses, where guessing the right combinations by inspection stops being possible and the machinery has to be built properly.

Exam-style problems for this module — the ones worth doing with a pen — are collected in the problem set.

Explain it in your own words¶

Answer in a few sentences each. No number will save you.

Explain to someone who has not done any of this why the first pendulum stops, given that there is no friction in the model at all. Where is its energy at that moment, and how do you know it is coming back?
Why does the symmetric mode oscillate at exactly the frequency of a single uncoupled pendulum? Answer without algebra — describe what the coupling spring is doing.
The exchange of energy between two pendulums and the beat you hear between two slightly mistuned guitar strings are the same phenomenon. Say in what sense, and say what plays the role of “the two strings” in the pendulum case.
What would you have to measure to decide whether a given pair is weakly or strongly coupled, and why is “the coupling spring feels stiff” not an answer?

Advanced: weak coupling, strong coupling, and the crossing that isn’t¶

Two regimes, one number. The splitting $\Delta\omega$ is the only thing that distinguishes coupled pairs, and comparing it with $\wnat$ sorts them into two worlds. When $\Delta\omega \ll \wnat$ — weak coupling — the exchange is slow compared with the oscillation itself, so a pendulum completes many clean swings while its amplitude drifts, and the handover is essentially complete. When $\Delta\omega$ approaches $\wnat$ , the two modes are so different in frequency that “an oscillation whose amplitude is slowly changing” stops being a useful description at all, and the transfer is partial: at $k_c = k$ , 7.4% of the energy never leaves the mass it started on.

The same distinction organises a great deal of physics under other names. In 52-quantum-optics an atom in a cavity is weakly coupled if it loses its excitation before it can trade it back, and strongly coupled if the trading wins — and the trading rate is the splitting between two states, exactly as here. The classical pendulums are not an analogy for that situation; they are the same mathematics with heavier objects.

Detuning, and the crossing that isn’t. Break the symmetry. Stiffen the second mass’s anchor spring alone, $k \to k(1 + \delta)$ , and sweep $\delta$ through zero. Uncoupled, the two frequencies would cross cleanly at $\delta = 0$ — one rising, one flat. Coupled, they do not. They approach, turn away from each other, and separate again, and at no value of $\delta$ are they ever equal.

The gap is narrowest exactly where the uncoupled frequencies would have met, and its minimum value is set by the coupling alone: at $k_c/k = 0.1$ the closest approach is 0.38 rad/s and never smaller, however carefully $\delta$ is tuned. This is an avoided crossing, and once you have seen one you will see it everywhere — in molecular energy levels, in the band structure of 54-photonic-crystals, and in a pair of optical resonators tuned through each other in 44-resonators. Two things that are coupled cannot be made degenerate merely by tuning them.

Where the linear model gives out. Huygens’ clocks did not do what the pendulums above do. His two clocks settled into permanent antiphase lock and stayed there, which no conservative linear system can do: this model’s motion is periodic and always returns. Real clocks are driven by escapements and lose energy to friction, and a driven, damped, weakly nonlinear pair can settle onto a stable synchronised state and stay on it. That belongs to the theory of coupled self-sustained oscillators, and it is a genuinely different subject — worth knowing that the famous observation at the top of this page is not, strictly, the phenomenon derived below.