4.2 The Pendulum and Its Energy

A pendulum — a mass on a rigid rod of length $L$ — obeys $\theta'' + (g/L)\sin\theta = 0$ . For small swings $\sin\theta \approx \theta$ , and this is Chapter 2's spring: oscillation at the natural frequency $\sqrt{g/L}$ . But a pendulum is not a spring. Push it hard enough and it goes over the top, something no linear equation ever does, so this section keeps the sine. Measuring time in units of $\sqrt{L/g}$ scrubs out the constants, leaving

$x'' + \sin x = 0,$

or, as a system, $x' = y$ , $y' = -\sin x$ , where $x$ is the angle from straight down and $y$ the angular velocity.

The system rests wherever $y = 0$ and $\sin x = 0$ : the points $(n\pi, 0)$ . At even multiples of $\pi$ the pendulum hangs straight down; at odd multiples it balances upside down. Near the bottom, $\sin x \approx x$ gives the center $x'' + x = 0$ — closed loops, dimensionless period $2\pi$ . Near the top, setting $u = x - \pi$ gives $u'' = -\sin(u + \pi) = \sin u \approx u$ : precisely the saddle in disguise from the last section, eigenvalues $\pm 1$ . So the portrait must alternate — center, saddle, center, saddle — forever along the $x$ -axis.

\begin{bmatrix}0 & 1 \\ -\cos\left(x\right) & 0\end{bmatrix}

The Jacobian of the pendulum system ✓ Computed · mojocas 0.1.0 ✓ Agrees with the text

The linearization of $x' = y$ , $y' = -\sin x$ at any point, computed from the system itself. The alternation is visible in one entry: $-\cos x$ is $-1$ at the hanging equilibria $x = 0, \pm 2\pi$ and $+1$ at the inverted ones $x = \pm\pi$ — the sign flip that turns each center into a saddle and back.

-1 + r^{2}

Characteristic polynomial at the inverted equilibrium ✓ Computed · mojocas 0.1.0 ✓ Agrees with the text

At $x = \pi$ the Jacobian is $\begin{bmatrix}0 & 1\\ 1 & 0\end{bmatrix}$ , so $\det(J - \lambda I)$ is the polynomial shown — whose roots are the eigenvalues $\pm 1$ named above, one positive and one negative, which is exactly what a saddle is. (The variable $r$ stands for $\lambda$ .)

What no linearization can say is how these local pictures knit together, and for that the pendulum volunteers a global gift. Multiply $x'' + \sin x = 0$ by $x'$ and recognize both terms as derivatives: $\frac{d}{dt}\left( \tfrac{1}{2} (x')^2 - \cos x \right) = 0$ . The energy

$H(x, y) = \frac{y^2}{2} - \cos x$

— kinetic plus potential — is constant along every solution. That is a confession of imprisonment: each solution is confined for all time to a single level curve of $H$ . To draw the complete phase portrait of an unsolvable nonlinear equation, no solving is required. Draw the contour map of $H$ .

A flat map of sky-blue level curves of the pendulum's energy, on a grid of one-unit squares, with the angle running horizontally and the angular velocity vertically. Nested closed ovals ring the origin, and the same ovals, cut off by the left and right edges, ring the points at plus and minus 2 pi on the horizontal axis. An amber curve crosses itself at the balance points at plus and minus pi, forming a cat's-eye that stands 2 units above and below the origin: the separatrix. Outside it, open wavy curves run left to right above and below. Each curve is one energy level of the pendulum, so each is one possible motion: back-and-forth swings inside the eye, over-the-top spinning outside. — Twelve levels of the energy $H(x, y) = y^2/2 - \cos x$ , drawn as sky-blue curves on $-6.5 \le x \le 6.5$ , $-3 \le y \le 3$ and viewed face-on, with $x$ horizontal, $y$ vertical and grid squares one unit — the complete phase portrait of the pendulum. Closed ovals ring the resting equilibrium at the origin, and the same ovals, cut off by the window's edges, ring $x = \pm 2\pi$ ; the amber self-crossing curve through $(\pm\pi, 0)$ is the separatrix, the level $H = 1$ ; the open wavy curves above and below it are over-the-top whirls.

Explore

Observe. Three families of curves, three kinds of motion. The nested ovals around $(0, 0)$ and, cut off at the edges, $(\pm 2\pi, 0)$ are back-and-forth swings; the open curves running the full width of the window are whirls, over the top in the same direction forever; and the amber cat's-eye curve crossing itself at $(\pm\pi, 0) \approx (\pm 3.14, 0)$ — the separatrix — is the frontier between them.
Predict. No arrows are drawn. Use $x' = y$ : in the upper half-plane $x$ must increase. Which way do the ovals circulate, and which way do the upper whirl curves run? Decide, then check that the lower half-plane tells a consistent story.
Verify. The separatrix carries the energy of the balanced-upright state, $H(\pm\pi, 0) = 1$ . Setting $x = 0$ in $y^2/2 - \cos x = 1$ gives $y = 2$ : the amber cat's-eye should stand exactly $2$ grid units above and below the origin. Check it against the grid.
Observe. The whirl curves undulate: farthest from the axis over $x = 0$ and $x = \pm 2\pi$ , closest over $x = \pm\pi$ . A whirling pendulum is fastest through the bottom and slowest over the top — readable straight off the map.
Predict, then verify. An oval reaching maximum angle $x_{\max}$ has $H = -\cos x_{\max}$ , so it crosses the $y$ -axis at height $y = \sqrt{2(1 - \cos x_{\max})} = 2\sin(x_{\max}/2)$ . For the oval reaching out near $x_{\max} \approx \pi/2$ , predict the height of its top — about $1.4$ — then check.

The Lift

Why should the level curves of a function be the orbits of a system? The most convincing answer is to stop flattening. $H$ is a function of two variables, so it is a landscape — a surface $z = H(x, y)$ standing over the phase plane — and a level curve is a horizontal slice of that landscape, dropped to the floor. The second figure performs the lift.

A half-transparent rolling surface shaped like an egg carton stretched along one direction - bowls over the origin and over the points plus and minus 2 pi along the angle axis, dipping to height minus 1, saddle passes of height 1 between them, walls climbing toward the front and back - floating above a flat floor at height minus 1.5, just below the deepest bowl. The floor carries the same sky-blue level curves as the contour map, with an amber curve crossing itself beneath each pass: the separatrix. Every floor curve is a horizontal slice of the surface dropped straight down, so the pendulum's phase portrait is revealed as the surface's contour map. — The energy landscape $z = H(x, y) = y^2/2 - \cos x$ , half-transparent, floating above its own twelve-level contour map, drawn on a floor at $z = -1.5$ just below the deepest bowls (the lowest value of $H$ is $-1$ ). Bowls sit over the resting equilibria at $x = 0$ and $x = \pm 2\pi$ ; mountain passes of height $1$ sit over the saddle points $(\pm\pi, 0)$ , and the amber floor curve beneath them is the separatrix; every orbit on the floor is a horizontal slice of the surface overhead.

Observe. The surface is an egg carton stretched along the $x$ -direction — bowls at the hanging equilibria, passes between them — with walls climbing like $y^2/2$ toward the front and back. Orbit the camera until you can see one floor curve and the surface slice directly above it at the same time.
Predict, then verify. A horizontal slice below pass height ( $-1 < H < 1$ ) is trapped inside one bowl; dropped to the floor it is a closed ring around a well — a swing. A slice above every pass ( $H > 1$ ) clears the whole range and runs unbroken from end to end — a whirl. Match each family on the floor to its slice height on the surface.
Observe. The slice at exactly pass height, $H = 1$ , pinches onto the passes themselves: dropped down, it is the amber separatrix, crossing itself precisely at the saddle points. The frontier between swinging and whirling is the altitude of a mountain pass.

Connect

One picture now settles questions that formulas struggle with. The bottom equilibrium is the floor of a bowl, and nearby slices are small closed rings: a pendulum nudged from rest swings near rest forever — stable, though not attracting, since nothing here dissipates. The top equilibrium is a mountain pass, terrain falling away on two sides, the slice through it crossing itself: the saddle instability of the last section, now with a topographic reason. Even timing is visible: ovals deep in a bowl are traversed with period near $2\pi$ , ovals near the separatrix take longer and longer, and the separatrix itself takes forever — released ever closer to upside down, the pendulum hangs ever longer near the top before committing.

The landscape also explains what conservation costs. Add a little friction, $x'' + c\,x' + \sin x = 0$ , and the energy obeys $\frac{d}{dt} H = -c\, y^2 \le 0$ : the state can no longer hold its altitude and slides downhill across the level sets, every swing and every whirl eventually spiraling into some well. The contour map stops being the portrait, but the landscape still governs — now as terrain being descended. That is Chapter 2's inward spiral, seen from above.

Example 1 Over the top

How fast must the pendulum move through its lowest point to go over the top?

Show solution

Launching from the bottom puts the state at $(0, y_0)$ , with energy $H = y_0^2/2 - \cos 0 = y_0^2/2 - 1$ . Clearing the top requires at least the energy of arriving there at rest, $H(\pm\pi, 0) = 1$ . So $y_0^2/2 - 1 \ge 1$ , that is, $y_0 \ge 2$ . The threshold case $y_0 = 2$ launches exactly onto the separatrix — the pendulum creeps toward the balance point forever without arriving — and any $y_0 > 2$ lands on a whirl curve and goes over. Restoring units, the critical angular speed is $2\sqrt{g/L}$ ; for a one-meter pendulum with $g = 9.8$ , about $6.3$ radians per second — almost exactly one full revolution per second. Check the threshold on the map: the separatrix meets the $y$ -axis precisely at height $2$ .

Crossing the Threshold

The contour map is an argument from conservation: orbits must lie on level curves, so drawing the levels draws the orbits. It is worth watching the same picture arrive by a completely different route — integrating the system forward in time, step by step, with no mention of energy at all. The figure below launches the pendulum from the bottom at three speeds, one either side of the threshold $y_0 = 2$ and one exactly on it, and integrates.

Four computed trajectories drawn over a grey copy of the pendulum's energy contour map, all launched from marked points on the vertical axis. From the lowest mark, at height 1, a red closed oval traces round and round the central well, reaching out to about 1.05 on either side - a swing. From the mark at height 2 an orange arch, traced both forward and backward in time, flattens toward the balance points near plus and minus 3.14 on the horizontal axis without reaching either - the upper half of the separatrix. From the highest mark, at about 2.65, a blue wavy curve runs the full width of the window above the separatrix - a whirl - and its mirror twin, launched downward, runs the full width below. Each computed curve lies along one of the grey contour lines beneath it. — Three launch speeds from the hanging position, integrated numerically over the grey energy contour map, with $x$ horizontal and $y$ vertical. Red starts at $y_0 = 1$ and closes into a swing; orange starts at exactly $y_0 = 2$ and, traced forward and backward in time, creeps toward the balance points at $x = \pm\pi$ ; blue starts at $y_0 = \sqrt{7} \approx 2.65$ and whirls over the top, with a mirror launch downward. The three launch energies, $H = -\tfrac{1}{2}$ , $1$ and $\tfrac{5}{2}$ , are among the map's levels.

Observe. Follow the red trajectory. It closes on itself — the pendulum returns to its starting state and repeats forever. Which family of the contour map does it belong to?
Predict. The orange launch is exactly at the threshold energy. Before looking closely: does it reach the balance point at $x = \pi$ , and if it does, what happens next?
Verify. Trace the orange curve toward $x \approx 3.14$ . It flattens and slows without arriving, because arriving would mean sitting at an equilibrium — and a solution that reaches an equilibrium in finite time would have to leave it again, which no solution of an autonomous system does.
Observe. The blue curve never returns to the axis: it runs off the right edge, having gone over the top. Its downward-launched twin runs off to the left. Compare their undulation with the map's whirl curves.
Predict, then verify. Each computed curve should lie exactly on top of one grey contour line, because energy is conserved along solutions. Check one, and say what a visible drift off the contour would have told you about the integrator rather than about the pendulum.

An original work of XYZ Homework, built around interactive XYZ 3D figures, following the standard first-course sequence (first-order equations through systems and the phase plane). Distinct from the LibreTexts-sourced Differential Equations for Engineers (Lebl) edition, which has its own attribution. License: CC-BY-NC-SA-4.0.