Anharmonic Pendulum and Amplitude Dependence

Legend has it that a bored Galileo, timing a swinging lamp in the Pisa cathedral against his own pulse, discovered that every swing took the same time whether the arc was wide or narrow. That claim — isochronism, the amplitude-independence of the period — became the founding principle of the pendulum clock. It is also, strictly speaking, false. The restoring torque on a pendulum goes as $\sin\theta$ , not $\theta$ , and the two agree only for small angles. Push the amplitude up and the period stretches: by one part in a thousand at $10°$ , by one percent at $23°$ , by eighteen percent at $90°$ — and without bound as the swing approaches the inverted position. Clockmakers knew this "circular error" a lifetime before anyone could compute it. Computing it is the business of this article, and the payoff is bigger than pendulums. Because the system conserves energy, the period reduces to a single integral over a quarter swing — and that integral is the complete elliptic integral $K(k)$ , with the amplitude hiding inside the modulus $k=\sin(\theta_0/2)$ . One dial, running from $0$ to $1$ , carries you from Galileo's perfect isochronism to the logarithmic blow-up at the separatrix, where swinging gives way to going over the top. The waveform itself stops being a sine and becomes a Jacobi elliptic function, sprouting odd harmonics as it flattens. Every step of the method — energy first integral, quarter-period quadrature, frequency expanded in powers of amplitude — works for any oscillator beyond the parabolic approximation. The pendulum is simply the cleanest place to learn nonlinearity honestly.

Key takeaways

The pendulum equation $\ddot{\theta}+\omega_0^2\sin\theta=0$ is nonlinear; only for small $\theta$ does $\sin\theta\approx\theta$ reduce it to simple harmonic motion with an amplitude-independent period.
Energy conservation turns the period into a quadrature, and the quadrature is exact: $T=\frac{4}{\omega_0}K(k)$ with modulus $k=\sin(\theta_0/2)$ — the complete elliptic integral of the first kind.
The period grows monotonically with amplitude: $T\approx T_0\left(1+\tfrac{1}{16}\theta_0^2+\tfrac{11}{3072}\theta_0^4+\cdots\right)$ . A $23°$ swing already runs $1\%$ slow; $90°$ runs $18\%$ slow.
As $\theta_0\to\pi$ the bob loiters near the inverted equilibrium and the period diverges — but only logarithmically: even at $179°$ it is just $3.9\,T_0$ .
The waveform is the Jacobi elliptic function $\sin(\theta/2)=k\,\mathrm{sn}(\omega_0 t,k)$ , whose flattened peaks carry odd harmonics ( $3\omega, 5\omega,\dots$ ) of relative size $\sim\theta_0^2/192$ .
The same machinery — first integral, quarter-period integral, amplitude-dependent frequency — governs every conservative one-dimensional anharmonic oscillator, from clock escapements to transmon qubits.
Key idea: for a conservative 1D system the period is a quadrature; for the pendulum that quadrature is the complete elliptic integral $T=\frac{4}{\omega_0}K(\sin\frac{\theta_0}{2})$ , which makes the period an explicit, monotonically increasing function of amplitude that diverges at the separatrix $\theta_0=\pi$ .

Where this lives today

The most consequential anharmonic pendulum of the 2020s is a few hundred nanometres across and has no bob. A superconducting transmon qubit is electrically a Josephson junction shunted by a capacitor, and its phase variable obeys exactly equation (1) — the Josephson cosine potential $-E_J\cos\varphi$ is the pendulum well. That is not a loose analogy; it is the design principle. A perfectly harmonic circuit would have equally spaced quantum levels, and a microwave pulse tuned to the $0\to1$ transition would climb the whole ladder indiscriminately. The pendulum's anharmonicity — the same physics as equation (7) — spaces the levels unequally, typically by an anharmonicity of $200$ – $300\ \mathrm{MHz}$ below a $4$ – $6\ \mathrm{GHz}$ transition, so the bottom two levels can be addressed as a clean two-level qubit. Every gate fired on today's superconducting quantum processors leans on the pendulum's refusal to be isochronous. Driven harder, the same junctions reproduce this article's other landmarks: past the critical current the phase crosses the separatrix and "rotates," which is precisely the junction switching into its voltage state. The classical story is equally alive. Precision pendulum gravimetry always needed the finite-amplitude correction of equation (6) subtracted before quoting $g$ ; modern MEMS resonators are engineered into or out of the Duffing regime on purpose — amplitude-frequency pulling is a bug in gyroscope scale factors and a feature in bifurcation-point amplifiers that sense attonewton forces from the fold of a bent resonance curve.

The physical picture

Picture the potential energy $U(\theta)=-mg\ell\cos\theta$ . Near the bottom it is a parabola, and a parabola means a perfect spring: restoring torque proportional to displacement, period blind to amplitude. But climb the well and it flares out wider than a parabola — the true torque $mg\ell\sin\theta$ falls ever further short of the linear estimate $mg\ell\,\theta$ , and a weaker-than-linear pull means a lazier return. The higher you swing, the more of each cycle you spend in the flared, weak-torque region near the turning points. That is the whole physics of the anharmonic pendulum; everything else is making it quantitative.

Figure

The exact period versus amplitude — from isochronism to the separatrix divergence

The single most useful mental device is the modulus dial. Define $k=\sin(\theta_0/2)$ . At $k=0$ the swing is infinitesimal, the elliptic integral collapses to $K=\pi/2$ , and Galileo is exactly right. At $k=1$ the amplitude is $180°$ , the bob creeps asymptotically toward balancing upside-down, and $K\to\infty$ . Everything in between — clocks, playground swings, ship roll — is one smooth function $K(k)$ interpolating between those poles. Learn to read amplitude through $k$ and every formula in this article becomes one line. The waveform tells the same story in the time domain. Because the well is wider than parabolic, the bob lingers near the turning points and rushes through the bottom. The peaks of $\theta(t)$ flatten; the zero crossings steepen. A flattened sine is no longer a sine:

Figure

A 170-degree pendulum swing against a sine of the same period — flattened tops, steepened crossings

Finally, zoom out to the phase plane, where the whole family of motions is visible at once. Small energies trace nested ovals around the bottom of the well — librations. Crank the energy past $E=mg\ell$ (measured from the bottom, exactly enough to reach the top) and the pendulum stops turning back: it circulates over the top forever, and the phase curve runs unbounded in $\theta$ . The boundary between the two families is the separatrix, the infinite-period orbit that approaches the inverted equilibrium but never arrives.

Figure

Phase portrait of the pendulum — nested librations, the separatrix, and over-the-top rotation

From restoring force to elliptic integral

Write Newton's second law for a point mass $m$ on a massless rigid rod of length $\ell$ : the only torque about the pivot is gravity's, $-mg\ell\sin\theta$ , and the moment of inertia is $m\ell^2$ . Dividing out $m\ell^2$ leaves an equation with a single parameter — and no linearization anywhere:

\ddot{\theta}+\omega_0^2\sin\theta=0,\qquad \omega_0^2=\frac{g}{\ell} \qquad (1)

where: $\theta$ — angular displacement from the downward vertical; $\omega_0=\sqrt{g/\ell}$ — the small-amplitude natural frequency; $g$ — gravitational acceleration; $\ell$ — pendulum length. Note what $\omega_0$ is and is not: it is a fixed parameter built from $g$ and $\ell$ , but it is the actual oscillation frequency only in the $\theta\to0$ limit. The whole subject lives in the gap between $\omega_0$ and the true frequency. To see the anharmonicity explicitly, expand the potential energy about the bottom of the well. The Taylor series of the cosine delivers the parabola first, then the correction that drives everything in this article:

U(\theta)=mg\ell(1-\cos\theta)=\tfrac12 mg\ell\,\theta^2-\tfrac{1}{24}mg\ell\,\theta^4+\cdots \qquad (2)

where: $U$ — gravitational potential energy, zero at the bottom; the quadratic term alone would give simple harmonic motion; the negative quartic term widens the well and weakens the restoring force. In the language of nonlinear oscillators, equation (1) truncated at cubic order, $\ddot\theta+\omega_0^2(\theta-\theta^3/6)=0$ , is a softening Duffing oscillator — the pendulum is the canonical physical realization of that textbook abstraction, and the sign of the $\theta^3$ term is why the period gets longer, never shorter.

Derivation AThe energy route to the exact period

We are deriving the exact period of the pendulum, equation (5), valid at any amplitude, starting from nothing but energy conservation. The strategy is the universal one for conservative 1D systems: use the energy first integral to express $\dot\theta$ as a function of $\theta$ , separate variables, and integrate over a quarter swing — the period becomes a quadrature, a single definite integral. The move that unlocks the pendulum specifically is the half-angle identity, which converts a difference of cosines into a difference of squared sines and puts the integral into the classical Legendre normal form. Watch two things: the constant of integration must be evaluated at the turning point, and the substitution in the final step must map the quarter swing onto exactly $\varphi\in[0,\pi/2]$ — get either wrong and the elliptic integral acquires spurious factors.

Step 1. Multiply equation (1) by $\dot\theta$ and recognize both terms as exact time derivatives: $\dot\theta\ddot\theta+\omega_0^2\sin\theta\,\dot\theta=\frac{\mathrm{d}}{\mathrm{d}t}\left(\tfrac12\dot\theta^2-\omega_0^2\cos\theta\right)=0$ . Why this works: multiplying by $\dot\theta$ is the mechanical way of dotting force with velocity — it converts the equation of motion into the statement that energy flows nowhere.

Step 2. Evaluate the conserved quantity at the turning point $\theta=\theta_0$ , where $\dot\theta=0$ :

\tfrac12\dot{\theta}^2-\omega_0^2\cos\theta=-\omega_0^2\cos\theta_0 \qquad (3)

Why the turning point: the amplitude is the one place where we know the velocity exactly — it vanishes — so the constant is fixed by a single measurable number, $\theta_0$ .

Step 3. Solve for the angular velocity and apply the half-angle identity $\cos\alpha=1-2\sin^2(\alpha/2)$ to both cosines:

\dot\theta=\pm\,2\omega_0\sqrt{\sin^2\frac{\theta_0}{2}-\sin^2\frac{\theta}{2}} \qquad (4)

Why half-angles: $\cos\theta-\cos\theta_0$ is awkward under a square root, but $2(\sin^2\frac{\theta_0}{2}-\sin^2\frac{\theta}{2})$ is a difference of squares — precisely the algebraic shape that elliptic integrals are built from. This identity is the pendulum's one indispensable trick.

Step 4. Separate variables and integrate over a quarter period — from the bottom $\theta=0$ to the turning point $\theta=\theta_0$ , taking the positive root: $\frac{T}{4}=\int_0^{\theta_0}\frac{\mathrm{d}\theta}{2\omega_0\sqrt{\sin^2(\theta_0/2)-\sin^2(\theta/2)}}$ . Why a quarter: the four legs of a full cycle — out, back, out the other side, back — take equal times by the symmetry $\theta\to-\theta$ of the potential.

Step 5. Substitute $\sin(\theta/2)=k\sin\varphi$ with modulus $k=\sin(\theta_0/2)$ . Differentiating gives $\tfrac12\cos(\theta/2)\,\mathrm{d}\theta=k\cos\varphi\,\mathrm{d}\varphi$ ; the radical becomes $\sqrt{k^2-k^2\sin^2\varphi}=k\cos\varphi$ , which cancels the $k\cos\varphi$ upstairs, while $\cos(\theta/2)=\sqrt{1-k^2\sin^2\varphi}$ migrates into the denominator. As $\theta$ runs $0\to\theta_0$ , $\varphi$ runs exactly $0\to\pi/2$ . Why this substitution: it is engineered so the turning-point singularity of the integrand is absorbed into a clean trigonometric boundary — the integral's endpoints become universal, and all amplitude dependence retreats into $k$ .

Step 6. What remains is the complete elliptic integral of the first kind:

T=\frac{4}{\omega_0}\,K\!\left(\sin\frac{\theta_0}{2}\right),\qquad K(k)\equiv\int_0^{\pi/2}\frac{\mathrm{d}\varphi}{\sqrt{1-k^2\sin^2\varphi}} \qquad (5)

where: $T$ — the exact period at amplitude $\theta_0$ ; $K(k)$ — a tabulated, monotonically increasing function with $K(0)=\pi/2$ and $K(k)\to\infty$ as $k\to1$ . Every property of the pendulum's period is now a property of one classical special function. Verified: at $k=0$ , $K=\pi/2$ gives $T=2\pi/\omega_0=T_0$ , the small-angle result; at $\theta_0=90°$ , $K(1/\sqrt2)=1.85407$ gives $T/T_0=2K/\pi=1.18034$ , matching the tabulated handbook value.

Which formula answers which question

you want	use	notes
the exact period at any amplitude	$T=\frac{4}{\omega_0}K(\sin\frac{\theta_0}{2})$ , eq. (5)	evaluate $K$ numerically or from tables
the period at moderate amplitude, fast	Borda series, eq. (6)	keep $\theta_0^2/16$ below $\sim60°$ ; add $\theta_0^4$ term to $\sim120°$
the actual oscillation frequency	$\omega(\theta_0)=\omega_0\,\pi/2K(k)$ , eq. (7)	softening: decreases with amplitude
the full waveform $\theta(t)$	$\sin(\theta/2)=k\,\mathrm{sn}(\omega_0t,k)$ , eq. (8)	Jacobi elliptic sine, quarter period $K/\omega_0$
harmonic content of the waveform	third-harmonic ratio $\approx\theta_0^2/192$ , eq. (9)	odd harmonics only
behavior near $\theta_0=\pi$	$T\approx\frac{4}{\omega_0}\ln\frac{8}{\pi-\theta_0}$ , eq. (10)	logarithmic divergence — use this, not the series

The period grows with amplitude

Equation (5) is exact but opaque: nobody's intuition speaks elliptic integrals natively. For amplitudes short of a right angle, the working formula is the series expansion — historically associated with Jean-Charles de Borda's precision pendulum measurements of gravity in the 1790s, which needed exactly this correction:

\frac{T}{T_0}=\frac{2}{\pi}K(k)=1+\frac{1}{16}\theta_0^2+\frac{11}{3072}\theta_0^4+\frac{173}{737280}\theta_0^6+\cdots \qquad (6)

where: $T_0=2\pi/\omega_0=2\pi\sqrt{\ell/g}$ — the small-angle period; $\theta_0$ — the amplitude in radians. The leading rule of thumb is worth memorizing: the fractional period increase is one-sixteenth of the squared amplitude. At $\theta_0=0.4\ \mathrm{rad}$ ( $23°$ ) that is $0.16/16=1.0\%$ ; at $10°$ it is $0.19\%$ ; at $90°$ the series (through $\theta_0^4$ ) gives $17.9\%$ against the exact $18.03\%$ .

Go deeper · Derivation B · 1 minThe Borda series and the one-sixteenth rule

Derivation BThe Borda series and the one-sixteenth rule

The goal is the amplitude expansion (6) — in particular its famous leading coefficient $1/16$ — extracted from the exact result (5) with nothing heavier than the binomial series. The strategy has two independent expansions that must not be conflated: first expand $K(k)$ in powers of the modulus $k$ , then expand the modulus $k=\sin(\theta_0/2)$ in powers of the amplitude $\theta_0$ . The watch-out is at second order and beyond: the $\theta_0^4$ coefficient receives contributions from both expansions, so truncating one of them too early silently corrupts the series even though the leading term survives.

Step 1. Binomial-expand the integrand of $K$ in the small quantity $k^2\sin^2\varphi<1$ : $\left(1-k^2\sin^2\varphi\right)^{-1/2}=1+\tfrac12k^2\sin^2\varphi+\tfrac38k^4\sin^4\varphi+\cdots$ Why this converges: for libration $k=\sin(\theta_0/2)<1$ , so the expansion parameter is genuinely small everywhere on the integration range.

Step 2. Integrate term by term using the Wallis integrals $\int_0^{\pi/2}\sin^2\varphi\,\mathrm{d}\varphi=\pi/4$ and $\int_0^{\pi/2}\sin^4\varphi\,\mathrm{d}\varphi=3\pi/16$ :

K(k)=\frac{\pi}{2}\left(1+\frac{1}{4}k^2+\frac{9}{64}k^4+\cdots\right)

Why the coefficients look familiar: the general term is $\left[\frac{(2n)!}{2^{2n}(n!)^2}\right]^2 k^{2n}$ — the squared central binomial coefficients that fingerprint $K$ in any formula collection.

Step 3. Expand the modulus in the amplitude: $k^2=\sin^2\frac{\theta_0}{2}=\frac{\theta_0^2}{4}-\frac{\theta_0^4}{48}+\cdots$ , hence $k^4=\frac{\theta_0^4}{16}+\cdots$ Why this second expansion is mandatory: the physics dial is $\theta_0$ , not $k$ ; quoting a series "in amplitude" while secretly meaning the modulus is a classic source of wrong fourth-order coefficients.

Step 4. Substitute and collect powers of $\theta_0$ . The $\theta_0^2$ term comes only from $\frac14k^2$ : $\frac14\cdot\frac{\theta_0^2}{4}=\frac{\theta_0^2}{16}$ . The $\theta_0^4$ term collects $\frac14\cdot(-\frac{\theta_0^4}{48})+\frac{9}{64}\cdot\frac{\theta_0^4}{16}=\left(-\frac{1}{192}+\frac{9}{1024}\right)\theta_0^4=\frac{11}{3072}\theta_0^4$ :

\frac{T}{T_0}=1+\frac{\theta_0^2}{16}+\frac{11}{3072}\theta_0^4+\cdots

Why the cancellation pattern matters: the two fourth-order contributions carry opposite signs, so both expansions must be kept to consistent order — exactly the trap flagged in the goal paragraph. *Verified: at $\theta_0=0.4$ rad the series gives $T/T_0=1.010093$ ; the exact $2K(\sin 0.2)/\pi=1.010093$ — agreement to six decimal places, the next term being $\sim10^{-6}$ .*

Since frequency is $2\pi/T$ , the softening reads directly off the series. This one-line corollary is the bridge to driven, nonlinear, and quantum oscillators:

\omega(\theta_0)=\omega_0\,\frac{\pi}{2K(k)}\approx\omega_0\left(1-\frac{\theta_0^2}{16}\right) \qquad (7)

where: $\omega(\theta_0)$ — the true oscillation frequency at amplitude $\theta_0$ . The frequency falls as the amplitude grows — a softening anharmonicity. Drive a pendulum near resonance and this is the term that bends the response curve toward lower frequencies, producing the hysteresis and amplitude jumps of nonlinear resonance: as the swing builds up, the oscillator detunes itself downward, away from the drive that created it.

The exact waveform

What does $\theta(t)$ actually look like when it is not a sine? The separation of variables that gave us the period, pushed one step further, gives the full time dependence — and introduces the elliptic generalization of the sine function:

\sin\frac{\theta(t)}{2}=k\,\mathrm{sn}(\omega_0 t,\,k) \qquad (8)

where: $\mathrm{sn}(u,k)$ — the Jacobi elliptic sine, a periodic function of $u$ with quarter period $K(k)$ that reduces exactly to $\sin u$ at $k=0$ ; time is measured from an equilibrium crossing. As $k$ grows, $\mathrm{sn}$ flattens at its extremes — the loitering near the turning points made analytic.

Go deeper · Derivation C · 2 minThe motion as a Jacobi elliptic function

Derivation CThe motion as a Jacobi elliptic function

We are deriving the exact waveform (8) by inverting the quadrature — the same integral that gave the period, now with a free upper limit. The strategy: separate variables in the velocity equation (4), change variables to $s=\sin(\theta/2)/k$ so the integral lands in a standard normal form, and then invert — recognizing that the resulting integral is, by definition, the inverse of the Jacobi function $\mathrm{sn}$ . Nothing here is harder than the period derivation; the only new idea is that a definite integral defines a function worth naming. The watch-out is bookkeeping: with time started at the bottom of the swing, the motion follows $\mathrm{sn}$ ; started at the turning point it would be the cousin function $\mathrm{cn}$ — mixing up the initial condition swaps the function.

Step 1. Separate variables in $\dot\theta=2\omega_0\sqrt{k^2-\sin^2(\theta/2)}$ (equation (4) with $k=\sin(\theta_0/2)$ ): $\omega_0\,\mathrm{d}t=\frac{\mathrm{d}\theta}{2\sqrt{k^2-\sin^2(\theta/2)}}$ . Why the positive root suffices: we follow the first quarter swing, bottom to turning point, where $\dot\theta>0$ ; the other legs follow by symmetry.

Step 2. Substitute $s=\sin(\theta/2)/k$ , so $\mathrm{d}s=\frac{1}{2k}\cos(\theta/2)\,\mathrm{d}\theta$ and $\cos(\theta/2)=\sqrt{1-k^2s^2}$ :

\omega_0\,\mathrm{d}t=\frac{\mathrm{d}s}{\sqrt{(1-s^2)(1-k^2s^2)}}

Why $s$ and not $\theta$ : the variable $s$ runs from 0 to 1 over the quarter swing regardless of amplitude, and the two-factor radical on the right is precisely the defining integrand of the incomplete elliptic integral of the first kind.

Step 3. Integrate from the equilibrium crossing ( $s=0$ at $t=0$ ) and invert. The integral $\int_0^s\frac{\mathrm{d}s'}{\sqrt{(1-s'^2)(1-k^2s'^2)}}$ defines $\mathrm{sn}^{-1}(s,k)$ , so $\omega_0 t=\mathrm{sn}^{-1}(s,k)$ , i.e. $s=\mathrm{sn}(\omega_0t,k)$ and $\sin\frac{\theta}{2}=k\,\mathrm{sn}(\omega_0 t,k)$ . Why inversion is legitimate: on the quarter swing the integral is strictly increasing in $s$ , so the inverse exists — the same logic by which $\arcsin$ inverts to $\sin$ , one modulus up the ladder.

Step 4. Check the clock: the quarter swing $s:0\to1$ takes $\omega_0 t=K(k)$ , so the full period is $T=4K(k)/\omega_0$ — the waveform derivation independently reproduces equation (5). Why this is a real check and not circularity: the period came from a definite integral, the waveform from an indefinite one; their agreement confirms the bookkeeping. *Verified: at $k\to0$ , $\mathrm{sn}\to\sin$ and (8) collapses to $\theta\approx\theta_0\sin\omega_0t$ , the SHM solution with the correct amplitude and phase convention.*

A flattened periodic waveform has a spectrum, and symmetry dictates its structure. Because the pendulum's potential is even, the motion obeys $\theta(t+T/2)=-\theta(t)$ — flip the sign after half a period and the trajectory repeats. Any Fourier component at an even multiple of the fundamental would violate that antisymmetry, so only odd harmonics survive. Perturbation theory on the softening Duffing equation (expand $\sin\theta\approx\theta-\theta^3/6$ , insert $\theta\approx\theta_0\cos\omega t$ , and collect the $\cos3\omega t$ that the cubic term generates) gives their leading size:

\frac{|\theta_{3\omega}|}{|\theta_{\omega}|}\;\approx\;\frac{\theta_0^2}{192} \qquad (9)

where: $\theta_{3\omega},\theta_\omega$ — the Fourier amplitudes at $3\omega$ and at the fundamental. The anharmonic contamination is quadratic in amplitude and numerically meek: at $\theta_0=1\ \mathrm{rad}$ ( $57°$ ) a direct numerical integration of equation (1) puts the third harmonic at $0.55\%$ of the fundamental, against $1/192=0.52\%$ from the leading-order formula — perturbation theory earning its keep even at a radian. If you strobe a real pendulum and Fourier-analyze the angle, this faint $3\omega$ line is the cleanest laboratory signature that you have left the harmonic world.

The separatrix

Every formula so far treats $\theta_0=\pi$ as a wall. Walk up to it. A pendulum released ever closer to the inverted position takes ever longer to complete a cycle — not because it moves slowly everywhere, but because it spends almost all its time creeping near the top, where the restoring torque $\propto\sin\theta$ dwindles to nothing. The divergence law is worth deriving because its form — logarithmic — explains why the wall is so hard to feel in practice:

T\;\approx\;\frac{4}{\omega_0}\,\ln\!\frac{8}{\pi-\theta_0}\qquad(\theta_0\to\pi) \qquad (10)

where: $\pi-\theta_0$ — the release angle's shortfall from fully inverted, in radians. Logarithms grow with legendary reluctance: even at $\theta_0=179°$ the period is only $3.9\,T_0$ .

Go deeper · Derivation D · 1 minLogarithmic divergence at the separatrix

Derivation DLogarithmic divergence at the separatrix

We are deriving the near-separatrix period law (10) from the standard large-modulus asymptotics of $K$ . The strategy: as $\theta_0\to\pi$ the modulus $k=\sin(\theta_0/2)\to1$ , where $K(k)$ diverges; the divergence is controlled by the complementary modulus $k'=\sqrt{1-k^2}$ , which for the pendulum has a beautifully simple geometric meaning. The watch-out is a widespread error in the literature of this very topic: carelessly "simplifying" $\frac{4}{\omega_0}\ln\frac{8}{\epsilon}$ into $\frac{2}{\omega_0}\ln\frac{16}{\epsilon}$ — the two are not equal (logarithms do not trade prefactors for arguments that way), and only the first matches the exact integral.

Step 1. Quote the standard asymptotic expansion of the complete elliptic integral near unit modulus: $K(k)\to\ln\frac{4}{k'}$ as $k'\to0$ , where $k'=\sqrt{1-k^2}$ . Why a logarithm appears: near $k=1$ the integrand of $K$ develops a $1/\cos\varphi$ singularity at the upper endpoint, and integrating $\sec\varphi$ is what manufactures logarithms.

Step 2. Compute the complementary modulus for the pendulum: $k'^2=1-\sin^2\frac{\theta_0}{2}=\cos^2\frac{\theta_0}{2}$ , so $k'=\cos\frac{\theta_0}{2}$ exactly. Why this is elegant: the complementary modulus is just the cosine of the half-amplitude — the modulus and its complement are the sine and cosine of one angle.

Step 3. Write $\theta_0=\pi-\epsilon$ for small shortfall $\epsilon$ : $k'=\cos\left(\frac{\pi}{2}-\frac{\epsilon}{2}\right)=\sin\frac{\epsilon}{2}\approx\frac{\epsilon}{2}$ . Why we may linearize: the log's argument only needs leading order in $\epsilon$ ; corrections shift $T$ by $O(\epsilon)$ , invisible against a divergence.

Step 4. Assemble the period: $T=\frac{4}{\omega_0}K\approx\frac{4}{\omega_0}\ln\frac{4}{\epsilon/2}=\frac{4}{\omega_0}\ln\frac{8}{\epsilon}$ , which is equation (10). Why the prefactor must stay 4: each of the four quarter-swings independently spends its logarithm of time near the top; halving the prefactor by fiat would claim the bob loiters only on two of them. *Verified: at $\theta_0=179°$ ( $\epsilon=0.01745$ rad), formula (10) gives $T/T_0=\frac{2}{\pi}\ln(8/\epsilon)=3.9010$ ; the exact elliptic integral gives $K(\sin 89.5°)=6.128$ and $T/T_0=2K/\pi=3.9011$ — the asymptotic form is accurate to four significant figures already one degree from the top.*

Past the wall, the physics changes branch rather than ending. With energy above $2mg\ell$ , $\dot\theta$ never vanishes: the bob circulates, $\theta$ advancing by $2\pi$ each finite period, described by the same elliptic functions with inverted modulus. On the phase portrait the two families — closed librations inside, running rotations outside — are walled off by the separatrix, the infinite-period orbit through the inverted equilibrium. Every swing pumped over the top and every Josephson junction driven past critical current crosses that wall.

Limiting cases

Case	Result	Meaning
$\theta_0\to0$	$T\to T_0=2\pi\sqrt{\ell/g}$ , waveform $\to$ sine	$k\to0$ , $K\to\pi/2$ , $\mathrm{sn}\to\sin$ : SHM and Galilean isochronism recovered exactly
$\theta_0=10°$	$T=1.0019\,T_0$	isochronism good to $0.19\%$ — why small-swing clocks work
$\theta_0=23°$ ( $0.4$ rad)	$T=1.010\,T_0$	the $1\%$ landmark; one-sixteenth rule: $\theta_0^2/16=0.01$
$\theta_0=90°$	$T=1.1803\,T_0$ exactly, from $K(1/\sqrt2)=1.85407$	the series through $\theta_0^4$ gives $1.179$ — still honest at a right angle
$\theta_0=179°$	$T=3.90\,T_0$	the divergence is only logarithmic; even near-inverted release is a few $T_0$
$\theta_0\to\pi$	$T\approx\frac{4}{\omega_0}\ln\frac{8}{\pi-\theta_0}\to\infty$	separatrix: the bob asymptotically approaches the inverted equilibrium
$E>2mg\ell$	rotation: $\theta$ advances $2\pi$ per period, no turning points	over-the-top branch, same elliptic functions with inverted modulus

Common trapA pendulum's period does not depend on its amplitude

That sentence is the single most repeated half-truth in introductory mechanics — true only in the $\theta_0\to0$ limit, and already wrong by $1\%$ at $23°$ . Three sharper traps hide behind it. First, the modulus convention: many tables and software libraries write $K(m)$ with the parameter $m=k^2=\sin^2(\theta_0/2)$ rather than the modulus $k$ (SciPy's `ellipk` takes $m$ , for instance); feeding one where the other belongs silently squares or unsquares the argument and wrecks the period. Second, the series domain: the Borda expansion (6) is a small-amplitude series — near $\theta_0=\pi$ no finite truncation works, and you must switch to the logarithmic law (10). Third, the reverse mistake: assuming the period diverges dramatically near the top. The divergence is logarithmic — at $179°$ the period is still under $4\,T_0$ — so an experiment that merely gets "close to inverted" will never see a spectacularly long period.

Worked examples

1 · A playground swing at 60 degrees (confidence builder)

A child on a swing of effective length $\ell=2.5\ \mathrm{m}$ swings with amplitude $\theta_0=60°$ . Find the small-angle period, the exact period, and the error a stopwatch-wielding parent makes by trusting the small-angle formula.

Step 1. Small-angle period: $T_0=2\pi\sqrt{\ell/g}=2\pi\sqrt{2.5/9.81}=3.17\ \mathrm{s}$ .

Step 2. Modulus: $k=\sin(30°)=0.500$ . From tables (or the AGM algorithm), $K(0.5)=1.6858$ .

Step 3. Exact period: $T=\frac{4}{\omega_0}K=T_0\cdot\frac{2K}{\pi}=3.17\times1.0732=\boxed{3.40\ \mathrm{s}}$ — the swing runs $\mathbf{7.3\%}$ slower than the small-angle formula predicts.

Step 4. Cross-check with the series: $\theta_0=\pi/3=1.047\ \mathrm{rad}$ , so $1+\frac{\theta_0^2}{16}+\frac{11\theta_0^4}{3072}=1+0.0685+0.0043=1.0728$ — within $4$ parts in $10^4$ of the exact factor. The one-sixteenth rule alone already captures $0.0685$ of the $0.0732$ .

2 · The circular error of a pendulum clock (exam level)

A seconds pendulum ( $T_0=2.000\ \mathrm{s}$ , one second per swing) in a wall clock normally runs at amplitude $\theta_0=4.0°$ . After a cleaning, the escapement delivers more impulse and the amplitude settles at $6.0°$ . How does the clock's daily rate change?

Step 1. The clock's gearing counts swings, so the timekeeping error is the fractional period excess $\delta(\theta_0)=T/T_0-1\approx\theta_0^2/16$ (radians!).

Step 2. At $4.0°=0.0698\ \mathrm{rad}$ : $\delta_1=(0.0698)^2/16=3.05\times10^{-4}$ . At $6.0°=0.1047\ \mathrm{rad}$ : $\delta_2=(0.1047)^2/16=6.85\times10^{-4}$ . (The exact elliptic values are $3.047\times10^{-4}$ and $6.858\times10^{-4}$ — at these amplitudes the one-sixteenth rule is essentially exact.)

Step 3. The change in rate is $\Delta\delta=\delta_2-\delta_1=3.81\times10^{-4}$ . Over a day: $86400\ \mathrm{s}\times3.81\times10^{-4}=\boxed{33\ \mathrm{s/day\ lost}}$ — each swing now takes longer, so the clock falls behind.

Step 4. Read the design lesson. The error scales as $\theta_0^2$ , so halving the amplitude quarters the circular error — which is why precision regulators swung tiny arcs of a degree or two, and why Huygens tried to cancel the effect entirely with cycloidal cheeks that make the effective pendulum isochronous. An amplitude stability of $\pm0.1°$ about $4°$ changes the rate by $2\theta_0\,\Delta\theta_0/16\approx1.5\times10^{-5}$ — about $1.3\ \mathrm{s/day}$ , which is why constancy of drive mattered more to clockmakers than smallness itself.

Origins · 1 minGeorg Duffing and how this was discovered

Georg Duffing

1861–1944 · engineer, Berlin

Georg Duffing was not a professor. A German industrial engineer preoccupied with vibrating machinery, he spent his evenings studying why real oscillators refuse to obey the linear textbook — and in 1918 published a slim monograph, Erzwungene Schwingungen bei veränderlicher Eigenfrequenz (Forced oscillations with variable natural frequency), that gave nonlinear vibration its canonical equation. The "Duffing oscillator," $\ddot x+\omega_0^2x+\beta x^3=F\cos\Omega t$ , is the pendulum's Taylor expansion with a drive attached: $\beta<0$ (the pendulum's case) softens the spring and bends resonance curves toward low frequency; $\beta>0$ hardens and bends them the other way. The bent curves overhang, giving two stable amplitudes at one drive frequency — the jumps and hysteresis engineers had seen in machines and could not explain. A century later his equation is the standard model of nonlinear resonance in MEMS gyroscopes, nanomechanical sensors, and superconducting circuits: an amateur's evening project that outlived nearly all the professional mechanics of its decade.

History

1602

— Galileo, in a letter to Guidobaldo del Monte, claims the pendulum's swings are isochronous at any amplitude — right enough to found horology, wrong in principle.

1673

— Christiaan Huygens publishes the Horologium Oscillatorium: he proves the circular pendulum is not isochronous and shows a cycloidal path would be, curing the circular error by geometry.

1825

— Adrien-Marie Legendre's Traité des fonctions elliptiques systematizes the integrals

K

and

E

, giving the pendulum's exact period its permanent home.

1829

— Carl Gustav Jacob Jacobi's Fundamenta Nova inverts the elliptic integrals into the functions

\mathrm{sn}

\mathrm{cn}

\mathrm{dn}

— the pendulum's exact waveform acquires a name.

1918

— Georg Duffing publishes his monograph on forced oscillations with variable natural frequency; the pendulum's cubic truncation becomes the master equation of nonlinear resonance.

Going further

References. Morin, Introduction to Classical Mechanics, ch. on oscillations (exact pendulum, elliptic-integral period) · Goldstein, Poole & Safko, Classical Mechanics 3rd ed., §6 (oscillations and the pendulum) · Landau & Lifshitz, Mechanics, Vol. 1, §11–12 (anharmonic oscillations) and §11 problems · Taylor, Classical Mechanics, ch. 5 (oscillations) and problems on the large-amplitude pendulum · Abramowitz & Stegun, Handbook of Mathematical Functions, ch. 17 (elliptic integrals and functions)

Provenance. A verifier should reproduce: (1) the energy first integral and quarter-period quadrature giving $T=\frac{4}{\omega_0}K(\sin\theta_0/2)$ (Morin, Taylor, Landau-Lifshitz §11); (2) the Borda series coefficients $1,\tfrac{1}{16},\tfrac{11}{3072},\tfrac{173}{737280}$ by expanding $K$ and the modulus (Abramowitz & Stegun ch.17, eq. 17.3.11); (3) the small-modulus expansion $K(k)=\tfrac{\pi}{2}(1+\tfrac14 k^2+\tfrac{9}{64}k^4+\cdots)$ and large-modulus form $K\sim\ln(4/k')$ ; (4) numeric checks: $K(1/\sqrt2)=1.8541$ so $T/T_0=1.1803$ at $90°$ ; $\theta_0=0.4$ rad gives $1.0\%$ ; (5) the Jacobi-sn solution and quarter-period $K/\omega_0$ (Goldstein §6, Landau-Lifshitz). All numbers are standard textbook/handbook values. Additional numbers on this page (the $3.90\,T_0$ period at $179°$ , the $33\ \mathrm{s/day}$ circular error, the $0.55\%$ third-harmonic content at $1$ rad, and all SVG curves) were recomputed directly from equations (5), (6), (9), (10) by AGM evaluation of $K$ and Runge-Kutta integration of equation (1).

Builds on: The Simple Pendulum · Simple Harmonic Motion · Energy in Simple Harmonic Motion · Small Oscillations About Equilibrium Leads to: The Physical Pendulum · Driven Oscillations · Resonance

Why does the period get longer with amplitude rather than shorter — the bob is moving faster, after all?

Both effects are real; the weak-torque effect wins. At larger amplitude the bob does gain speed through the bottom, but it must also travel farther, and — decisively — it enters territory near the turning points where the restoring torque $mg\ell\sin\theta$ falls increasingly short of the linear $mg\ell\,\theta$ . Out there the bob is barely being pushed home, so it dawdles. Quantitatively the well $U=mg\ell(1-\cos\theta)$ is wider than its parabolic approximation (the quartic correction in equation (2) is negative), and a wider well at the same curvature always means a longer period. In the extreme case the effect is total: released near $\theta_0=\pi$ , the bob hangs almost motionless near the top for most of its cycle.

At what amplitude does the small-angle formula actually start to matter in a real experiment?

Use the one-sixteenth rule: the fractional error is $\theta_0^2/16$ with $\theta_0$ in radians. A lab pendulum at $5°$ is off by $0.05\%$ — invisible with a stopwatch. At $10°$ it is $0.19\%$ , marginal. At $23°$ you cross the $1\%$ line, which a hundred timed swings resolve easily; at $60°$ the error is $7.3\%$ and unmissable; at $90°$ it is $18\%$ . So the folklore rule "keep it under ten degrees" is a $0.2\%$ -accuracy statement. If your timing precision is $\Delta T/T\sim10^{-3}$ , the isochronous approximation is defensible up to about $\theta_0\approx\sqrt{16\times10^{-3}}\approx0.13$ rad $\approx7°$ .

Where does the sin(θ0/2) — the half-angle — in the modulus come from?

From the half-angle identity, and it is less mysterious than it looks. The velocity-squared is proportional to $\cos\theta-\cos\theta_0$ , and the identity $\cos\alpha=1-2\sin^2(\alpha/2)$ converts that difference of cosines into $2[\sin^2(\theta_0/2)-\sin^2(\theta/2)]$ — a difference of squares, which is the algebraic mold elliptic integrals are cast from. The substitution $\sin(\theta/2)=k\sin\varphi$ then maps the swing onto a standard quarter-circle of $\varphi$ , and the amplitude survives only inside $k=\sin(\theta_0/2)$ . Geometrically, $k$ runs from $0$ to $1$ as the amplitude runs from $0$ to $\pi$ — the natural "how nonlinear are we" dial, saturating exactly at the separatrix.

My software's elliptic-integral function gives me the wrong period. What convention trap did I hit?

Almost certainly the modulus-versus-parameter trap. This article writes $K(k)$ with modulus $k=\sin(\theta_0/2)$ ; many references and most numerical libraries use the parameter $m=k^2=\sin^2(\theta_0/2)$ — SciPy's `ellipk(m)`, Mathematica's `EllipticK[m]`, and Abramowitz & Stegun all take $m$ . Feeding $k$ where $m$ is expected effectively takes a square root of the argument and underestimates the period; the reverse overestimates it. Sanity check before trusting any output: at $\theta_0=90°$ you must get $T/T_0=1.1803$ , i.e. $K=1.85407$ from $k=1/\sqrt2\approx0.7071$ or equivalently $m=0.5$ .

Does a pendulum released just below the top really take forever to swing?

No — and the slowness of the divergence is the surprise. The period grows only logarithmically, $T\approx\frac{4}{\omega_0}\ln\frac{8}{\pi-\theta_0}$ : at $179°$ it is $3.9\,T_0$ , and even one arcsecond from inverted only about $9.7\,T_0$ . A period of ten $T_0$ demands release-angle control near $10^{-6}$ rad, far below mechanical repeatability — the logarithm eats your experimental heroism, which is why "infinitely slow pendulum" demonstrations always disappoint. True infinite period exists only exactly on the separatrix, a measure-zero orbit destroyed by any friction anyway.

What actually happens past 180 degrees — is there a period at all?

Yes, but of a different kind of motion. Amplitude beyond $\pi$ is not a bigger swing — it is no swing at all. With energy above $2mg\ell$ , $\dot\theta$ never vanishes: the pendulum circulates, $\theta$ advancing by $2\pi$ every period instead of oscillating. The rotation period is finite and decreases with energy, given by the same elliptic machinery with the modulus replaced by its reciprocal; the separatrix through the inverted equilibrium walls the two families apart in phase space. A child pumping a swing over the bar and a Josephson junction pushed past critical current both cross from libration to rotation.

Why do the corrections start at θ0² — why is there no linear-in-amplitude correction to the period?

Symmetry. The potential is even, $U(\theta)=U(-\theta)$ , so the dynamics cannot know the sign of the amplitude: the period must be an even function of $\theta_0$ , and its expansion can contain only $\theta_0^2,\theta_0^4,\dots$ The same evenness kills the even harmonics in the waveform. A linear-in-amplitude period shift can occur, but only in asymmetric wells — a pendulum with a magnet pulling sideways, or the cubic $\alpha x^2$ anharmonicity of interatomic potentials, whose asymmetry is what produces thermal expansion. The pendulum's perfectly symmetric well is precisely why its leading anharmonic effects are so tidy: quadratic in amplitude, odd in harmonics.

Anharmonic Pendulum and Amplitude Dependence

Where this lives today

The physical picture

From restoring force to elliptic integral

Which formula answers which question

The period grows with amplitude

The exact waveform

The separatrix

Limiting cases

Worked examples

1 · A playground swing at 60 degrees (confidence builder)

2 · The circular error of a pendulum clock (exam level)

Going further

Before this

Where this leads