The Real Pendulum

The hook

Every physics student learns that a pendulum’s period is \(2\pi\sqrt{l/g}\), independent of the size of the swing. Galileo believed this, supposedly after timing a swinging lamp in Pisa cathedral against his pulse, and it is the reason pendulum clocks work. But it is not quite true. The formula comes from the approximation \(\sin\theta \approx \theta\), and a pendulum swinging through a wide arc takes longer. At \(90^\circ\) the period is \(18\%\) longer; a pendulum released from almost upside down can take as long as you like.

This question finds the exact period as an integral, and then the first correction for moderate swings. The integral cannot be evaluated in elementary functions: it is an elliptic integral, and the story of how such integrals were tamed runs from Gauss’s arithmetic–geometric mean to the theory of elliptic curves.

Before you start

The speed is \(l\dot\theta\), and the height above the lowest point is \(l(1 - \cos\theta)\). Since the string does no work (its tension is always perpendicular to the motion), energy is conserved.

Working it through

Stage 1 Energy

At the extreme angle \(\gamma\) the bob is momentarily at rest. Conservation of energy gives \[\tfrac12ml^2\dot\theta^2 + mgl(1 - \cos\theta) = mgl(1 - \cos\gamma),\] and dividing by \(ml\), \(\frac12l\dot\theta^2 = g(\cos\theta - \cos\gamma)\).■

Stage 2 The period as an integral

Using \(\cos\theta = 1 - 2\sin^2\frac\theta2\), we have \(\cos\theta - \cos\gamma = 2\bigl(\sin^2\frac\gamma2 - \sin^2\frac\theta2\bigr)\), so \[\dot\theta = 2\sqrt{\frac gl}\,\Bigl(\sin^2\frac\gamma2 - \sin^2\frac\theta2\Bigr)^{1/2}.\] By symmetry the swing from \(0\) to \(\gamma\) takes a quarter of the period, so \[P = 4\int_0^\gamma\frac{\mathrm{d}\theta}{\dot\theta} = 2\sqrt{\frac lg}\int_0^\gamma\Bigl(\sin^2\frac\gamma2 - \sin^2\frac\theta2\Bigr)^{-1/2}\mathrm{d}\theta. \qed\]

Stage 3 Removing the singularity

As \(\theta\) goes from \(0\) to \(\gamma\), \(\phi\) goes from \(0\) to \(\frac\pi2\). Differentiating, \(\frac12\cos\frac\theta2\,\mathrm{d}\theta = k\cos\phi\,\mathrm{d}\phi\), so \[\mathrm{d}\theta = \frac{2k\cos\phi\,\mathrm{d}\phi}{\cos\frac\theta2} = \frac{2k\cos\phi\,\mathrm{d}\phi}{\sqrt{1 - k^2\sin^2\phi}}.\] Meanwhile \(\sin^2\frac\gamma2 - \sin^2\frac\theta2 = k^2(1 - \sin^2\phi) = k^2\cos^2\phi\). The awkward factors cancel: \[P = 2\sqrt{\frac lg}\int_0^{\pi/2}\frac{1}{k\cos\phi}\cdot\frac{2k\cos\phi}{\sqrt{1 - k^2\sin^2\phi}}\,\mathrm{d}\phi = 4\sqrt{\frac lg}\int_0^{\pi/2}\frac{\mathrm{d}\phi}{\sqrt{1 - k^2\sin^2\phi}}.\] The new integrand is perfectly well behaved, since \(k < 1\).

Stage 4 The first correction

To first order, \((1 - k^2\sin^2\phi)^{-1/2} \approx 1 + \frac12k^2\sin^2\phi\), and \(\int_0^{\pi/2}\sin^2\phi\,\mathrm{d}\phi = \frac\pi4\). So \[P \approx 4\sqrt{\frac lg}\Bigl(\frac\pi2 + \frac{k^2}{2}\cdot\frac\pi4\Bigr) = 2\pi\sqrt{\frac lg}\Bigl(1 + \frac{k^2}{4}\Bigr).\] For small \(\gamma\), \(k = \sin\frac\gamma2 \approx \frac\gamma2\), so \(k^2 \approx \frac{\gamma^2}{4}\) and \[P \approx 2\pi\sqrt{\frac lg}\Bigl(1 + \frac{\gamma^2}{16}\Bigr). \qed\] For a swing of \(20^\circ\) (\(\gamma \approx 0.35\)) the correction is less than \(1\%\); at \(60^\circ\) it is about \(7\%\) (Figure 11.3.1).

Looking back

• Energy gives a first integral. A second-order equation became a first-order one, and the period became an integral.

• A substitution tames a singularity. The integrand blew up at the turning point; the substitution \(\sin\frac\theta2 = k\sin\phi\) removed the problem.

• Exact first, then approximate. We found an exact formula and expanded it, which also tells us how good the approximation is.

What’s really going on

Elliptic integrals

The integral \[K(k) = \int_0^{\pi/2}\frac{\mathrm{d}\phi}{\sqrt{1 - k^2\sin^2\phi}}\] is the complete elliptic integral of the first kind, so called because a close relative gives the perimeter of an ellipse. The exact period is \(P = 4\sqrt{l/g}\,K\bigl(\sin\frac\gamma2\bigr)\). Expanding the integrand fully and integrating term by term using the Wallis integrals \(\int_0^{\pi/2}\sin^{2n}\phi\,\mathrm{d}\phi\) of Section 4.4, one finds \[P = 2\pi\sqrt{\frac lg}\Bigl(1 + \bigl(\tfrac12\bigr)^2k^2 + \bigl(\tfrac{1\cdot3}{2\cdot4}\bigr)^2k^4 + \bigl(\tfrac{1\cdot3\cdot5}{2\cdot4\cdot6}\bigr)^2k^6 + \cdots\Bigr).\] No finite combination of familiar functions gives \(K(k)\); it is a genuinely new function.

Gauss’s remarkable formula

There is a far better way to compute \(K\). In Section 2.6 you met the arithmetic–geometric mean: start with \(a\) and \(b\), and repeatedly replace them by their arithmetic and geometric means. Gauss discovered in 1799 that \[K(k) = \frac{\pi}{2\,\mathrm{AGM}\bigl(1, \sqrt{1 - k^2}\bigr)},\] so the period of a pendulum with amplitude \(\gamma\) is exactly \[P = \frac{2\pi\sqrt{l/g}}{\mathrm{AGM}\bigl(1, \cos\frac\gamma2\bigr)}.\] The AGM converges so fast that three or four steps give the period to ten significant figures; the values in Figure 11.3.1 were computed this way. Gauss wrote in his diary that this discovery “opens an entirely new field of analysis”, and he was right: inverting elliptic integrals led Abel and Jacobi to elliptic functions, and those led to elliptic curves, which are now central to number theory (they were the key to the proof of Fermat’s Last Theorem) and to the cryptography that secures the internet.

The thinking skill

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Further reading

• David Morin, Introduction to Classical Mechanics (Cambridge University Press), Chapter 4, which treats the large-amplitude pendulum.

• Jonathan Borwein & Peter Borwein, Pi and the AGM (Wiley). Gauss’s discovery and its uses, including the fastest ways to compute \(\pi\).

• Dava Sobel, Longitude (Fourth Estate). The story of timekeeping at sea, beyond the pendulum.