Pendulum Phase Portrait

A 1 m pendulum is released from rest at 30°. Find its period exactly — the textbook formula is not the answer — and then find how close to vertical you would have to release it to stretch one full swing to 10 s. Take g = 9.81 m/s², no damping.

1. Two quantities fix the natural frequency, and the mass is not one of them: it multiplies every term of the equation of motion and cancels before anything else happens.

   $${\omega }_0=\sqrt{g/L}=\sqrt{9.81}=3.1321rad/s$$

2. Replace sin θ by θ and what remains is a harmonic oscillator, whose period is 2π/ω₀. Read what dropped out: θ₀. Amplitude-independence is a property of that replacement, not of the pendulum.

   $$T_{\mathrm{linear}}=\frac{2\pi }{{\omega }_0}=2.0061\mathrm{s}$$

3. Now do it without the replacement. Energy conservation needs no small angle — the kinetic energy at angle θ is exactly the potential energy given up in falling from θ₀ — so solve it for the angular velocity instead.

   $$\frac{1}{2}{L}^2{\dot{\theta }}^2=gL(\cos \theta -\cos {\theta }_0)⟹\dot{\theta }={\omega }_0\sqrt{2(\cos \theta -\cos {\theta }_0)}$$

4. Time is distance over speed, integrated, and a quarter period is the swing from 0 out to θ₀. The release angle is back, sitting in the integrand and in the upper limit at once, so the period cannot be independent of it.

   $$T=\frac{4}{{\omega }_0}{\int }_0^{{\theta }_0}\frac{\mathrm{d}\theta }{\sqrt{2(\cos \theta -\cos {\theta }_0)}}$$

5. The integrand blows up at θ = θ₀, and the identity cos θ = 1 − 2sin²(θ/2) shows the blow-up is only a square root of a simple zero: the difference of cosines is a difference of squares. Write k for the sine of half the amplitude, and let φ sweep 0 to π/2 while θ sweeps out to θ₀.

   $$k=\sin \frac{{\theta }_0}{2},\sin \frac{\theta }{2}=k\sin \phi ⟹\cos \theta -\cos {\theta }_0=2k^2{\cos }^2\phi$$

6. The awkward root now cancels against dθ, and what is left is the complete elliptic integral of the first kind — the formula the panel above displays, which writes its argument as sin²(θ₀/2), that is k², not k.

   $$T=\frac{4}{{\omega }_0}{\int }_0^{\pi /2}\frac{\mathrm{d}\phi }{\sqrt{1-k^2{\sin }^2\phi }}=\frac{4}{{\omega }_0}K(k)$$

7. K has no elementary closed form, but its integrand does: a binomial series in u = k²sin²φ, whose coefficients are the central binomial ones.

   $$(1-u{)}^{-1/2}=\sum _{n\ge 0}{c}_n{u}^n,c_n=\frac{(2n)!}{4^n(n!{)}^2}$$

8. Integrate that series term by term and Wallis' integral hands back the very same c_n the binomial expansion just produced. That coincidence is the whole trick.

   $${\int }_0^{\pi /2}{\sin }^{2n}\phi \mathrm{d}\phi =\frac{\pi }{2}{c}_n$$

9. Each term therefore picks c_n up twice, so every coefficient of the period series is a binomial coefficient squared: 1/4, 9/64 and 25/256 are the squares of 1/2, 3/8 and 5/16.

   $$\frac{T}{T_{\mathrm{linear}}}=\sum _{n\ge 0}{c}_n^2{k}^{2n}=1+\frac{k^2}{4}+\frac{9k^4}{64}+\frac{25k^6}{256}+\cdots$$

10. At 30° the half-amplitude is 15°, so k² is only 0.067 and each term is more than 20 times smaller than the one before it. Stop at 3 terms and you get 1.017378, which multiplies out to 2.0409 s; the 4th term is what moves the final digit.

    $$k^2={\sin }^{2}15^{\circ }=0.066987:1+0.016747+0.000631+0.000029=1.017407$$

11. Multiply. This is the exact period the panel prints, and 1.0174 is the ratio printed beside it — both of them reached without evaluating K numerically once.

    $$T=2.0061\times 1.017407=2.0410\mathrm{s}$$