Simple Pendulum / SHM

Find the period of a 1 m pendulum released at 20°, and find out how wrong the textbook formula is — because the textbook formula is not the answer. Take g = 9.81 m/s².

1. The restoring torque goes as sin θ, which does not give simple harmonic motion. The famous formula only appears after replacing sin θ with θ — an approximation, not a derivation, and everything that follows inherits it.

   $$m\ell \ddot{\theta }=-mg\sin \theta \stackrel{\sin \theta \approx \theta }{\to }\ddot{\theta }=-\frac{g}{\ell }\theta$$

2. With that swap the equation is the harmonic oscillator and the period drops out. Note what it does not contain: the mass, and the amplitude. Both absences are consequences of the approximation, not of the pendulum.

   $$T_0=2\pi \sqrt{\frac{\ell }{g}}=2\pi \sqrt{\frac{1}{9.81}}=2.006\mathrm{s}$$

3. The exact period is an elliptic integral, but its series expansion says how fast the approximation decays. The correction is quadratic in the release angle, in radians.

   $$T=T_0\left(1+\frac{{\theta }_0^2}{16}+\cdots \right),{\theta }_0=20^{\circ }=0.349\mathrm{rad}$$

4. Evaluate it. This is the value the simulator above prints, because it uses the corrected period rather than the textbook one.

   $$T=2.006\left(1+\frac{0.349^2}{16}\right)=2.006\times 1.00762=2.021\mathrm{s}$$

5. Maximum speed comes from energy alone — drop height ℓ(1−cos θ₀) converted to kinetic energy — and needs no approximation at all.

   $$v_{\max }=\sqrt{2g\ell (1-\cos {\theta }_0)}=\sqrt{2(9.81)(1)(1-\cos 20^{\circ })}=1.088m/s$$