Problem — Non-isochronous pendulum

Problem

For large oscillations, the period of the pendulum depends on the amplitude. The approximation $\sin\theta \approx \theta$ does not hold. Show that for amplitude $\theta_0 = 60^\circ$ the period is about $7\%$ greater than the value $T_0 = 2\pi\sqrt{L/g}$ of small oscillations. Explain why it increases and not decreases.

Solution

The expansion of the exact period for finite amplitude, to first order, is:

$T \approx T_0\left(1 + \frac{\theta_0^2}{16}\right)$

with $\theta_0$ in radians. For $\theta_0 = 60^\circ = \pi/3 \approx 1{,}047\;\text{rad}$:

$\frac{\theta_0^2}{16} = \frac{1{,}097}{16} \approx 0{,}0685$

$\ev{T/T_0 \approx 1 + \theta_0^2/16 \approx 1{,}07}$

i.e. about $+7\%$.

Why it increases: the true restoring force is $-mg\sin\theta$, weaker than the linear one $-mg\theta$ because $\sin\theta < \theta$. A weaker restoring force makes the oscillation slower, so the period grows with amplitude.

Links

Topics: Oscillations and simple harmonic motion Concepts: Simple pendulum · Period and frequency Skills: Analysis of limiting cases and thought experiments Objects: Simple pendulum