homework and exercises - Show bigger amplitude of physical pendulum means bigger period

Suppose you have a physical pendulum. It is true that as amplitude increases, the period increases. Can we demonstrate this fact without explicitly finding the period (which is pretty involved and pretty messy) in:

1. 
   

   an intuitive fashion,

   
   


2. 
   

   rigorously?

   
   

Answer

Let's draw our pendulum:

The equation of motion is:

$$ F\ell = -I\frac{d^2\theta}{dt^2} $$

This may seem a bit unfamiliar, but it's just the circular motion equivalent of $F = ma$. We replace the force by the torque, $F\ell$, the mass by the moment of inertia $I$ and the acceleration by the angular acceleration $\ddot{\theta}$. A bit of quick geometry gives us $F = mg\sin\theta$, so our equation becomes:

$$ mg\sin\theta\ell = -I\frac{d^2\theta}{dt^2} $$

Assuming our mass is a point, the moment of intertia is just $I = m\ell^2$, and with a quick rearrangement we get:

$$ \frac{d^2\theta}{dt^2} = -\frac{g}{\ell}\sin\theta $$

What your physics teacher will do next is point out that $\sin\theta$ can be expanded as a power series:

$$ \sin\theta = \theta - \frac{\theta^3}{3!} + \frac{\theta^5}{5!} - ... $$

and if $\theta$ is small then the higher powers of $\theta$ are very small and we get $\sin\theta \approx \theta$. Substitute this for $\sin\theta$ in our equation above and we get:

$$ \frac{d^2\theta}{dt^2} = -\frac{g}{\ell}\theta \tag{1} $$

which is our good old simple harmonic motion equation.

Now we can answer your question, because if we keep increasing the angle of swing we're going to get to a point where the $\theta^3$ term is too large to be ignored. In that case our equation (1) becomes:

$$ \frac{d^2\theta}{dt^2} = -\frac{g}{\ell}\left(\theta - \frac{\theta^3}{3!}\right) \tag{2} $$

Now take two pendulums (penduli?), one described by the simple harmonic equation (1) and one described my our more accurate equation (2), and start them at some initial angle $\theta_0$. The angular acceleration calculated by equation (2) is less than the angular acceleration calculated by equation (1) for all values of $\theta$ (except at $\theta = 0$). So if both penduli start at the same place, $\theta_0$, pendulum 2 must take longer to get to $\theta = 0$ than pendulum 1 will. But this time is just a quarter of the period, and that means the period of pendulum 2 must be greater than the period of pendulum 1. So for a real pendulum the period must increase with increasing amplitude of swing.