# Period of pendulum motion

1. Jan 28, 2017

### StrangelyQuarky

1. The problem statement, all variables and given/known data

A pendulum obeys the equation $$\ddot{\theta} = -\sin(\theta)$$ and has amplitude $$\theta_0$$. I have to show that the period is
$$T = 4 \int_{0}^{\frac{\pi}{2}} \frac{d\phi}{\sqrt{1-\alpha \sin^2(\phi)}}$$ where $$\alpha = \sin^2(\frac{\theta_0}{2})$$

2. The attempt at a solution

I derived an expression for time:

$$\dot{\theta}\frac{d\dot{\theta}}{d\theta} = -\sin(\theta)$$

I said that the pendulum starts out at the height of its amplitude $$\theta = \theta_0$$ where it also has zero velocity

$$\int_{0}^{\dot{\theta}} \dot{\theta}d\dot{\theta} = \int_{\theta_0}^{\theta} -\sin(\theta)d\theta$$

$$\dot{\theta} = \frac{d\theta}{dt}= \pm \sqrt{2(\cos(\theta)-\cos(\theta_0))}$$

So for the period we can integrate from $$\theta_0 \text{ to } 0$$, which is a quarter of the period, then multiply by 4 to get the whole period.

$$T = 4 \int_{\theta_0}^{0} \frac{d\theta}{\sqrt{2(\cos(\theta)-\cos(\theta_0))}}$$

By the half-angle identity,

$$T = 2 \int_{\theta_0}^{0} \frac{d\theta}{\sqrt{\sin^2(\theta_0 /2)-\sin^2(\theta /2)}} = 2 \int_{\theta_0}^{0} \frac{d\theta}{\sqrt{\alpha-\sin^2(\theta /2)}}$$

And this is where I'm stuck. It looks similar to the desired answer, but I can't think of any identity or substitution that would give the right integrand and limits.

Last edited: Jan 28, 2017
2. Jan 29, 2017