newtonian mechanics - Period of a pendulum

In the book 'Calculus the Early Transcendetals' at page 776 (7th edition) they give that the period of a pendulum with length $\text{L}$ that makes a maximum angle $\theta_0$ with the vertical is:

$$\text{T}_{\left[\text{s}\right]}=4\sqrt{\frac{\text{L}}{\text{g}}}\int_{0}^{\frac{\pi}{2}}\frac{1}{\sqrt{1-\sin^2\left(\frac{\theta_0}{2}\right)\sin^2(x)}}\space\text{d}x$$

Questions:

• Does this formula work, for any pendulum?



• How did they get this formula (hint to derive the given formula)?

Answer

That formula holds for a simple pendulum of length $L$ in a gravitational field $g$ and released from rest with an amplitude $\theta_0$.

Since this system is conservative its mechanical energy is constant and equals the gravitational potential energy when it is released. Setting the zero of potential energy at the fixed point of the pendulum, the mechanical energy is

$$E=-mgL\cos\theta_0=\frac 12 mL^2\dot\theta^2-mgL\cos\theta.$$ Solving for $\frac{d\theta}{dt}$, separating the variables and integrating you should get $$\int_0^t dt=\sqrt{\frac{L}{2g}}\int_{\theta_0}^\theta \frac{d\theta}{\sqrt{\cos\theta-\cos\theta_0}}.$$ The idea now is to make a couple of variable changes in order to arrive to a known integral. What you have to do is to write $\cos a=1-2\sin^2(a/2)$ and then you define $\sin(\theta/2)=\sin(\theta_0/2)\sin x$, such that $$d\theta=\frac{2\sin(\theta_0/2)\sqrt{1-\sin^2x}}{\sqrt{1-\sin^2(\theta_0/2)\sin^2x}}dx.$$ The integral becomes $$t=\sqrt{\frac{L}{g}}\int_{x_0}^x\frac{dx}{\sqrt{1-\sin^2(\theta_0/2)\sin^2x}}.$$ The period equals four times the time the pendulum takes going from $\theta=0$ to $\theta=\theta_0$. Therefore $$T=4\sqrt{\frac{L}{g}}\int_{0}^{\pi/2}\frac{dx}{\sqrt{1-\sin^2(\theta_0/2)\sin^2x}}.$$ This is a complete elliptical integral of the first kind. A nice description of the pedulum including some nice plots can be found at wikipedia