In one derivation of the corrected period of a pendulum, we started off like so:

The mass has a height $y$ given by $l(1-\cos \theta )$. $E = K + E \rightarrow \frac{1}{2}ml^2 \dot{\theta}^2 + mgl(1-\cos \theta)$

The next step introduces $\theta _0$, and I've got no idea where this came from.

$$\frac{1}{2}ml^2 \dot{\theta}^2 + mgl(1-\cos \theta)= mgl(1-\cos \theta _0)$$

Now we just solve for $\dot{\theta}$ and solve the DE.W I'm interested in the theta side of the equation.

$$\int \frac{d\theta}{\sqrt{\cos \theta - \cos \theta _0 }}$$

We go through a bunch of subs and changes of vairble to arrive at

$$\int ^{2\pi} _0 \frac{du}{\sqrt{1-K^2 \sin ^2 u }}$$

So my two questions are

  • Why are we involving two $\theta$ values? The text didn't make it clear why we needed an extra $\theta _0$. It appears that $mgl(1-\cos \theta _0)$ is the total energy of the system. Our total energy cannot surpass the initial gravitational potential energy, this is clear. My only thought as to what $\theta$ means is the instantaneous position of the angle.
  • My text mentioned that this is an elliptical integral. Mathematically speaking, what is an elliptical integral? Can this integral be solved exactly, or does it always require approximations from the expansion?
  • 1
    $\begingroup$ The variable $\theta$ represents the angle of the pendulum as a function of time. $\theta_0$ is just the initial angle at $t=0$. $\endgroup$ – David H Dec 20 '13 at 2:50
  • $\begingroup$ $\theta_0$ does have to be the angle at $t=0$, it would be the maximum angle at which the velocity and therefore also the kinetic energy are zero. $\endgroup$ – fibonatic Dec 20 '13 at 5:26
  • $\begingroup$ Which textbook are you referring to? $\endgroup$ – Nanashi No Gombe Jun 24 '19 at 11:53

Since the total mechanical energy is conserved, from which we get $KE_i+PE_i=KE_f+PE_f$, it should be more clear now that $\theta_0$ is the angle at which you raise the pendulum (i.e., that term represents the $PE_i$). The terms on the left side your first centered equation are the final kinetic and potential energies, after some time $t$. In effect, you are right that $mgl\left(1-\cos\theta_0\right)$ is the total energy of the system.

An elliptic integral is one in which the function you are integrating, $f$, depends on the the variable and it's square root: $f\to f\left(x,\,\sqrt{x}\right)$. Since your function is $$ f(u,\sqrt{u})=\frac{1}{\sqrt{1-k^2\sin^2u}} $$ This is an elliptic integral. Wikipedia's article on elliptic integrals is pretty good too (and even includes your integral as an example of an elliptic integral).


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