The energy derivation of the pendulum equation will be used in this post: Please read the energy derivation of the pendulum equation here (scroll down to ' ''Energy'' derivation of (Eq. 1) ' )

As a small addition to the derivation on wikipedia:

When equating change in kinetic energy with change in gravitational potential energy, it is important to realize that the $\displaystyle v$ in $\displaystyle (\frac{1}{2})(mv^2)$ is $speed$, not velocity.

This makes sense, because if $\displaystyle v$ were velocity, squaring it would not make any sense

Now, with this in mind, $\displaystyle v$ is always $positive$. It is the $magnitude$, of velocity at any given time. Now, $\displaystyle v = l(| \frac{d\theta}{dt} |)$

Using $\displaystyle v = l(| \frac{d\theta}{dt} |) $, we have $\displaystyle | \frac{d\theta}{dt} | = \sqrt{(\frac{2g}{l})(cos(\theta) - cos(\theta_0))}$

For $0\leq \theta \leq \theta_0$, (the first quarter period of the pendulums motion), it is clear that $\frac{d\theta}{dt} < 0 $ (the pendulum starts at $\theta_0$ and $DECREASES$ to $0$)

Therefore for the first quarter period, $\displaystyle \frac{d\theta}{dt} = (-1)(\sqrt{(\frac{2g}{l})(cos(\theta) - cos(\theta_0))}$

$\square$

$\square$

There are several assumptions that are being made during the energy derivation of the pendulum equation:

(1) All the mass in the system is concentrated in the pendulum bob, therefore the string connecting the pendulum to the stand is assumed to be mass-less.

(2) The string connecting the pendulum to the stand is assumed to be in-extensible and in-compressible. This is why $|\vec{r(t)}|$ does not change during motion, therefore we are assuming that the pendulum is travelling in a circular arc.

(3) The pendulum is in a completely uniform gravitational field (What this means is that the gravitational force acting on the pendulum is assumed to be constant regardless of the pendulums position during its motion - In reality this is not the case (the pendulum does experience a larger gravitational force at the bottom of its swing and a smaller gravitational force at the very top of its swing, but the difference is so small that we ignore it completely)

(4) There are zero additional forces within the system

$\square$

From the energy derivation of the pendulum equation, we know that for the first quarter time period:

$\displaystyle \frac{d\theta}{dt} = (-1)( \sqrt{(\frac{2g}{l})(cos(\theta) - cos(\theta_0))}) $

Since $\displaystyle \frac{d\theta}{dt} =(-1)( \sqrt{(\frac{2g}{l})(cos(\theta) - cos(\theta_0))})$ ,

$\displaystyle \frac{dt}{d\theta} =(-1) (\sqrt{\frac{l}{2g}})(\frac{1}{\sqrt{cos(\theta) - cos(\theta_0)}})$

Therefore, effectively , time is a function of the angle the pendulum creates with the vertical.

Now, by the fundamental theorem of calculus , we know that $\displaystyle \int_{\theta_0}^0 (-1) (\sqrt{\frac{l}{2g}})(\frac{1}{\sqrt{cos(\theta) - cos(\theta_0)}}) = \int_{\theta_0}^0 \frac{dt}{d\theta} = t(0) - t(\theta_0)$

Where $\displaystyle t$ is time as a function of $\theta$.

It is thus clear that $t(0) - t(\theta_0)$ is the time taken for the pendulum to swing from $\theta = \theta_0$ to $\theta = 0$ , or $\frac{T}{4}$ where $T$ is the full time period of the pendulum.

Now, we also know that $\displaystyle \int_{\theta_0}^0 (-1) (\sqrt{\frac{l}{2g}})(\frac{1}{\sqrt{cos(\theta) - cos(\theta_0)}}) = \displaystyle \int_{0}^{\theta_0} (\sqrt{\frac{l}{2g}})(\frac{1}{\sqrt{cos(\theta) - cos(\theta_0)}})$

This is because we can swap the order of integration by just multiplying the integrand by $-1$.

So, $ \displaystyle \int_{0}^{\theta_0} (\sqrt{\frac{l}{2g}})(\frac{1}{\sqrt{cos(\theta) - cos(\theta_0)}}) = \frac{T}{4} $,

this implies that:

$\displaystyle T = 4(\sqrt{\frac{l}{2g}})\int_{0}^{\theta_0} (\frac{1}{\sqrt{cos(\theta) - cos(\theta_0)}})$

Lets consider the integral $\displaystyle \int_{0}^{\theta_0} (\frac{1}{\sqrt{cos(\theta) + 1}})$

First, we will show that $\displaystyle (\frac{1}{\sqrt{cos(\theta) - cos(\theta_0)}}) \geq (\frac{1}{\sqrt{cos(\theta) + 1}})$

To do this we will prove something that is equivalent to this, namely that:

$\displaystyle \sqrt{cos(\theta) - cos(\theta_0)} \leq \sqrt{cos(\theta) + 1}$

Squaring both sides, we have $\displaystyle cos(\theta) - cos(\theta_0) \leq cos(\theta) + 1$

Subtracting both sides by $\displaystyle cos(\theta)$ , we have $\displaystyle -cos(\theta_0) \leq 1$, implying $\displaystyle cos(\theta_0) \geq -1$, which is always true for $\displaystyle cos(x)$.

Therefore, we have proved that $\displaystyle (\frac{1}{\sqrt{cos(\theta) - cos(\theta_0)}}) \geq (\frac{1}{\sqrt{cos(\theta) + 1}})$

Now, lets try and evaluate $\displaystyle \int_{0}^{\theta_0} (\frac{1}{\sqrt{cos(\theta) + 1}})$

We should start by trying to evaluate the indefinite integral $ \displaystyle \int (\frac{1}{\sqrt{cos(\theta) + 1}})$

This integral can be reduced to $\displaystyle \frac{1}{\sqrt2}\int sec(\frac{\theta}{2})$ by noticing that $\displaystyle cos(\theta) = cos(\frac{\theta}{2} + \frac{\theta}{2}) = cos^2(\frac{\theta}{2}) - sin^2(\frac{\theta}{2}) = 2cos^2(\frac{\theta}{2}) $

There are several different ways to evaluate integrals of the form $\displaystyle \int secx$, we will use one particular method in order to evaluate $\displaystyle \frac{1}{\sqrt2}\int sec(\frac{\theta}{2})$

I evaluated $\displaystyle \frac{1}{\sqrt2}\int sec(\frac{\theta}{2})$ using the fact that $\displaystyle cos(\theta) = \frac{1 - tan^2(\frac{\theta}{2})}{1 + tan^2(\frac{\theta}{2})}$

With this, I found that $\displaystyle \frac{1}{\sqrt2}\int sec(\frac{\theta}{2}) = (\sqrt2) ln(tan(\frac{\pi}{4} + \frac{\theta}{4}))$

Therefore, $\displaystyle \int_{0}^{\theta_0} (\frac{1}{\sqrt{cos(\theta) + 1}}) = \int_{0}^{\theta_0} \frac{1}{\sqrt2}\int sec(\frac{\theta}{2}) = (\sqrt2) ln(tan(\frac{\pi}{4} + \frac{\theta_0}{4})) $

$\displaystyle ln(x)$ is a strictly increasing function that increases without bound, that is $\displaystyle \lim_{x \to \infty} ln(x) = \infty$

Now, the argument of $\displaystyle ln$ in $\displaystyle ln(tan(\frac{\pi}{4} + \frac{\theta_0}{4})) $ is $\displaystyle tan(\frac{\pi}{4} + \frac{\theta_0}{4})$ , and as $\displaystyle \theta_0 \to \pi$, $\displaystyle tan(\frac{\pi}{4} + \frac{\theta_0}{4}) \to \infty$.

Therefore, as $\displaystyle \theta_0 \to \pi$, $\displaystyle tan(\frac{\pi}{4} + \frac{\theta_0}{4}) \to \infty$, which means that $\displaystyle ln(tan(\frac{\pi}{4} + \frac{\theta_0}{4})) \to \infty$.

But this means that

$\displaystyle \int_{0}^{\theta_0} (\frac{1}{\sqrt{cos(\theta) + 1}}) =

\int_{0}^{\theta_0} \frac{1}{\sqrt2}\int sec(\frac{\theta}{2}) =

(\sqrt2) ln(tan(\frac{\pi}{4} + \frac{\theta_0}{4}))$ $\to \infty$ as $\theta_0 \to \pi$

$\displaystyle \int_{0}^{\theta_0} (\frac{1}{\sqrt{cos(\theta) + 1}}) =

\int_{0}^{\theta_0} \frac{1}{\sqrt2}\int sec(\frac{\theta}{2}) =

(\sqrt2) ln(tan(\frac{\pi}{4} + \frac{\theta_0}{4}))$ $\to \infty$ as $\theta_0 \to \pi$

However, we know that $\displaystyle (\frac{1}{\sqrt{cos(\theta) - cos(\theta_0)}}) \geq (\frac{1}{\sqrt{cos(\theta) + 1}}) $

Which means that $\displaystyle T = \int_{0}^{\theta_0} \displaystyle (\frac{1}{\sqrt{cos(\theta) - cos(\theta_0)}}) \geq \int_{0}^{\theta_0} (\frac{1}{\sqrt{cos(\theta) + 1}}) $

As we know that $\displaystyle \int_{0}^{\theta_0} (\frac{1}{\sqrt{cos(\theta) + 1}}) \to \infty$ as $\displaystyle \theta_0 \to \pi$, this suffices to show that $\displaystyle \lim_{\theta_0 \to \pi} T = \infty $

This is a nice result, as it implies that it can take arbitrarily long for a pendulum to complete a full swing given we incline the pendulum close enough to the vertical (under ideal conditions)

This is a nice result, as it implies that it can take arbitrarily long for a pendulum to complete a full swing given we incline the pendulum close enough to the vertical (under ideal conditions)