First, you're overcomplicating things. The total energy of the pendulum will be equal at all points of its swing, so the simplest way to find it is to calculate the PE at the ends of its swing, when the KE is equal to zero.
You already did this:
$$PE = mgL(1-\cos\theta) \approx 1.26 \times 10^{-3} \text{J}$$
which is the answer to the question.
You cannot calculate the KE at the bottom of the swing and then add it to the PE at the top of the swing as you're doing because the two energies are being calculated at different points in the swing.
You could just calculate the KE at the bottom of the swing (when the PE will be zero) to get the same result, but in your calculations for KE you're making a couple of mistakes:
- Your calculation $\omega = \frac{2 \pi}{T}$ only applies to uniform circular motion, which isn't the case for a pendulum.
- When using $v = A \omega$, the $A$ isn't the amplitude of the pendulum but rather the radius of the circular arc, i.e. the length of the pendulum.
To calculate the KE at the bottom of the swing, start with the simple harmonic motion equation
$$\theta (t) = \theta_0 \cos \left( \frac{2\pi}{T} t \right)$$
Deriving this gives
$$\omega (t) = \frac{-2\pi\theta_0}{T} \sin \left( \frac{2\pi}{T} t \right)$$
Now using the fact that at the bottom of the first swing, $t = \frac{T}{4}$, we get
$$\begin{align}
\omega_{\text{max}} (t) &= \rule[-25px]{1px}{55px} \frac{-2\pi\theta_0}{T} \sin \left[ \frac{2\pi}{T} \left( \frac{T}{4} \right) \right] \;\rule[-25px]{1px}{55px} \\[2mm]
&= \rule[-20px]{1px}{50px} \frac{-2\pi\theta_0}{T} \sin \left( \frac{\pi}{2} \right) \; \rule[-20px]{1px}{50px} \\[2mm]
&= \frac{2\pi\theta_0}{T}
\end{align}$$
and finally
$$\begin{align}
v_{\text{max}} &= L \omega_{\text{max}} \\[2mm]
&= \frac{2\pi\theta_0 L}{T}
\end{align}$$
Using $\theta_0 = 4^{\circ} = 0.0698$ rads and your calculated value $T=1.331$ yields
$$\begin{align}
v_{\text{max}} &= \frac{2\pi (0.0698)(0.44)}{1.331} \\[2mm]
&\approx 0.145 \text{m/sec}
\end{align}$$
and so finally the KE at the bottom of the swing will be
$$\begin{align}
KE &= \frac{1}{2}mv^2 \\[2mm]
&\approx 1.26 \times 10^{-3} \text{J}
\end{align}$$
which agrees with the previously calculated PE value.
Best Answer
First off, determine how many independent dimensionless quantities there are. You have three independent dimensions and eight quantities (six parameters, time, and the angle), so you have five independent dimensionless quantities.
Naturally one of them is just $\theta$.
Since you have some natural frequencies of oscillation, it makes sense to have another dimensionless quantity be either $\omega t$ (with $\omega$ the natural frequency of the unforced, undamped linear pendulum) or $\Omega t$. You seem to have chosen it to be $\omega t$.
It is also natural to have a dimensionless parameter be a nondimensionalized version of $\omega-\Omega$ because you are specifically interested in beating. You wrote this as $\frac{\omega-\Omega}{\omega}$. This makes sense.
Now we need two more. The setup you're going for here compares the external forcing to the gravitational forcing, and the damping to the gravitational forcing. This is a choice, although it is a pretty reasonable one.
But you need the right units for $k$. The way you wrote it, the units for $k$ are $M L^2 T^{-1}$. This is a slightly weird convention (different from the convention for the damping coefficient in the usual spring-mass system with linear damping). But it doesn't hurt anything as long as you're aware of it.
Anyway, since the only source of a time that you're "allowed" to use in this term is $g$, you divide by $\sqrt{g}$ to cancel $T^{-1}$ and are now looking at $\frac{k}{\sqrt{g}}$, which has $ML^{3/2}$ units. Then you cancel what's left over with the simplest thing that has $ML^{3/2}$ units available which is $m L^{3/2}$.