Derivation of 2$\pi$ in the period of a simple pendulum Most of us know that the period of a simple pendulum is given by

$$2 \pi \sqrt{\dfrac{\ell}{g}}.$$

But how did the $2\pi$ term get into that argument. From dimensional analysis, we can find the period but not the constant.

Is there a calculus argument as some limit is taken? Is it based on an energy equation? Surely there is a way to derive that $2 \pi$.
The pendulum movement equation is given by
\begin{align*}
\frac{\mathrm{d}^{2}\theta}{\mathrm{d}t^{2}} + \frac{g}{\ell}\sin(\theta) = 0
\end{align*} 

For small values of $\theta$, we can make the approximation $\sin(\theta) \approx \theta$, from whence we obtain the equation
\begin{align*}
\ddot{\theta} + \frac{g}{\ell}\theta = 0
\end{align*}

whose associated characteristic equation is
\begin{align*}
x^{2} + \frac{g}{\ell} = 0 \Longrightarrow x = \pm i\sqrt{\frac{g}{\ell}}
\end{align*}

Thus the solutions are $\theta(t) = c_{1}\sin\left(\displaystyle t\sqrt{\frac{g}{\ell}}\right) + c_{2}\cos\left(\displaystyle t\sqrt{\frac{g}{\ell}}\right)$. Finally, for $T = 2\pi\sqrt{\displaystyle\frac{\ell}{g}}$, one has
\begin{align*}
\theta\left(t + 2\pi\sqrt{\frac{\ell}{g}}\right) = c_{1}\sin\left(t\sqrt{\frac{g}{\ell}} + 2\pi\right) + c_{2}\cos\left(t\sqrt{\frac{g}{\ell}} + 2\pi\right) = \theta(t)
\end{align*}
Therefore we conclude that $T$ is the period indeed.
The expression $\sqrt{\frac{g}{l}}$ represents the frequency in radians per second. Divide by $2\pi$ radians per cycle to get $\frac{1}{2\pi}\sqrt{\frac{g}{l}}$ cycles per second.

Now take the reciprocal to get $2\pi\sqrt{\frac{l}{g}}$ seconds per cycle which, by definition, is the period.