I am trying to intuitively understand why $\pi$ is used when calculating the value of $g$ using the harmonic motion of a pendulum:
$$g ~=~\frac{4\pi^2L}{T^2}.$$
Does it have something to do with the curvature? I am thinking something along the lines of that, aswell as the fact that the oscillation of a pendulum would follow a circular path. The squared value of it would be because it was performed in 3d space.
I am just looking for a mathematically intuitive understanding of this.
Best Answer
You'll probably groan when you read this answer since it isn't nearly as complicated as you might think.
Essentially, there is factor of $\pi$ since the angular frequency $\omega = 2\pi f = \frac{2\pi}{T}$
A well know result from the linearized pendulum problem is that, for small angular displacements, the angular frequency is
$$\omega = \sqrt{\frac{g}{L}}$$
which follows from the differential equation in the angular displacement $\theta$:
$$\ddot \theta + \frac{g}{L}\sin \theta = 0 \approx \ddot \theta + \frac{g}{L}\theta\;,\quad \theta \ll 1$$
Thus,
$$\left(\frac{2\pi}{T}\right)^2 = \frac{g}{L} \Rightarrow g = \frac{4 \pi^2 L}{T^2}$$