show that mgh = 1/2mv^2 for a simple pendulum

Answer 1

PE at the top of the swing is all HE at the bottom.

Answer 2

length L
angle from vertical = theta
height above bottom = L * (1 - cos theta)
for small angles cos theta = 1 - theta^2/2
so for small A height = L * theta^2/2
so potential energy at top = m g L * theta^2/2
if max theta = a then max pe = m g L a^2/2 =( ma^2/2) g L
if theta = a sin 2 pi f t
omega =d theta/dt = a (2 pi f) cos wt
max omega = a (2 pi f) when t = 0 at bottom
max velocity = L * a * 2 pi f
max Ke = (1/2) m ( L * a * 2 pi f)^2 = (ma^2 / 2) L^2 (2pif)^2
now
does g L = L^2(2 pi f)^2
or (2 pi f)^2 = g/L ?????? I think so :)