Understanding the Period of a Pendulum: How is T = 2pi * sqrt(L/g) Derived?

[Okazaki]

Homework Statement

Prove the equation: T = 2 pi * sqrt(L/g), then determine the period of a "seconds pendulum" (period = 2 sec on Earth) on the surface of the moon.

Homework Equations

T = 2pi * r/v
T = 2pi/ω
a_r = rω²

The Attempt at a Solution

Do you assume a_r is a_g?

T = 2pi/ω
= 2 * pi / sqrt(a/r)
= 2 * pi * sqrt(r/a)

...Apparently, L is supposed to be the length of the pendulum, so somewhere along the line, I completely screwed up.

I can solve the second part of the question just fine, but since I've never actually seen T = 2 pi * sqrt(L/g) used or proved, I'd rather not work with it until I understand where it comes from and how it works.

 

[PWiz]

Start with the component of gravity which causes the bob to move towards the mean position. See how this relates to the SHM equation. Remember that for small displacements from the mean position, the motion is SH. You should be able to derive it now.
Hint: Use angular displacement.

 

[AlephNumbers]

If you get stuck, keep in mind that most introductory physics textbooks will have a derivation of this.