Mechanical Vibrations (Pendulums)

[dipset24]

Homework Statement

Assume that the differential equation of a simple pendulum of length L is L$$\Theta$$'' + g$$\Theta$$=0 where g=GM/R$$^2$$ is the gravitational acceleration at the location of the pendulum.

Two pendulums are of lengths L1 and L2 and when located at the respective distances R1 and R2 from the center of the earth-have periods p1 and p2. Show that:

p1/p2=R1$$\sqrt{L1}$$/R2$$\sqrt{L2}$$

The Attempt at a Solution

I do not know where to begin. If someone could help me out.

 

[Astronuc]

One is looking for the ratio of the periods. Remember ω = 2 pi/T, where T is the period, and one is looking for a system described by $$\ddot{\Theta} + \omega^2\Theta = 0$$.

Please refer to - http://hyperphysics.phy-astr.gsu.edu/hbase/pend.html#c1

 

[RoyalCat]

Do you remember what the period, T, is equal to in SHM?
EDIT: Ah, ninja'd. :P