swinging pendulum question, please help?

Question

The oscillation period of a pendulum varies as the square root of its length and inversely as the square root of the acceleration due to gravity. Given that the moon has 1/6 the earth's gravity, how much shorter should a pendulum on the moon be in order to have a period that is twice as long as a pendulums period on earth?

Answer 1

I gave a similar problem to my college level remedial algebra students.

Let
T = period
L = pendulum length
gm or ge = gravity of moon or earth

Then

T = sqrt(L/g) , right?

You want Tm = 2Te , that is, moon's pend. per.twice earth's

Just substitute:

Tm = sqrt(Lm/gm) = 2sqrt(Le/ge)
and
gm = ge/6

You want to solve for a ratio:
Lm/Le = ... I'll let you do the rest....
( I get Lm/Le = 4/6)...

Answer 2

2/6