# Determine the length of a pendulum on the moon

So I'm working on homework, and encountered this problem. I thought I understood what to do, but I'm not getting the right answer. I'm so frustrated I decided to create an account, so this is my first post. I'll do my best to adhere to the PF format of questions.

The Question is:
A pendulum oscillating on the moon has the same period as a(n) 3.66 m pendulum oscillating on Earth. If the moon’s gravity is one-sixth of Earth’s gravity, ﬁnd the length of the pendulum on
the moon.

Attempt:
Tmoon = Tearth

Tearth = 2∏√(L/g)
Tearth = 2∏√(3.66m/9.8m/s) = 3.8397891 s-1 = Tmoon

gmoon(T/2∏)2 = L
(9.8/6)( 3.8397891/2∏)2 = L = 59.419563 m

Somewhere, I've gone wrong. Any guidance would be much appreciated

## Answers and Replies

gneill
Mentor
Perhaps it would help to put off plugging in numbers until after a little bit of massaging of the formulas.

You have a formula for the period of a pendulum that relates the period to the length of the pendulum and the local acceleration due to gravity. Since the idea is to have the pendulum on the Moon have the same period as the one on the Earth, assign variable names to the lengths and accelerations and equate the expressions. You should be able to simplify the expression to extract a suitable ratio.