# What will happend for simple pendulum over the moon? [closed]

What will happend for this:( gravity - time - length - frequency - number of oscillations) if we do the simple pendulum experiment over the moon?

• Hi and welcome to the Physics SE! What are your thoughts? Please note that you are expected to have thoroughly searched for an answer before asking your question. And it's important to detail where you're stuck and why, in order to attract good answers. You can consider checking the advice on writing good questions. – stafusa Jan 19 '18 at 17:31
• Hi and welcome to physics.SE! Please note that homework-like questions and check-my-work questions are generally considered off-topic here. We intend our questions to be potentially useful to a broader set of users than just the one asking, and prefer conceptual questions over those just asking for a specific computation. – ACuriousMind Jan 19 '18 at 17:32

In moon, if you take a pendulum of earth［keeping the system unchanged,you will see that the pendulum swings slowly...

The system is unchanged, that is the possible friction at the knot of support and thread hanging from it is unchanged, so from the formula of time period , you can say easily that "L" remains constant and so "g" must decrease...

Besides, the air of moon is not "strong" enough to hold back the slowly-moving pendulum...This it's air resistance is negligible or there is hardly any air...

Nothing spectacular. The same physics apply to the pendulum as on earth.

So

T = 2 pi * sqrt(L/g)

So the swing period would increase and the momentum would be less, but that is about it.

Try it out here http://hyperphysics.phy-astr.gsu.edu/hbase/pend.html

as acceleration due to gravity is 1/6th of that of Earth on the moon so the time period of pendulum should increase T=2*pi*sqrt(l/g)