What Would a Pendulum's Period Be on the Moon?

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Homework Help Overview

The discussion revolves around determining the period of a pendulum on the Moon, given its period on Earth. The problem involves understanding the relationship between gravitational acceleration and pendulum motion, specifically using the formula T=2π√(l/g).

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the ratio of periods on the Moon and Earth, questioning how to approach the problem without knowing the length of the pendulum. There are discussions about the gravitational acceleration on the Moon and its relation to the period calculation.

Discussion Status

The conversation includes attempts to derive the period on the Moon by setting up equations and simplifying them. Some participants express confusion over the calculations, while others suggest careful step-by-step analysis. There is no explicit consensus on the final answer, but multiple interpretations and approaches are being explored.

Contextual Notes

Participants reference the gravitational acceleration on the Moon, noting it is approximately 1/6 that of Earth's gravity. There are also mentions of using external resources for gravitational values, raising questions about the appropriateness of such references in the context of the homework.

Coco12
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Homework Statement



if a pendulum has a period of .36s on Earth, what would its period be on the moon

Homework Equations


T=2pi sqrt l/g


The Attempt at a Solution


How do u go about solving thAt without length?
 
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Set up the ratio $$\frac{T_{\rm moon}}{T_{\rm earth}}$$ and see what cancels.
 
Hi Coco12! :wink:
Coco12 said:
How do u go about solving thAt without length?

This is a dimensions question …

you can solve it without knowing the things that don't change.​

The length doesn't change, so call it "l", and write out the two equations for the Earth and the moon …

what do you get? :smile:
 
Is the gravity of the moon 1/6 of the earth?
 
yep. roughly that. (sorry for butting in).
 
So you are saying: I have to put 2pi sqrt l/1/6g over 2pisqrt l/g

2pi cancel..
 
yeah, what else cancels?
 
you square them so l cancel leaving you with 1/6g/g .. However I'm not getting the right answer(.88s)
 
Last edited:
  • #10
I too got the same answer and the answer seems right :)
 
  • #11
Coco12 said:
you square them so l cancel leaving you with 1/6g/g .. However I'm not getting the right answer(.88s)
You forgot the square root?
 
  • #12
nasu said:
You forgot the square root?

No, the answer comes 0.88s.
 
  • #13
Coco12, write it out carefully as:

Tearth = 2π(√L)/√(gearth)

Tmoon = 2π(√L)/√(gmoon)

now divide …

what do you get? :smile:
 
  • #14
nil1996 said:
No, the answer comes 0.88s.
Yeah I think Coco is saying that the textbook answer is 0.88s but Coco does not get this answer him/her? self. As tiny-tim is saying, Coco should go through the steps carefully to get the right answer. Also, squaring it is not necessary. It is possible to get the answer by taking stuff all under the same square root.
 
  • #15
I got it thanks
 

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