Homework Help: Simple harmonic motion period

1. Nov 5, 2013

Coco12

1. The problem statement, all variables and given/known data

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

2. Relevant equations
T=2pi sqrt l/g

3. The attempt at a solution
How do u go about solving thAt without length?

2. Nov 5, 2013

TSny

Set up the ratio $$\frac{T_{\rm moon}}{T_{\rm earth}}$$ and see what cancels.

3. Nov 5, 2013

tiny-tim

Hi Coco12!
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?

4. Nov 5, 2013

Coco12

Is the gravity of the moon 1/6 of the earth?

5. Nov 5, 2013

tiny-tim

approx

according to http://en.wikipedia.org/wiki/Moon it's 1.622 m/s2

(are you allowed to use that?)

6. Nov 5, 2013

BruceW

yep. roughly that. (sorry for butting in).

7. Nov 5, 2013

Coco12

So you are saying: I have to put 2pi sqrt l/1/6g over 2pisqrt l/g

2pi cancel..

8. Nov 5, 2013

BruceW

yeah, what else cancels?

9. Nov 5, 2013

Coco12

you square them so l cancel leaving you with 1/6g/g .. However I'm not getting the right answer(.88s)

Last edited: Nov 5, 2013
10. Nov 5, 2013

nil1996

I too got the same answer and the answer seems right :)

11. Nov 5, 2013

nasu

You forgot the square root?

12. Nov 5, 2013

nil1996

13. Nov 6, 2013

tiny-tim

Coco12, write it out carefully as:

Tearth = 2π(√L)/√(gearth)

Tmoon = 2π(√L)/√(gmoon)

now divide …

what do you get?

14. Nov 6, 2013

BruceW

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. Nov 6, 2013

Coco12

I got it thanks