Simple harmonic motion period

[Coco12]

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?

 

[TSny]

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

 

[tiny-tim]

Hi Coco12!

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? ​

 

[Coco12]

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

 

[tiny-tim]

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

approx

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

(are you allowed to use that?)

 

[BruceW]

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

 

[Coco12]

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

2pi cancel..

 

[BruceW]

yeah, what else cancels?

 

[Coco12]

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

 

[nil1996]

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

 

[nasu]

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?

 

[nil1996]

You forgot the square root?

No, the answer comes 0.88s.

 

[tiny-tim]

Coco12, write it out carefully as:

T_earth = 2π(√L)/√(g_earth)

T_moon = 2π(√L)/√(g_moon)

now divide …

what do you get? ​

 

[BruceW]

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.

 

[Coco12]

I got it thanks