Determine an expression for the period of motion.

Click For Summary

Homework Help Overview

The discussion revolves around determining an expression for the period of motion in the context of simple harmonic motion (SHM) and gravitational forces. Participants are exploring the relationship between force, mass, and the period of oscillation.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the form of the force equation and its relation to SHM, questioning how to express the period in terms of the force constant and mass. There is an exploration of the equivalent values for constants in the equations and how they relate to the period of motion.

Discussion Status

Several participants have provided insights into the relationship between the force constant and the period of motion. There is ongoing exploration of how to simplify the derived expressions, with some participants expressing uncertainty about rearranging equations and the role of mass in the final expression.

Contextual Notes

Participants note that they are required to find an algebraic expression without specific numerical values. There is also a discussion about the implications of the mass in the equations and how it factors into the final expression for the period.

  • #31
borobeauty66 said:
Just realized that mass isn't part of the 4Gpi p/3 equation, it comes later. :-P
Not quite sure what you mean. In any case, when you simplify your expression in post #25, the mass cancels out.
 
Physics news on Phys.org
  • #32
Doc Al said:
Not quite sure what you mean. In any case, when you simplify your expression in post #25, the mass cancels out.

What I mean is the equation give at the start is F = -(4Gpi p/ 3) m r r^

It's not that included in the brackets. So I suppose it's not included in the value of K
 
  • #33
gif.latex?F%20=%20-(\frac{4G\pi%20\rho%20}{}3)%20m%20r%20\hat{r}.gif


This is the original equation.

I'm now wondering if

F = -k x is infact equal to the above?
 
  • #34
borobeauty66 said:
gif.latex?F%20=%20-(\frac{4G\pi%20\rho%20}{}3)%20m%20r%20\hat{r}.gif


This is the original equation.

I'm now wondering if

F = -k x is infact equal to the above?
They both have the same form: Force = -(some constant)x, where x is the displacement from some equilibrium point. That's all that matters.
 
  • #35
Ah ok.

So final equation is t = 2π/(√4Gπρ/3)/m

Still not sure how I can simplify this but thanks.
 
  • #36
The mass should cancel and thus not appear in your final version.
 
  • #37
What I meant by saying the equation was wrong was in your #3 post you put the equation wrong. In actual fact the mass should come after the division. Thus

gif.latex?F%20=%20\frac{4G\pi%20\rho%20}{3}%20m%20r%20\hat{r}.gif


and not

gif.latex?F%20=%20\frac{4G\pi%20\rho%20m}{3}%20r.gif


which would make k = 4Gπρ and not k = 4Gπρm

Surely this means the m isn't canceled out at the end?
 
  • #38
borobeauty66 said:
What I meant by saying the equation was wrong was in your #3 post you put the equation wrong. In actual fact the mass should come after the division. Thus

gif.latex?F%20=%20\frac{4G\pi%20\rho%20}{3}%20m%20r%20\hat{r}.gif


and not

gif.latex?F%20=%20\frac{4G\pi%20\rho%20m}{3}%20r.gif
Those expressions are identical! (Except for the unneeded unit vector.) For the same reason that (a/b)x is the same as (ax/b).
 
  • #39
Doc Al said:
Those expressions are identical! (Except for the unneeded unit vector.) For the same reason that (a/b)x is the same as (ax/b).

ah ok!

thats what confused me, i thought they were different. :-P
 

Similar threads

  • · Replies 10 ·
Replies
10
Views
2K
Replies
25
Views
1K
Replies
25
Views
3K
  • · Replies 10 ·
Replies
10
Views
2K
Replies
2
Views
2K
  • · Replies 51 ·
2
Replies
51
Views
4K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
6
Views
2K
  • · Replies 14 ·
Replies
14
Views
3K
  • · Replies 7 ·
Replies
7
Views
2K