Determine an expression for the period of motion.

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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.

borobeauty66
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problem solved, thanks
 
Last edited:
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borobeauty66 said:
...4Gπρ
F =−_______ mrrˆ (this second r^, being a unit vector)
...3

Fx = -kx
I'm not understanding the problem. You have the equation you need, just change r to x. What would k be?

The real question is: Can you solve it to find the period? (Compare it to the equations for simple harmonic motion.)

FYI: Writing your equations using Latex will make things easier:
[tex]F = - \frac{4G\pi\rho m}{3} r[/tex]
 
Last edited:
Doc Al said:
What would k be?

k is the force constant.
 
borobeauty66 said:
k is the force constant.
Sure, so in your equation what would it be?
 
Doc Al said:
Sure, so in your equation what would it be?

k = mω^2

We haven't been given the values for anything, we're supposed to find the expression in a format of algerbra only.
 
borobeauty66 said:
k = mω^2
That's the solution for the usual simple harmonic motion equation. Since your equation (for gravity) has the same form, it will have the same solution. Compare the two equations you wrote before: What will take the place of 'k' in your equation?
We haven't been given the values for anything, we're supposed to find the expression in a format of algerbra only.
Right.
 
Doc Al said:
That's the solution for the usual simple harmonic motion equation. Since your equation (for gravity) has the same form, it will have the same solution. Compare the two equations you wrote before: What will take the place of 'k' in your equation?

Right.
Do you mean F = -kx

and

k = mω^2

Therefore

F = -mw^2 x
 
borobeauty66 said:
Do you mean F = -kx
I mean to take that equation and compare it to the one you wrote for your gravity problem. What would be the equivalent to k?
 
Doc Al said:
I mean to take that equation and compare it to the one you wrote for your gravity problem. What would be the equivalent to k?

OK give me a few minutes. I'm not very quite when it comes to rearranging and combing equations.
 
  • #10
borobeauty66 said:
OK give me a few minutes. I'm not very quite when it comes to rearranging and combing equations.
You shouldn't have to rearrange anything, just read it off.

Your equation has the form: Force = (some constant) x, what's that constant? (It's much easier than you think!)
 
  • #11
Doc Al said:
You shouldn't have to rearrange anything, just read it off.

Your equation has the form: Force = (some constant) x, what's that constant? (It's much easier than you think!)

I'm over-complcating things then?

I've just got to pop out anyway, so I'll have a think about it.
 
  • #12
borobeauty66 said:
I'm over-complcating things then?
I'd say so. You know the solution for the standard SHM problem of a mass on a spring. So you want to compare the equations, since the same equations have the same solutions.
 
  • #13
Not sure if my thinking is correct.

k = 4Gπρ/3
 
  • #14
borobeauty66 said:
Not sure if my thinking is correct.

k = 4Gπρ/3
Almost! (You left out the mass.) But that's exactly the right idea.
 
  • #15
Doc Al said:
Almost! (You left out the mass.) But that's exactly the right idea.

Oh yes, mass sorry, so,

k = 4Gπρm/3

Ok, I'm with you up to hear, just unsure how this applies to period of motion, which is what the question asks for.
 
  • #16
borobeauty66 said:
Oh yes, mass sorry, so,

k = 4Gπρm/3

Ok, I'm with you up to hear, just unsure how this applies to period of motion, which is what the question asks for.
Well, what's the period for the standard SHM equation? (In terms of k.)
 
  • #17
Doc Al said:
Well, what's the period for the standard SHM equation? (In terms of k.)

You mean the frequency? ω?
 
  • #18
borobeauty66 said:
You mean the frequency? ω?
Well, you want the period. But you can express the period in terms of the frequency.
 
  • #19
Doc Al said:
Well, you want the period. But you can express the period in terms of the frequency.

Oh I see, the period, t.

So, since ω = 2π/t

t =2π/ω (is this correct? as i say, I'm terrible as rearranging equations)

Not sure where to go from here.
 
  • #20
borobeauty66 said:
Oh I see, the period, t.

So, since ω = 2π/t

t =2π/ω (is this correct? as i say, I'm terrible as rearranging equations)
Good.

Not sure where to go from here.
Now use the expression for ω.
 
  • #21
Hmmm, If I were to make F = -k x
into F= - mw^2 x
and then
F = -m (2π/t) x

and then rearrange to make t the subject, wouldn't this give me the period?
 
  • #22
borobeauty66 said:
Hmmm, If I were to make F = -k x
into F= - mw^2 x
and then
F = -m (2π/t) x
You forgot the square.
and then rearrange to make t the subject, wouldn't this give me the period?
Lets keep it real simple. Forget your gravity problem for the moment. You have the SHM relationship F = -kx. Tell me the period in terms of k and m.
 
  • #23
Doc Al said:
You forgot the square.

Lets keep it real simple. Forget your gravity problem for the moment. You have the SHM relationship F = -kx. Tell me the period in terms of k and m.

So, F = -k x

-k = F/x

That makes

-mw^2 = f/x

-w = square root of f/x +m

Am i going in the wrong direction?
 
  • #24
borobeauty66 said:
So, F = -k x

-k = F/x

That makes

-mw^2 = f/x

-w = square root of f/x +m

Am i going in the wrong direction?
Yes. You already have the solution. From post #5, solve for ω in terms of m and k. And then use what you told me in post #19 to get period in terms of ω and then m and k. When all is said and done, I want to see an expression for period in terms of m and k only.

Then all you have to do to apply this standard solution to your problem is to replace k by what you told me it equaled in post # 15.
 
  • #25
OK,

so,

k = mω^2
ω = 2π/t = √k/m

2π/t = √k/m

makes

t=2π/√k/m

So replacing k

t = 2π/√(4Gπρm/3)/m
 
  • #26
Good. Now simplify that as much as you can.
 
  • #27
Doc Al said:
Good. Now simplify that as much as you can.

Now, there's a challenge!
 
  • #28
Anyway, I think I got it from here (minus simplifying it) Many thanks for your help!
 
  • #29
Good! And you're quite welcome.
 
  • #30
Doc Al said:
Almost! (You left out the mass.) But that's exactly the right idea.

Just realized that mass isn't part of the 4Gpi p/3 equation, it comes later. :-P
 

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