Determine an expression for the period of motion.

[borobeauty66]

problem solved, thanks

 

[Doc Al]

...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:
$$F = - \frac{4G\pi\rho m}{3} r$$

 

[borobeauty66]

What would k be?

k is the force constant.

 

[Doc Al]

k is the force constant.

Sure, so in your equation what would it be?

 

[borobeauty66]

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.

 

[Doc Al]

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.

 

[borobeauty66]

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

 

[Doc Al]

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?

 

[borobeauty66]

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.

 

[Doc Al]

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!)

 

[borobeauty66]

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.

 

[Doc Al]

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.

 

[borobeauty66]

Not sure if my thinking is correct.

k = 4Gπρ/3

 

[Doc Al]

Not sure if my thinking is correct.

k = 4Gπρ/3

Almost! (You left out the mass.) But that's exactly the right idea.

 

[borobeauty66]

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.

 

[Doc Al]

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.)

 

[borobeauty66]

Well, what's the period for the standard SHM equation? (In terms of k.)

You mean the frequency? ω?

 

[Doc Al]

You mean the frequency? ω?

Well, you want the period. But you can express the period in terms of the frequency.

 

[borobeauty66]

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.

 

[Doc Al]

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.

 

[borobeauty66]

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?

 

[Doc Al]

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.

 

[borobeauty66]

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?

 

[Doc Al]

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.

 

[borobeauty66]

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

 

[Doc Al]

Good. Now simplify that as much as you can.

 

[borobeauty66]

Good. Now simplify that as much as you can.

Now, there's a challenge!

 

[borobeauty66]

Anyway, I think I got it from here (minus simplifying it) Many thanks for your help!

 

[Doc Al]

Good! And you're quite welcome.

 

[borobeauty66]

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.

 

[Doc Al]

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

Not quite sure what you mean. In any case, when you simplify your expression in post #25, the mass cancels out.

 

[borobeauty66]

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

 

[borobeauty66]

This is the original equation.

I'm now wondering if

F = -k x is infact equal to the above?

 

[Doc Al]

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.

 

[borobeauty66]

Ah ok.

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

Still not sure how I can simplify this but thanks.