Determine an expression for the period of motion.

[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.

 

[Doc Al]

The mass should cancel and thus not appear in your final version.

 

[borobeauty66]

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

and not

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

Surely this means the m isn't canceled out at the end?

 

[Doc Al]

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

and not

Those expressions are identical! (Except for the unneeded unit vector.) For the same reason that (a/b)x is the same as (ax/b).

 

[borobeauty66]

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.