What Constitutes Circular Motion in Classical Dynamics?

[Tangent87]

Hi, I'm doing this Classical Dynamics section II question which can be found here (http://www.maths.cam.ac.uk/undergrad/pastpapers/2008/Part_2/list_II.pdf ) on page 27.

I have done most of the question but am unsure about the last part. Specifically using Hamilton's equations to show there's circular motion of radius r with the angular frequency given. I am basically just unsure of what you actually have to show, I mean they've already told us that r^2=x_1^2+x_2^2 so what actually constitutes circular motion? I can show from Hamilton's equations that

\stackrel{..}{x_i}=(p_3-\frac{eF}{c})\frac{e}{m^{2}cr}\frac{dF}{dr}x_i


for i=1,2 getting someway towards the expression for the angular frequency but don't really know where to go from here seeing as r depends on BOTH x_1 and x_2 so that differential equation isn't of SHM form!

 

[Tangent87]

I'm still stuck on this, I know that two SHM systems acting perpendicular to each other produce circular motion but that's not what we have in this case (or is it?) since the r depends on the x_i!

 

[zzzoak]

We have

<br /> x_i&#039;&#039;=-\Omega^2(r) \; x_i<br />

Making a square and adding

<br /> x_1&#039;&#039;^2+x_2&#039;&#039;^2=\Omega^4(r) \; (x^2_1+x^2_2)=\Omega^4(r) \; r^2.<br />

RHS depends only on r, it follows and LHS too

<br /> a^2(r)=\Omega^4(r) \; r^2.<br />

And we get

<br /> a(r)=-\Omega^2(r) \; r.<br />

 

[Tangent87]

I see what you've done, thanks.