Question

Help with picture.

Answer 1

Help me set up the problem please.

Answer 2

@John_ES maybe?

Answer 3

Doyou know Lagrangian equations?

Answer 4

Anyways, I put here the solution obtaind with Lagrange equations of motion,
\[L=T-U=\frac{1}{2}m(\dot{x}^2+\dot{y}^2)+8(x^2+4y^2)\]
\[\frac{d}{dt}\left(\frac{\partial L}{\partial x}\right)-\frac{\partial L}{\partial x}=0\\
\frac{d}{dt}\left(\frac{\partial L}{\partial y}\right)-\frac{\partial L}{\partial y}=0\]
\[\ddot{x}+\frac{16}{m}x=0\Rightarrow x=A_1\cos(\omega_1t+\phi_1)\]
\[\ddot{y}+\frac{64}{m}y=0\Rightarrow y=A_2\cos(\omega_2t+\phi_2)\]
Where,
\[\omega_1=2\sqrt{2}\\
\omega_2=4\sqrt{2}\]
Then you only need to apply the initial conditions to find the A's and phi's.

Answer 5

Also the path is closed, because,
\[\omega_2/\omega_1=2\] is an entire number (it will be closed if it is a rational number, in general). I attach you the picture.

Answer 6

Ohhhh I see now. Gosh these things are never simple anymore. Thanks John! I do have a few more I'm a bit stuck with. :/

Answer 7

Post them, and we'll see if it is possible to solve them.

Answer 8

I'm really sorry but I think I am doing the initial conditions wrong. I am thinking of using t=0 in the x equation but that doesn't really give me A nor phi. Should I use r0 and v0 in the homogeneous equations instead? I am confused.