Find the path of a particle given a potential function.

[693Janu]

I am tasked with finding the path a particle takes through this potential field.
$$U(x,y) = x^2+xy+y^2$$
I then took the gradient, and this produced a pair of differential equations.
$$\frac{d^2x}{dt^2}=\frac{1}{m}(-2x-y)$$
$$\frac{d^2x}{dt^2}=\frac{1}{m}(-2y-x)$$
I have yet to encounter set of coupled differential equations in this form. It looks simple, but it has proven to be very tricky to solve.

Any thoughts? Thank you!

 

[mfb]

There is a different set of coordinates where the differential equations become independent.

 

[693Janu]

There is a different set of coordinates where the differential equations become independent.

I was given initial coordinates $$(3,4)$$

 

[mfb]

My comment was independent of initial conditions.

 

[Charles Link]

@693Janu Your second equation should read ## d^2 y/dt^2 =... ##. And if I understood the hint by @mfb , try adding the two differential equations together. The necessary change of coordinates becomes obvious, I believe. Also, subtract one equation from the other. This will give you two sets of solvable equations, from which you can then solve for ## x ## and ## y ##.

 

[693Janu]

Great advice, thank you so much. However, I have never done this discussed "change of coordinates before". I do see how after adding the two equations together I can say... let $$u=x+y$$ and that gives a nice SHM solution. But how do I extract x(t) from that answer? Or do I even need to?

 

[Charles Link]

Great advice, thank you so much. However, I have never done this discussed "change of coordinates before". I do see how after adding the two equations together I can say... let $$u=x+y$$ and that gives a nice SHM solution. But how do I extract x(t) from that answer? Or do I even need to?

Also subtract the two equations and let ## v=x-y ##. Solve for ## v ##. You have solutions for ## u ## and ## v ##. Solve for ## x ## and ## y ##. (This part I edited/added to my post 5, maybe after you already looked at it.)

 

[693Janu]

I had that intuition before I even read your post (personal high five), and it worked out! I have two solutions and I can now solve for Initial conditions. That's great! However, the last tricky step is finding the period. How would I find the period with two different omega values? Because I have $${w_1}^2 = 3/m $$ $${w_2}^2 = 1/m $$

 

[Charles Link]

I had that intuition before I even read your post (personal high five), and it worked out! I have two solutions and I can now solve for Initial conditions. That's great! However, the last tricky step is finding the period. How would I find the period with two different omega values? Because I have $${w_1}^2 = 3/m $$ $${w_2}^2 = 1/m $$

Believe it would be the slower frequency, given that the x motion is a multiple (3) of the slower one (y). The motion will be periodic at the slower one=it will return to the same place at the slower rate. Did I interpret that correctly, or does ## u ## have a frequency that is 3x that of ## v ##?

 

[mfb]

The squared frequencies have a a factor 3, that doesn't give a common frequency for the motion.

 

[693Janu]

This was fun! Thanks so much! Fantastic finish to a great weekend.

 

[Charles Link]

Additional comment: The function ## u ## has frequency ##\omega_1=3 \omega_2 ##, where ## \omega_2 ## is the frequency of v. i.e. ## \omega=2 \pi f ##. If ## u ## and ## v ## are both periodic with this slower ## f_2 ##, then ## x ## and ## y ## will also repeat at this same frequency, because ## x=(u+v)/2 ##, and ## y=(u-v)/2 ##. The period is ## T=1/f_2=2 \pi/\omega_2 ##. ## \\ ## And additional note: The complete solution with initial coordinates will also require an initial velocity.

 

[mfb]

@Charles Link: The factor 3 should appear between the squared frequencies, not between the frequencies.

 

[Charles Link]

@Charles Link: The factor 3 should appear between the squared frequencies, not between the frequencies.

Thank you, my mistake. That does make it a little more difficult finding the period I think. Perhaps it then becomes aperiodic.

 

[693Janu]

Additional comment: The function ## u ## has frequency ##\omega_1=3 \omega_2 ##, where ## \omega_2 ## is the frequency of v. i.e. ## \omega=2 \pi f ##. If ## u ## and ## v ## are both periodic with this slower ## f_2 ##, then ## x ## and ## y ## will also repeat at this same frequency, because ## x=(u+v)/2 ##, and ## y=(u-v)/2 ##. The period is ## T=1/f_2=2 \pi/\omega_2 ##. ## \\ ## And additional note: The complete solution with initial coordinates will also require an initial velocity.

Imagine that, I was given initial velocity.

 

[Charles Link]

Imagine that, I was given initial velocity.

The solutions you have should be of the form ## u=A \cos(\omega_1 t+\phi_1 ) ## and ## v=B \cos(\omega_2 t+\phi_2) ##. It remains to solve for ## A ## , ## B ##, ## \phi_1 ##, and ## \phi_2 ## from the initial conditions. Meanwhile @mfb had an important input, catching my mistake so that ## \omega_1=\sqrt{3} \omega_2 ##. My first instincts is that this thing is aperiodic, but I could be wrong.

 

[mfb]

It is aperiodic as ##\sqrt{3}## is not a rational number.