Path Equation for 2D weakly-anisotropic harmonic oscillator

  • #1

Homework Statement


[tex]\omega_{x}[/tex] = [tex]\omega[/tex]

[tex]\omega_{y}[/tex] = [tex]\omega[/tex] + [tex]\epsilon[/tex]

where 0 < [tex]\epsilon[/tex]<<[tex]\omega[/tex]

Question: Find the path equation.

Homework Equations



I started with the 2D equations:

x(t) = A[tex]_{x}[/tex]cos([tex]\omega_{x}[/tex]t + [tex]\phi_{x}[/tex])
y(t) = A[tex]_{y}[/tex]cos([tex]\omega_{y}[/tex]t + [tex]\phi_{y}[/tex])

The Attempt at a Solution



by inverting x(t) to get t(x), I then substituted the result into y(t). The result is as below:

y(x) = A[tex]_{y}[/tex]cos[([tex]\omega_{x}[/tex]/[tex]\omega_{y}[/tex])cos[tex]^{-1}[/tex](x/A[tex]_{x}[/tex]) - ([tex]\omega_{x}[/tex]/[tex]\omega_{y}[/tex])[tex]\phi_{x}[/tex] + [tex]\phi_{y}[/tex]]

I guess it becomes more of a mathematical problem. How do I simplify this equation, hopefully to find something familiar? An idea I had was to use double-angle trig formulas but I am not sure how it would help.

I personally think this is quite challenging. I have been thinking about it for days now. I tried all the trig identities I know off to manipulate the equation but I can't seem to get it. Unless I am missing something.

Thanks in advance.

PS: A^x is actually A(subscript)x and similarly, A^y is actually A(subscript)y. I don't know what is wrong with the formatting.
 
Last edited:

Answers and Replies

  • #2
maybe I should define what the symbols are:

[tex]\omega_{x}[/tex] is the angular frequency in the x-axis
[tex]\omega_{y}[/tex] is the angular frequency in the y-axis
 
  • #3
211
0
I'd try to substitute the [tex]\omega[/tex] you have and then try to expand it by small parameter [tex]\epsilon[/tex]...
 

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