If we had two waves perpendicular to each other, with equations:

$x=α \sin(ωt)$ (1)

$y=β \sin(ωt+π/2) ==> y=β \cos(ωt)$(2)


$x^2/α^2+y^2/β^2=1$ is an equation of an ellipse and when $α=β$, we have a circle. enter image description here

Now how can we find the path equation of a point when one wave interfere in an angle $π/4$ with the other?.

I found that the equation of a $\sin()$ function in $θ$ rad can be found here: http://mathman.biz/html/rotatingsine.html


Expand $\sin(\omega t + \phi) = \sin(\omega t) \cos(\phi) + \cos(\omega t) \sin(\phi)$. Substituting $x$ into it we get

$\frac{y}{\beta} = \frac{x}{\alpha} \cos(\phi) + \sqrt{1-(\frac{x}{\alpha})^2} \sin(\phi)$

$(\frac{y}{\beta} - \frac{x}{\alpha} \cos(\phi))^2 = (1-(\frac{x}{\alpha})^2) \sin(\phi)^2$

After simplification

$(\frac{y}{\beta})^2 + (\frac{x}{\alpha})^2 -2 \frac{x y}{\alpha \beta}\cos(\phi) - \sin(\phi)^2 = 0$

which is an ellipse (might be degenerate).

This is not the nicest way to do it, but straightforward.


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