# Combining two simple harmonic motion in perpendicular directions

So, I was studying about the combination of two simple harmonic motions (S.H.M.) in the perpendicular direction withstand frequency, but with different amplitude and phases; and, I also studied the graphs in the X-Y plane, depicting the motion of a particle under the influence of forces corresponding to the above S.H.M., which looked like this

Assuming, that the values of the parameters is not needed here, as it is for representation purpose only.

I know that this absolutely doesn't look like an ellipse, but being a mathematics student, I would be only satisfied by a formal proof that this graph is not an ellipse.

Hence, my question is: how would you go on proving that the above graph is not an ellipse?

• What if it's an ellipse? Related: physics.stackexchange.com/q/146477/226902 physics.stackexchange.com/a/633043/226902 Mar 8 at 9:56
• Your answer is here: farside.ph.utexas.edu/teaching/336k/Newton/node28.html (see Figure 10). Mar 8 at 10:03
• If the frequencies are the same, then it is an ellipse. It’s the 2D projection of a circle in the 4D phase space.
– LPZ
Mar 8 at 10:06
• @Quillo thanks for the reference of that site because that's the only one I needed. Mar 8 at 10:28
• @lpz could you please elaborate your answer as I think from your answer I would get a new way of thinking for this problem. Mar 8 at 10:29

This is an ellipse. Suppose for simplicity that the phase shift is $$\pi/2$$: $$x(t)=A\cos(\omega t + \phi),\\ y(t)=B\sin(\omega t +\phi),$$ then $$\frac{x^2}{A^2}+\frac{y^2}{B^2}=1,$$ which is an equation for an ellipse. I suggest doing the case of unequal phases as an exercise (hint: express $$\sin(\omega t),\cos(\omega t)$$ in terms of $$x,y$$, and employ again the main trigonometric identity.)
Making everything dimensionless, your system of ODE's are: $$\ddot x +x=0 \\ \ddot y +y=0$$ in the $$\mathbb R^2$$ configuration space. Naturally, you go to $$\mathbb R^4$$ phase space: $$\dot x = p \\ \dot p = -x \\ \dot y = q \\ \dot q = -y$$ It is more convenient to identify $$\mathbb R^4$$ with $$\mathbb C^2$$ with: $$z = x+ip \\ w = y+iq$$ giving: $$\dot z = -iz \\ \dot w = -iw$$ This makes more apparent the conservation laws. With $$z_0,w_0$$ the initial conditions, there is conservation of energy: $$|z_0|^2+|w_0|^2 = |z|^2+|w|^2$$ so geometrically, trajectories are in a hypersphere $$\mathbb S^3$$. There is also: $$w_0z-z_0w = 0$$ so geometrically, trajectories are in a plane going through the origin. Trajectories lie in the intersection of a hypersphere and plane going through its origin, they are therefore circles of the same radius. You then project them onto the $$x,y$$ plane, which gives an ellipse.
If you're into math, this is exactly the Hopf fibration. You are essentially decomposing the energy hypersphere into circular trajectories. The "quotient" can be identified algebraically with the ratio $$z_0/w_0$$ with an extra point for the case $$w_0 = 0$$, which, topologically, is the same as the sphere $$\mathbb S^2$$.