# Projection of elliptical motion

SETUP :- Here I have a line that rotates with a constant angular velocity and intersects a circle and an ellipse. The ellipse's major axis is equal in length to the diameter of the circle. The intersection points are $$A$$ and $$B$$. Now we take the projection of $$A$$ and $$B$$ on the $$x$$-axis to get $$C$$ and $$D$$ respectively (see the figure below)

We can easily recognize th motion of point $$D$$ as Simple Harmonic Motion (SHM). And the motion of point $$C$$ is something that I have never encountered before.

QUESTION :- Is there a name for the type of motion that the point $$C$$ undergoes? How do we describe the motion of point $$C$$? Also, where in physics does this type of motion take place?

• We have the tag harmonic-oscillator, so creating the simple harmonic motion one is superfluous since it is covered already. Dec 27, 2019 at 11:52

Using the semi-major axis of $$a$$, the semi-minor axis of $$b$$ the polar coordinates of the ellipse are $$r(\theta)$$ where $$\theta$$ is the angle the rotating line makes to the horizontal

$$r = \frac{a b}{\sqrt{a^2 + (b^2-a^2) \cos^2 \theta }} \tag{1}$$

Point C has $$x$$-coordinate of

$$x_C = r \cos \theta = \frac{a b \cos\theta}{\sqrt{a^2 + (b^2-a^2) \cos^2 \theta }} \tag{2}$$

The resulting motion, not only has a main harmonic that varies with $$\cos \theta$$, but also has higher-order harmonics of $$\cos 3\theta$$ and $$\cos 5\theta$$, etc

$$x_C \approx \left( \frac{b (11 a^2-3 b^2)}{8 a^3} \right) \cos \theta + \left( \frac{ b ( a^2-b^2 )}{8 a^3} \right) \cos 3\theta + \ldots \tag{3}$$

You are free to do a Fourier analysis on $$x_C$$ to discover the higher-order harmonics yourself.

You should also plot the function for various eccentricity values $$\epsilon = \sqrt{1 - \left( \tfrac{b}{a} \right)^2}$$ between 0 and 0.99999

For example the eccentricity $$\epsilon=0.8$$ looks like this:

It follows the general harmonic motion of $$\cos \theta$$ but with extra lumps in-between.

Now one can take $$x_C(\theta)$$, assume $$\theta = \omega t$$ and differentiate to get

$$\ddot{x}_C = - \omega^2 \left( \frac{1-2 \epsilon^2 \sin^2 (\omega t) - \epsilon^2}{(1-\epsilon^2 \cos^2 (\omega t))^2} \right) x_C \tag{4}$$

This ODE is more complex than S.H.M. of $$\ddot{x}_C = -\omega^2 x_C$$ and does not have a specific name, as it is not a common situation as far as I can tell.

Appendix

To find the polar coordinates of the ellipse use $$(x=r \cos\theta, \, y=r \sin \theta)$$ in the equation of the ellipse $$\left( \frac{x}{a} \right)^2 + \left( \frac{y}{b} \right)^2 = 1 \tag{5}$$ and solve for $$r$$.

• Thanks a lot for the answer but where does this type of motion come up in the real world?
– user243267
Nov 29, 2019 at 7:51
• Whenever something traverses an ellipse with constant angular velocity. Nov 29, 2019 at 7:51
• But that is just restating the problem. Does this motion have any physical significance?
– user243267
Nov 29, 2019 at 7:53
• I think what you are trying to ask is what is the ODE look like to get this motion, and is it seen anywhere in science. It does not ring a bell to be, but I cannot speak for the entirety of science. Nov 29, 2019 at 7:59
– user243267
Nov 29, 2019 at 8:01

$$\underline {\text {Motion of Point C}}$$

The equation of ellipse is given by $$(1)$$ and one can parametrize the curve by $$(2)$$.

$$\frac {x^2}{a^2} + \frac {y^2}{b^2} =1 \tag {1}$$ $$x=a \cos \alpha \;|\; y=b \sin \alpha \tag {2}$$

We can determine the equation of motion by using $$(2)$$ and $$(3)$$. $$\tan \theta = \frac {b}{a} \tan \alpha \text { where } \theta = \omega t \tag{3}$$

$$\underline {\text {Any Special Name?}}$$ Names are usually given to things only when the thing is of some interest. The SHM was given a name because the acceleration of point $$D$$ obeys $$(4)$$ and $$(4)$$ is encountered in various physical phenomena, like the spring-mass problem and the pendulum problem (with small angle approximation).

$$\ddot {x} = - \omega^2 x \tag {4}$$

I don't know whether the motion of point $$C$$ has a special name (my guess : most probably not). If "I just think it's a cool math problem" is your sole reason for expressing interest in the motion of point $$C$$, then I'd advise you to not be concerned about its name.

Pick a constant k to represent how far from a circle the orbit is.

Then for any angle $$\theta$$, $$C = k \cos \theta$$.

That isn't so special.

But if instead of constant angular velocity it rotated like an orbit, with the angular rotation varying, then it gets more complicated and I don't have such an easy answer. It's all simple when the angle is the independent variable and time is a dependent variable. When you make time the independent variable then it looks uglier.

• Why would $k$ be a constant? It should be a function of $\theta$.
– user243267
Nov 29, 2019 at 7:04
• $k$ just happens to be constant. You could use a plotting function to see it. Look at the ellipse that C is on. If you squeeze it by a constant amount it turns into a circle. In the example, $k \approx \frac{1}{2.8}$ Nov 29, 2019 at 7:31
• Oh so, $k$ is a multiplicative factor. I thought it as the difference.
– user243267
Nov 29, 2019 at 7:46