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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)

Projection of circular and elliptical motion

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?

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    $\begingroup$ We have the tag harmonic-oscillator, so creating the simple harmonic motion one is superfluous since it is covered already. $\endgroup$ – Kyle Kanos Dec 27 '19 at 11:52
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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:

Wolfram

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 $r$.

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  • $\begingroup$ Thanks a lot for the answer but where does this type of motion come up in the real world? $\endgroup$ – user243267 Nov 29 '19 at 7:51
  • $\begingroup$ Whenever something traverses an ellipse with constant angular velocity. $\endgroup$ – ja72 Nov 29 '19 at 7:51
  • $\begingroup$ But that is just restating the problem. Does this motion have any physical significance? $\endgroup$ – user243267 Nov 29 '19 at 7:53
  • $\begingroup$ 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. $\endgroup$ – ja72 Nov 29 '19 at 7:59
  • $\begingroup$ I understand. Again thanks for your answer. $\endgroup$ – user243267 Nov 29 '19 at 8:01
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$$\underline {\text {Motion of Point $C$}} $$

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.

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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.

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  • $\begingroup$ Why would $k$ be a constant? It should be a function of $\theta$. $\endgroup$ – user243267 Nov 29 '19 at 7:04
  • $\begingroup$ $k$ 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, $k \approx \frac{1}{2.8}$ $\endgroup$ – J Thomas Nov 29 '19 at 7:31
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    $\begingroup$ Oh so, $k$ is a multiplicative factor. I thought it as the difference. $\endgroup$ – user243267 Nov 29 '19 at 7:46

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