Does the Sum of Two Spinning Sinusoidal Vectors Trace out a Limaçon?

Question

The Question:

Picture two vectors in the complex plane spinning at different angular frequencies, one spinning at an angular frequency twice of the other.

$$Z_1=e^{i\omega t} \ \ \ \& \ \ \ \ Z_2=e^{i2\omega t}$$

Alternatively...

$$Z_1=\cos(\omega t)+i \sin (\omega t) \ \ \ \& \ \ \ \ Z_2=\cos(2\omega t)+i \sin (2\omega t) $$

I was playing around with this Java application I downloaded, made primarily to understand the concept of "beat frequencies," (linked at end, in case you want to download) and started looking at the sum of these two spinning vectors.

In the picture above, the green vector attached to the origin is $Z_2$, the blue vector attached to the origin is $Z_1$, and the red vector is their sum. The dot indicates the projection of their sum onto the real axis.

I'm interested in the figure which the red phasor traces as a function of time. I've captured a gif - you can access it with the following link:

https://gyazo.com/97807fe65da1666e4271ce5906541d0a

It seems to me, although I may be wrong, that the endpoint of the red phasor is tracing out a limaçon...or at least, something similar to it.

After looking around on the web for a little, I could not find any sources to confirm or deny this...or in fact, any sources talking about the shape which the sum of spinning vectors form at all.

I'd appreciate any help! Both information, and direction to any resources you may know!

Thanks.

Link to Java Application Download:

https://www.compadre.org/introphys/items/detail.cfm?ID=8174

Answer 1

It does not trace out a limacon but rather a shape whose radius at any point is the square root of the radius of a limacon. It looks rather like a limacon but is not in reality.

To see why this is we can find an expression for the radius of the complex number $Z_1+Z_2$ where $Z_1 = e^{i \omega t}$ and $Z_2 = e^{i 2\omega t}$.

$Z_1 + Z_2 = e^{i \omega t} + e^{i 2\omega t} = (\cos(\omega t) + \cos(2\omega t))+i(\sin(\omega t) + \sin(2\omega t))$

$r^2 = (Re(Z_1+Z_2))^2 + (Im(Z_1+Z_2))^2 = (\cos(\omega t) + \cos(2\omega t))^2 + (\sin(\omega t) + \sin(2\omega t))^2$

$= \cos^2(\omega t) + 2\cos(\omega t)\cos(2\omega t) + \cos^2(2\omega t) + \sin^2(\omega t) + 2\sin(\omega t)\sin(2\omega t) + \sin^2(2\omega t)$

$ = 2+2\cos(\omega t)\cos(2\omega t)+2\sin(\omega t)\sin(2\omega t)$

$=2(1+\cos(\omega t)\cos(2\omega t)+\sin(\omega t)\sin(2\omega t))$

$=2(1+\cos(\omega t)(\cos^2(\omega t) - \sin^2(\omega t)) + \sin(\omega t)(2\sin(\omega t)\cos(\omega t)) )$

$= 2(1+\cos^3(\omega t) - \cos(\omega t)\sin^2(\omega t) + 2\cos(\omega t)\sin^2(\omega t))$

$=2(1+\cos^3(\omega t) + \cos(\omega t)\sin^2(\omega t)) = 2(1+\cos^3(\omega t) + \cos(\omega t)(1-\cos^2(\omega t)))$

$ = 2(1+\cos^3(\omega t) + \cos(\omega t)-\cos^3(\omega t)) = 2(1+\cos(\omega t))$

So we have that $r^2 = 2(1+\cos(\omega t))$ which means $r = \sqrt{2+2\cos(\omega t)}$.

Now $\omega t$ is the angle $\theta$ at any point in time as the sum of the two phasors/complex numbers rotates. So the equation for the radius of the curve in polar coordinates is $r = \sqrt{2+2\cos(\theta)}$.

The general equation of a limacon is $r = a + b\cos(\theta)$. So the curve traced out is not a limacon but rather a curve whose radius at any angle $\theta$ is the square root of the radius of the limacon $r = 2 + 2\cos(\theta)$ at the same angle $\theta$.

Here is a graph from Desmos showing the limacon $r = 2 + 2\cos(\theta)$ in blue and the curve traced out by your two phasors in red. As you can see the radius at any angle of the red curve is the square root of the radius at that angle of the blue curve.

Answer 2

I have had a look at the previous answer by @mihirb that I have upvoted in particular for its graphics. Nevertheless

1. The final answer $r=\sqrt{2+2\cos(\theta)}$ can be given the simplified form :

$$r = 2\left|\cos(\tfrac12 \theta)\right|$$

providing IMHO an easier understanding of the curve, in particular the presence of a unique cusp and its symmetry which respect to the abscissas' axis.

2. Moreover, the proof can be considerably shortened by considering that $e^{i \omega \theta}+(e^{i \omega \theta})^2$ is the image of the unit circle by function

$$z \mapsto z+z^2=(z+\tfrac12)^2-\tfrac14,$$

otherwise said, the image of a translated circle by function $z \mapsto z^2$ followed by a second translation (this latter one doesn't affect of course the shape of the image).

Remark : function $f(z)=z^2$ is essential for the study of the cardioid and its very interesting connection with Mandelbrot set.