Is there a simple geometric proof of why two simple harmonic oscillators draw a circle?

Question

We all know that a circle can be drawn with the trigonometric functions $x=\cos(t), y=\sin(t)$. If we define the sine and cosine functions in terms of triangles (like we do in high school), then this is quite obvious.

But then later on in our education, we learn that the solution to a simple harmonic oscillator is the sine function. A weight on an undamped spring goes back and forth following a sine wave over time, and that this is the intuition behind a lot of wave motion (like sound waves).

However, it's not generally taught why the sine wave solution to simple harmonic motion is the same function as the sine wave as defined by triangles or circles. Or in other words, when you take two harmonic oscillators and plot their outputs as $x=\cos(t), y=\sin(t)$, why should they make a perfect circle?

More specifically: it's intuitive enough that they must form a shape that makes a full loop of some kind. But why does it happen to be a perfect circle, as opposed to an alternative shape like a larger "squircle", or something smaller similar to a rhombus with rounded corners?

After doing a bit of research, the best answer I've found is that if you independently derive the Taylor series for each of:

• $\sin(t)$ as defined by simple harmonic oscillator (as the negative of its second derivative)
• $\sin(x)$ as defined trigonometrically (as the opposite side of a right triangle over the hypoteneuse)

Then you discover they're the same Taylor series and therefore the same function and therefore simple harmonic oscillators can be used to draw a circle.

But I'm wondering two things.

First, is there some kind of more intuitive, direct, ideally geometric explanation? Resorting to Taylor series certainly works as a formal proof, but it doesn't build any kind of intuitive understanding. And for something so foundational to math and engineering as circles and sine waves, it seems like there should be a more obvious link.

And second, if there is no such intuitive explanation and we're forced to resort to the Taylor series equivalence, is there an easy yet rigorous way to derive that that doesn't involve mixing up the two origins/definitions of the sine function to do so? E.g. every example I can find of deriving the Taylor series expansion of a simple harmonic oscillator involves using a trigonometric identity at some point, e.g. to show that the derivative of sine is cosine. But this defeats the whole purpose because it assumes what I'm trying to prove -- that the simple harmonic oscillator sine wave is the same as the trigonometric sine wave in the first place.

Answer 1

Nice question! You might reverse this, asking why an object with uniform circular motion looks like a harmonic oscillator when you project it to one dimension (as Greg Martin alludes to in the comments.) One way to see this is to imagine a ball tethered to a post moving in a circle. The tension force from the tether acts as centripetal force on the ball pointing toward the post. So if the ball is at point $(x,y)$ and the post is at the origin, the force acts in direction $(-x, -y)$, and so the acceleration is proportional to $(-x, -y)$; that is, $\frac{\partial^2}{\partial t^2}(x,y) = (-cx, -cy)$. If we look at just the $x$-component, we get $\frac{\partial^2}{\partial t^2}x = -cx$, the equation for a harmonic oscillator.

Answer 2

One way to see this is from conservation of energy. Potential energy of a harmonic spring is proportional to elongation squared $x^2$, and kinetic energy is proportional to velocity squared, $v^2=\dot x ^2$, so for some constants representing the spring tension and the mass, $$a x^2 + b \dot x ^2$$ is conserved, so the velocity and the position are on an ellipse, and thus a good parametrization is to consider $x \sim \sin(\theta)$. A little algebra will convince you that $\theta$ is linear with time.

Answer 3

Here's a geometric argument to complement the existing (+1) answers:

Suppose a point (the black dot at the end of the gray arrow) moves at unit speed counterclockwise around the unit circle. When the point is at $(x, y)$, a vector of unit length, its velocity $(x', y')$ is

1. A unit vector,
2. Orthogonal to the position (because we're moving on a circle), and
3. Rotated counterclockwise from the position (because the point moves counterclockwise).

On the other hand, by elementary geometry in Cartesian coordinates, $(-y, x)$ is the unique vector satisfying all three conditions. We conclude \begin{align*} x' &= -y, \\ y' &= \phantom{-}x. \end{align*} Differentiating the first equation and using the second, we deduce $$ x'' = (x')' = (-y)' = -y' = -x; $$ similarly, $y'' = -y$.

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This amounts to the identity $$ e^{it} = \cos t + i\sin t, $$ from which we formally deduce $$ \frac{d^{2}}{dt^{2}}(e^{it}) = (i^{2})e^{it} = -e^{it}. $$

Answer 4

Is there a simple geometric proof of why two simple harmonic oscillators draw a circle?

No, there is no geometric proof, because it is not always true. If the two oscillators are oscillating with different amplitudes, the resulting figure is an ellipse, not a circle. If the two oscillators are oscillating at to different frequencies related by a ratio of natural numbers, the resulting figure is a Lissajous curve.