2D and 3D Isotropic Harmonic Oscillators and Lissajous Figures

Megaton ultra speed and we’re already on Chapter 4!!!!! hehh

And not just Chapter 4 — like the middle-to-late part of Chapter 4??? It’s the isotropic harmonic oscillator in 2D and 3D!!! Everything before this was high school Physics 2 level stuff,

plus a tiny bit of diff. eq. nothing, so I just skipped all of it!!!

OK then, let’s get into it.

The book says: consider “the motion of a particle that experiences a linear restoring force directed only toward one fixed point,” and

gives this kind of setup as the example,

and I was also picturing something like this!

I mean — a similar phenomenon could happen in a setup like this too!!!

But…. the force here (electric force) isn’t linear…. It’d get more complicated… heh heh?

Let’s just go with the book’s setup ^^*

Since the force points only toward one fixed point (isotropic harmonic oscillator),

$$\overrightarrow {F}\quad =\quad m\overrightarrow {\ddot {r}}\quad =\quad -k\overrightarrow {r}\quad \text{we can write it like this, right?} \\ \overrightarrow {r}\quad =\quad x\hat {i}\quad +\quad y\hat {j}\quad \text{since} \\ m\frac {d^{2}}{dt^{2}}(x\hat {i}\quad +\quad y\hat {j})\quad =\quad -k(x\hat {i}\quad +\quad y\hat {j}) \\ m\ddot {x}\quad =\quad -kx\quad m\ddot {y}\quad =\quad -ky$$

We can ‘separate’ the force like this!!! heh

Now, this kind of diff. eq.~ is a piece of cake lol lol lol

$$x\quad =\quad Acos(wt\quad +\quad \alpha )\\ y\quad =\quad Bcos(wt\quad +\quad \beta )\quad w\quad =\quad \sqrt {\frac {k}{m}}$$

Are we done??!?!?! Nope ~~~

We analyzed x and y separately~ but the two aren’t… totally unrelated, right?!?!?!

And since x is a function of t, and y is also a function of t, somehow if we kill that t, it feels like we should be able to couple x and y together….

OK let’s go!!!

First — A, B, α, β are all set by the initial conditions!

Let me try writing α and β bundled together at least.

$$\beta \quad =\quad \alpha \quad +\quad \text{something}$$

Then the equation for y becomes

$$y\quad =\quad Bcos(wt\quad +\quad \alpha \quad +\quad \Delta )\quad \text{write it like this and bust out the high school trig. (cos-cos minus sin-sin)}\\ \quad =\quad B\left\{cos(wt+\alpha )cos\Delta \quad -\quad sin(wt+\alpha )sin\Delta \right\} \\ \quad =\quad B\left\{cos(wt+\alpha )cos\Delta \quad -\quad \sqrt {1-cos^{2}(wt+\alpha )}\cdot sin\Delta \right\} \\ \text{Find the hidden}\quad x\quad \text{go go}!! \\ \quad =\quad B\left\{\frac {x}{A}cos\Delta \quad -\quad \sqrt {1-\left( \frac {x}{A} \right)^{2}}\cdot sin\Delta \right\}$$

Now let me clean this up a bit~~

$$\frac {y}{B}\quad -\quad \frac {x}{A}cos\Delta \quad =\quad -\sqrt {1-\left( \frac {x}{A} \right)^{2}}sin\Delta \\ \frac {y^{2}}{B^{2}}\quad +\quad \frac {x^{2}}{A^{2}}cos^{2}\Delta \quad -\frac {2cos\Delta}{AB}xy\quad =\quad \left( 1-\left( \frac {x}{A} \right)^{2} \right) sin^{2}\Delta \\ \quad =\quad sin^{2}\Delta \quad -\quad \frac {x^{2}}{A^{2}}sin^{2}\Delta$$$$\frac {y^{2}}{B^{2}}\quad -\frac {2cos\Delta}{AB}xy\quad +\quad \frac {x^{2}}{A^{2}}\quad =\quad sin^{2}\Delta$$

We’ve got it down to this!!!!

For the general quadratic

$$ax^{2}\quad +\quad bxy\quad +\quad cy^{2}\quad +\quad dx\quad +\quad ey\quad =\quad f$$

whether

$$b^{2}\quad -\quad 4ac$$

is less than 0, equal to 0, or greater than 0 tells you ellipse, parabola, or hyperbola,

and in our equation the discriminant works out to

$$\left( -\frac {2cos\Delta}{AB} \right)^{2}\quad -4\frac {1}{A^{2}}\cdot \frac {1}{B^{2}}\quad =\quad \frac {4cos^{2}\Delta}{A^{2}B^{2}}-\frac {4}{A^{2}B^{2}}\quad =\quad -\frac {sin^{2}\Delta}{A^{2}B^{2}} \\ -\left( \frac {sin\Delta}{AB} \right)^{2}\quad =\quad -(\text{positive number})\quad =\quad \text{negative number}\quad <\quad 0$$

So — discriminant less than 0, which means our x and y trace out an elliptical orbit.

A little bonus: if Δ is π/2, then cosΔ=0 and sinΔ=1, so the xy term drops out and you get an ellipse sitting cleanly along the x and y axes,

and if Δ is 0 or π, sinΔ=0 and cosΔ=1, and you can spot that now it’s not tracing an ellipse anymore — it’s just sliding back and forth along a straight line~~~!!!

And lastly — about the angle ψ that the ellipse’s axis makes with the x-axis, I worked that one out over at http://gdpresent.blog.me/221133339181 here!

Classical Mechanics I Studied #13. The angle ψ between the ellipse’s axis and the x-axis

Back when I was an underclassman, I worked out the relationship for the angle ψ that the major axis makes with the x-axis when you’ve got an ellipse like this;;;;; classical mechanics…

gdpresent.blog.me

The idea is simple. You find the relationship between ψ, Δ, and A, B such that doing a linear transformation by angle ψ kills the xy term,

and if you’re curious, just go check out that link!

OK then — done with 2D to that extent, let’s keep moving. What about the 3D isotropic harmonic oscillator????

Let’s blast through it lol the content is exactly the same as 2D heh

Well it’d be something like this kind of setup, well heh heh

Because it’s an isotropic harmonic oscillator!!!!!

$$\overrightarrow {F}\quad =\quad m\overrightarrow {\ddot {r}}\quad =\quad -k\overrightarrow {r}\quad \text{since this is the case} \\ m\frac {d^{2}}{dt^{2}}(x\hat {i}\quad +\quad y\hat {j}\quad +z\hat {k})\quad =\quad -k(x\hat {i}\quad +\quad y\hat {j}\quad +\quad z\hat {k}) \\ m\ddot {x}\quad =\quad -kx\quad m\ddot {y}\quad =\quad -ky\quad m\ddot {z}\quad =\quad -kz \\ r\text{ - just one extra term tacked on, that's literally it}!$$$$\text{Then, like before} \\ x\quad =\quad Acos(wt+\alpha )\\ y\quad =\quad Bcos(wt+\beta )\\ z\quad =\quad Ccos(wt+\gamma )\quad \\ \text{would work, but here}$$$$x\quad =\quad A_{1}sin(wt)\quad +\quad B_{1}cos(wt)\\ y\quad =\quad A_{2}sin(wt)\quad +\quad B_{2}cos(wt)\\ z\quad =\quad A_{3}sin(wt)\quad +\quad B_{3}cos(wt) \\ \text{it's nicer to write it like this, because}$$$$\overrightarrow {r}\quad =\quad x\hat {i}\quad +\quad y\hat {j}\quad +z\hat {k}\\ \quad \\ =\quad \left\{A_{1}sin(wt)\quad +\quad B_{1}cos(wt) \right\} \hat {i}\quad +\quad \left\{A_{2}sin(wt)\quad +\quad B_{2}cos(wt) \right\} \hat {j}\quad +\left\{A_{3}sin(wt)\quad +\quad B_{3}cos(wt) \right\} \hat {k} \\ =\quad \overrightarrow {A}sin(wt)\quad +\quad \overrightarrow {B}cos(wt)$$

Written this way, you can see at a glance that the vector r lives entirely in the plane spanned by vector A and vector B,

and then it becomes super easy to shift your perspective to “OK, we’re in that plane~~~,”

and once you’re there you can knock the dimension down to 2D,

because the analysis from there is exactly the method we just did!! heh heh

Conti~~~~~~~~~~nuing on — up to now we’ve been saying isotropic isotropic isotropic over and over,

so then what’s anisotropic ~~~~~~~~~~~~~~~~~~~~

It just means not isotropic!!!!

Concretely — if the spring constant is different in the x, y, and z directions, the thing is anisotropic!?!? heh heh heh

So

$$m\ddot {x}\quad =\quad -k_{1}x\\ m\ddot {y}\quad =\quad -k_{2}y\\ m\ddot {z}\quad =\quad -k_{3}z\quad \text{we write it like this, and}$$$$x\quad =\quad Acos(w_{1}t\quad +\quad \alpha )\\ y\quad =\quad Acos(w_{2}t\quad +\quad \beta )\\ z\quad =\quad Acos(w_{3}t\quad +\quad \gamma )\quad w_{n}\quad =\quad \sqrt {\frac {k_{n}}{m}}$$

Now, if there are integers n such that

$$\frac {w_{1}}{n_{1}}\quad =\quad \frac {w_{2}}{n_{2}}\quad =\quad \frac {w_{3}}{n_{3}}$$

this kind of relationship is satisfied — i.e. integers n actually ’exist’ that make it work —

then the trajectory this thing traces out as it oscillates is called a Lissajous figure.

Saying that the time

$$\frac {2\pi n_{1}}{w_{1}}\quad =\quad \frac {2\pi n_{2}}{w_{2}}\quad =\quad \frac {2\pi n_{3}}{w_{3}}$$

satisfying this kind of relationship ’exists’

means that somehow, one way or another, it eventually comes back to where it started!!!!!!!!!!!!!! heh heh

(This is a reference image of a Lissajous pattern heh heh heh)

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Originally written in Korean on my Naver blog (2015-06). Translated to English for gdpark.blog.