Oscillations and Damped Harmonic Motion
Zoom in on literally any potential with a Taylor expansion and BAM โ it's a spring, which is why oscillations pop up absolutely everywhere in physics.
OK so I’ve been flying through this stuff at warp speed… because honestly? It’s the exact same content I did in high school. The integrals are juuust a tiny bit chunkier now, that’s it. So yeah โ of course it feels easy. No shame in that.
BUT (BAM!!!!) โ now we’re doing oscillations!!! And this part? This is where it’s gonna feel unfamiliar.
In high school you only did the simple pendulum, right? Maybe?? That was it??
I studied oscillations so so so hard because everyone kept telling me it’s so so so important. Why~~~ is it so important?
Because an absurd number of natural phenomena in the universe can be approximated with a spring.
Like,
Why does a wall make a sound when you hit it? โ spring, go go
Why does the sky look blue? โ spring, go go
How do we draw eeeeeeeeeverything we see around us?
โ spring…. heh
OK, enough intro. Let’s go.
Here’s the same thing I just said, dressed up a little fancier:
There are all kinds of wild and complicated potentials out there in the world.
Say we’ve got a potential like that. BUT!!! If we’re not trying to look at the whole range of $x$ โ
if we just zoom in on one specific position~~~~~~~ and Taylor-expand the thing, swoosh โ
โ then it’s totally fine to write $V(x)$ as that red quadratic!! Because right around the $x$ we actually care about,
the red curve and the black curve are almooost identical!!!!!!
$$V(x)\text{'s Taylor expansion at }x_{0}\\ V(x)\quad =\quad V(x-x_{0})\quad +\quad V'(x-x_{0})x\quad +\quad \frac {1}{2!}V''(x-x_{0})x^{2}\quad +\quad \{\cdot \cdot \cdot \cdot \cdot \cdot \\ \text{Red}\quad :\quad \text{let's set that point as the reference, so }0\\ \text{Purple}\quad :\quad \text{slope is }0\text{, so }0\\ \text{Green}\quad :\quad \text{let's discard (approximation)}$$Why does it matter so much that $V(x)$ is now quadratic in $x$?
Uhhh… hold on… we’ve seen something like this before, haven’t we?
Remember โ differentiate $V(x)$ once with respect to $x$ and you get $F(x)$? Right?????
$$\text{Now }V(x)\text{ is} \\ V(x)\cong \frac {1}{2!}V''(x-x_{0})x^{2}\quad \text{then} \\ -\frac {dV(x)}{dx}\quad =\quad -V''(x-x_{0})x\quad \text{if we view }V''(x-x_{0})\text{ as the constant }k\\ -\frac {dV(x)}{dx}\quad =\quad -kx\quad =\quad F(x)\quad \text{Oh my!!!}\quad \text{What a surprise!!}\quad \text{What is this,}\quad \text{it's a spring!!!}$$Zoom in on any one point we picked, and the system looks like a spring.
Quick look at the equation of motion for a spring:
$$Motion\quad Equation\\ m\frac {d^{2}x}{dt^{2}}\quad =\quad -kx\quad :\quad \text{a linear second-order differential equation with constant coefficients}\quad (\text{see the differential equations special post, go go})\\ m\frac {d^{2}x}{dt^{2}}\quad +\quad kx\quad =\quad 0\\ (D^{2}+\frac {k}{m})x\quad =\quad 0\\ (D\quad +\quad iw)(D\quad -\quad iw)x\quad =\quad 0\\ \therefore \quad x\quad =\quad Ae^{iwt}\quad +\quad Be^{-iwt}$$http://gdpresent.blog.me/220341671707
That’s the diff eq special post.
I’m planning to write up the “(linear) differential equation” โ you know, the thing every science/engineering undergrad runs into roughly ten thousand times…
gdpresent.blog.me
Alright! Now let’s nudge this just a little closer to the real world.
Let’s put in some resistance โ friction, air resistance, whatever. This is the thing!!
Going with what we did before, let’s assume the resistance force is linear in $v$. (Nonlinear is a nightmare, trust me;;)
OK so, equation of motion. I’ll just call the coefficient $c$ this time instead of $C_1$~~~~~
$$m\frac {d^{2}x}{dt^{2}}\quad =\quad -kx\quad -cv\quad =\quad -kx\quad -c\frac {dx}{dt}$$Boom โ that’s the equation of motion.
And quick notation note: differentiation with respect to time gets a dot on top of the head, and we call it a “dot.” Differentiation with respect to space gets a little tick mark on the side, by the ear, and we call it a “prime.”
$$m\ddot {x}\quad =\quad -kx\quad -c\dot {x}\\ m\ddot {x}\quad +\quad c\dot {x}\quad +\quad kx\quad =\quad 0$$Solving this is, again, a second-order linear ODE with constant coefficients!!
Which โ yeah โ the diff eq special post would really help here!!
But I know there will absolutely be people who do not want to read that!!!
In that case: go look at Boas, Mathematical Physics, Chapter 8, Section 6!!
(And to actually follow Section 6, you should start at Section 3 and walk through each one. It’s not hard at all, promise.)
$$m\ddot {x}\quad +\quad c\dot {x}\quad +\quad kx\quad =\quad 0\\ \ddot {x}\quad +\quad \frac {c}{m}\dot {x}\quad +\quad \frac {k}{m}x\quad =\quad 0\\ \frac {c}{m}\quad \equiv \quad 2\gamma \quad ,\quad \frac {k}{m}\equiv \quad w_{0}^{2} \\ \ddot {x}\quad +\quad 2\gamma \dot {x}\quad +\quad w_{0}^{2}x\quad =\quad 0\\ (D^{2}\quad +\quad 2\gamma D\quad +\quad w_{0}^{2})x\quad =\quad 0\\ (D-(-\gamma +\sqrt {\gamma^{2}-w_{0}^{2}}))(D-(-\gamma -\sqrt {\gamma^{2}-w_{0}^{2}}))x\quad =\quad 0\\ \therefore \quad x\quad =\quad Ae^{\left( -\gamma +\sqrt {\gamma^{2}-w_{0}^{2}} \right) t}\quad +\quad Be^{\left( -\gamma -\sqrt {\gamma^{2}-w_{0}^{2}} \right) t}$$<WAIT WAIT WAIT WAIT WAIT!!!!!!!
$$\gamma^{2}-w_{0}^{2}\quad <\quad 0$$what do we do?!!>
Relax โ easy.
$$\sqrt {\gamma^{2}-w_{0}^{2}}\quad =\quad \sqrt {-(-\gamma^{2}+w_{0}^{2})}\quad =\quad i\sqrt {w_{0}^{2}-\gamma^{2}} \\ \therefore \quad x\quad =\quad Ae^{\left( -\gamma +i\sqrt {w_{0}^{2}-\gamma^{2}} \right) t}\quad +\quad Be^{\left( -\gamma -i\sqrt {w_{0}^{2}-\gamma^{2}} \right) t}\\ (w_{0}^{2}-\gamma^{2}\equiv w_{d}^{2})\\ \quad =\quad Ae^{-\gamma t}e^{iw_{d}t}\quad +\quad Be^{-\gamma t}e^{-iw_{d}t}$$$$x\text{ here has to be real, right??}\\ \text{So what if the right-hand side comes out complex? (Just throw that part out!)}\\ \text{But since we're throwing it out anyway, it's actually easier to just set }A\text{ and }B\text{ as complex conjugates from the start.}\\ \text{So: let }A\quad =\quad \frac {A'}{2}e^{i\theta_{0}}\quad ,\quad B\quad =\quad \frac {A'}{2}e^{-i\theta_{0}}\text{ and work it from there โ this makes forcing the RHS real} \\ \text{way easier! (Not saying you *have* to do it this way!!!! Just saying it's smoother.)}$$<WAIT WAIT WAIT WAIT WAIT!!!!!!!
$$\gamma^{2}-w_{0}^{2}\quad =\quad 0$$what do we do?!!>
$$\text{Pulling from the diff eq special,}\\ (D+\gamma )(D-\gamma )x\quad =\quad \\ \therefore \quad x\quad =\quad (At\quad +\quad B)e^{-\gamma t}$$
So โ this one:
$$x\quad =\quad Ae^{-\gamma t}e^{iw_{d}t}\quad +\quad Be^{-\gamma t}e^{-iw_{d}t}$$That’s damped harmonic motion โ the everyday, garden-variety vibration you see all the time, right?
Overdamping:
$$x\quad =\quad Ae^{\left( -\gamma +\sqrt {\gamma^{2}-w_{0}^{2}} \right) t}\quad +\quad Be^{\left( -\gamma -\sqrt {\gamma^{2}-w_{0}^{2}} \right) t}$$And critical damping:
$$x\quad =\quad (At\quad +\quad B)e^{-\gamma t}$$Now! Here’s something worth pausing on. It’s nothing crazy though;; heh
Overdamping โ that’s a “vibration” that eeeeeases back to its resting position slowly over time. Where would you actually want that?
Like… a luxury sedan? When you hit a pothole, you want it to go whoosh and glide back down smooth. No bouncing. That’s nice. That’s overdamping.
OK but overdamping sounds unambiguously great, right?? When would you ever pick critical damping over that??
Where would you want critical damping? A race car. A ridiculous high-performance supercar. You do not want shock absorbers that go boing-boing-boing-whoooosh โ you want shock absorbers that snap back to neutral as fast as possible, so the car regains stability and gets back to flooring it.
lol
One more thing โ let’s look at this from an energy point of view real quick.
$$\text{Total }E\quad =\quad \frac {1}{2}mv^{2}\quad +\quad \frac {1}{2}kx^{2}\quad \\ \text{Differentiate both sides with respect to }t\text{, i.e. look at how }E\text{ changes over time} \\ \frac {dE}{dt}\quad =\quad \frac {1}{2}m(2v)\dot {v}\quad +\quad \frac {1}{2}k(2x)\dot {x}\\ \quad =\quad mv\dot {v}\quad +\quad kx\dot {x}\quad =\quad m\dot {x}\ddot {x}\quad +\quad kx\dot {x}\quad =\quad \dot {x}(m\ddot {x}+kx)\quad =\quad \dot {x}(-c\dot {x})\\ \quad =-c\dot {x}^{2}\quad \neq \quad 0\quad$$The “rate of change of energy” is not zero โ and that makes so so so much sense!!!
We set this up as a damped oscillation with resistance!! Common sense says it’s eventually gonna stop oscillating โ so of course $E$ is decreasing. Nothing weird about that at all!!
Originally written in Korean on my Naver blog (2015-06). Translated to English for gdpark.blog.