My favorite topic in an introductory differential equations course is mechanical and electrical vibrations. I enjoyed learning about it as a student and I enjoyed teaching it later. (Or more accurately, I enjoyed being exposed to it as a student and really learning it later when I had to teach it.)

I find this subject interesting for three reasons.

- The same equations describe a variety of mechanical and electrical systems.
- You can get practical use out of some relatively simple math.
- The solutions display wide variety of behavior as you vary the coefficients.

This is the first of a four-part series of posts on mechanical vibrations. The posts won’t be consecutive: I’ll write about other things in between.

Simple mechanical vibrations satisfy the following differential equation:

We could simply write down the general solution be done with it. But the focus here won’t be *finding* the solutions but rather *understanding* how the solutions behave.

We’ll think of our equation as modeling a system with a mass attached to a spring and a dash pot. All coefficients are constant. *m* is the mass, γ is the damping from the dash pot, and *k* is the restoring force from the spring. The driving force has amplitude *F* and frequency ω. The solution *u*(*t*) gives the position of the mass at time *t*.

More complicated vibrations, such as a tall building swaying in the wind, can be approximated by this simple setting.

The same differential equation could model an electrical circuit with an inductor, resistor, and capacitor. In that case replace mass *m* with the inductance *L*, damping γ with resistance *R*, and spring constant *k* with the reciprocal of capacitance *C*. Then the equation gives the charge on the capacitor at time *t*.

We will assume *m* and *k* are positive. The four blog posts will correspond to γ zero or positive, and *F* zero or non-zero. Since γ represents damping, the system is called undamped when γ = 0 and damped when γ is greater than 0. And since *F* is the amplitude of the forcing function, the system is called free when *F* = 0 and forced otherwise. So the plan for the four posts is

- γ = 0,
*F*= 0. Free, undamped vibrations (this post) - γ > 0,
*F*= 0. Free, damped vibrations - γ = 0,
*F*> 0. Forced, undamped vibrations - γ > 0,
*F*> 0. Forced, damped vibrations

## Free, undamped vibrations

With no damping and no forcing, our equation is simply

*m u*`''`

+ *k u* = 0

and we can write down the solution

*u*(*t*) = *A* sin ω_{0}*t* + *B* cos ω_{0}*t*

where

ω_{0}^{2} = *k*/*m*.

The value ω_{0} is called the natural frequency of the system because it gives the frequency of vibration when there is no forcing function.

The values of *A* and *B *are determined by the initial conditions, i.e. *u*(0) = *B* and *u*`’`

(0) = *A* ω_{0}.

Since the sine and cosine components have the same frequency ω_{0}, we can use a trig identity to combine them into a single function

*u*(*t*) = *R* cos(ω_{0}*t* – φ)

The amplitude *R* and phase φ are related to the parameters *A* and *B* by

*A* = *R* cos φ, *B* = *R* sin φ.

That’s it for undamped free vibrations: the solutions are just sine waves. The next post in the series will make things more realistic and more interesting by adding damping.

**Update**: Here’s an animation of undamped free oscillation using code by Stéfan van der Walt described here.

Bonus education: not only some DiffEq, but now I know what ‘dashpot’ means. Nice.

Likewise, one of my favorite subjects. Looking forward to some chaos!

Why does the LHS stop with u” and not u”’ or u””? I mean, why does the system take into account displacement, rate of change of displacement and rate of rate of change of displacement but not go further? Sorry for the n00b question, my intuitive understanding of math is rather weak.

Vijay: Most differential equations in application are second order because Newton’s laws are second order differential equations.

John: Mathematically, I can see how the equation is derived but when I imagine a physical body, a spring for instance, and try to see how the math maps to the physical properties of the spring, things get fuzzy. Like, why does the rate of rate of rate of change of displacement (u”’) not come into picture? Mathematically, that’s clear because mass, damping and restoring force are the only three factors, without a fourth factor, a u”’ can’t make an appearance. But physically it’s hard for me to visualize why mass affects the system exactly in the proportion of the rate of rate of change of displacement (u”) and not rate of rate of rate of change of displacement(u”’) and so on.

If I throw a stone in the air, I can see it physically describe the parabolic arc and that makes the underlying math extremely tangible for me. I guess I’m looking for a similarly tangible explanation of why these systems are second order differentials.

My nephew just started calculus and physics two weeks ago. This is essentially what we discussed over a dinner this weekend. I’m looking forward to the next installment!

Vijay: There can be higher order effects, a spring stretched too far or a dash pot subjected to a very high (or low) velocity, but (at least for engineering) we generally try to stay in the region where the 2nd order equations are a good representation. Things get messy quickly when you stray away.

If one were modeling a proto- solar system the changing mass of the planets would give you u”’ to deal with, as might loss of water ice for a comet.

Newton’s second law is F=p’ = (mu’)’ = m’u’ + mu”. Changing mass can be modeled by the m’u’ term. The terms in John’s equation other than mu” represent forces. A u”’ term allows solutions where the object spontaneously starts accelerating.

Ed, Tom: Thanks guys. I still haven’t reached the level of visceral understanding that I’m after, but the additional info in your comments could lead me to solving the puzzle eventually.

/dampled/damped/

Though “dampled” has rather a nice flavor, sorry to encourage removing it.

According to Google it’s an occasional typo for “sample” (probably due to the proximity of the “d” and “s” keys), as well as for “damped” (more speculatively due to a nearby mental groove for “sampled”?).

More on topic, I look forward to more on the math of vibrations. As you point out, being exposed to it and really getting it are often two entirely different events.

The parallelism between mass, spring, and damping and inductance, capacitance, and resistance is indeed elegant. There was quite an industry in analog computers to model mechanical systems, once upon a time.

lens: Thanks. I kinda like the sound of “dampled” too. 🙂

Reminds me of “dappled” as in the poem that starts out “Glory be to God for dappled things …”

Tom, in case of variable mass, the equation F = p’ = (mu’)’ = m’u’ + mu” is wrong, for it is not invariant under Galilean transformations. In Newtonian mechanics, mass does not actually change: at most, it can be transferred from one region to another. Thus, a variable-mass system should be always considered as a subsystem of a larger system with constant total mass, and the equation F = p’ can only be applied to the whole system.

For more information, see for example:

A. R. Plastino & J. C. Muzzio, “On the use and abuse of Newton’s second law for variable mass problems” (http://articles.adsabs.harvard.edu/full/1992CeMDA..53..227P)

and

J. Peraire & S. Widnal, “Variable Mass Systems: The Rocket Equation” in Dynamics, MIT Course Number 16.07 (http://ocw.mit.edu/courses/aeronautics-and-astronautics/16-07-dynamics-fall-2009/lecture-notes/MIT16_07F09_Lec14.pdf)

@Ed, @Vijay

Regarding the “how many derivatives do you go” question, this is related to problems of overfitting — the more terms in hand, the harder it is to get good bounds on their weighting coefficients from a fixed set of data — but I also think the main is a game, very enjoyable. How much behavior can be described by a simpler set of equations than a more powerful?