This is the second post in a series on vibrations determine by the equation

m u'' + γ u' + k u = F cos ωt

The first post in the series looked at the simplest case, γ = 0 and F = 0. Now we’ll look at the more realistic and more interesting case of damping γ > 0. We won’t consider forcing yet, so F = 0 and our equation reduces to

m u'' + γ u' + k u = 0.

The effect of damping obviously depends on the amount of damping, i.e. the size of γ relative to other terms. Damping removes energy from the system and so the amplitude of the oscillations goes to zero over time, regardless of the amount of damping. However, the system can have three qualitatively different behaviors: under-damping, critical damping, and over-damping.

The characteristic polynomial of the equation is

m x² + γ x + k

This polynomial has either two complex roots, one repeated real root, or two real roots depending on whether the discriminant γ² – 4mk is negative, zero, or positive. These three states correspond to under-damping, critical damping, and over-damping.

Under-damping

When damping is small, the system vibrates at first approximately as if there were no damping, but the amplitude of the solutions decreases exponentially. As long as γ² < 4mk the system is under-damped and the solution is

u(t) = R exp( –γt/2m ) cos( μt– φ )

where

μ = √(4mk – γ²) / 2m.

The amplitude R and the phase φ are determined by the initial conditions.

As an example, let γ =1, m = 1, and k = 5. This implies μ =  √19/2. Set the initial conditions so that R = 6 and φ = 1. The solid blue line below is the solution 6 exp(-t/2) cos(t– 1), and the dotted green lines above and below are the exponential envelope 6 exp(-t/2).

And here is a video of the mass-spring-dashpot system in action made using the code discussed here.

Note that the solution is not simply an oscillation at the natural frequency multiplied by an exponential term. The damping changes the frequency of the periodic term, though the change is small when the damping is small.

The factor μ is called the quasi-frequency. Recall from the previous post that the natural frequency, the frequency of the solution with no damping, is denoted ω₀. The quasi-frequency is related to the natural frequency by

μ = √(1 – γ²/4mk) ω₀ ≈ (1 – γ²/8mk) ω₀.

The approximation above holds for small γ since a Taylor expansion shows that √(1 – x) ≈ 1 – x/2 for small x.

This says that the quasi-frequency decreases the natural frequency by a factor of approximately γ²/8mk. So when γ is small there is little change to the natural frequency, but the amount of change grows quadratically as γ increases.

Critical damping

Critical damping occurs when γ² = 4mk. In this case the solution is given by

u(t) = (A + Bt) exp( – γt/2m).

The position of the mass will asymptotically approach 0. Depending on the initial velocity, it will either go monotonically to zero or reach some maximum displacement before it turns around and goes to 0.

For an example, let m = 1 and k = 5 as before. For critical damping, γ must equal √20. Choose initial conditions so that A = 1 and B = 10.

Over-damping

When γ² > 4mk the system is over-damped. When damping is that large, the system does not oscillate at all.

The solution is

u(t) = A exp(r₁ t) + B exp(r₂ t)

where r₁ and r₂ are the two roots of

m x² + γ x + k = 0.

Because γ > 0, both roots are negative and solutions decay exponentially to 0.

Coming up

The next two posts in the series will add a forcing term, i.e. F > 0 in our differential equation. The next post will look at undamped driven vibrations, and the final post will look at damped driven vibrations.

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.

1. The same equations describe a variety of mechanical and electrical systems.
2. You can get practical use out of some relatively simple math.
3. 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 ω₀t + B cos ω₀t

where

ω₀² = k/m.

The value ω₀ 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 ω₀.

Since the sine and cosine components have the same frequency ω₀, we can use a trig identity to combine them into a single function

u(t) = R cos(ω₀t – 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.

Consulting in differential equations