Damping · Differential Equations

Intuition

Add friction and the energy leaks away. Whether the characteristic roots are real or complex now depends on the square of the friction against four times the mass times the stiffness, and that comparison sorts the motion into three kinds: underdamped, which still oscillates while it decays; critically damped, the quickest return without oscillating; and overdamped, a slower creep back. All three die away.

A door with a closer. Too little damping and it swings past the frame and back; too much and it creeps shut; set just right, it closes quickly without swinging past. Critical damping is that setting.

Released from $x=1$ at rest: $x''+\tfrac12x'+4x=0$ is underdamped and oscillates as it decays; $x''+4x'+4x=0$ is critically damped and returns fastest without crossing; $x''+10x'+4x=0$ is overdamped and creeps back.

Three regimes, one discriminant

For $mx''+cx'+kx=0$ with : if the motion is underdamped, with ; if it is critically damped, ; if it is overdamped, a combination of two decaying exponentials. In every case .

\begin{aligned}c^{2}<4mk&:\ \text{underdamped}\\c^{2}=4mk&:\ \text{critically damped}\\c^{2}>4mk&:\ \text{overdamped}\end{aligned}

What each regime does

Underdamped motion oscillates at $\omega_{d}<\omega_{0}$ : friction slows the oscillation as well as shrinking it.
Critical damping, $c=2\sqrt{mk}$ , is the boundary: the quickest return to rest without oscillating.

Damped motion always dies away

\begin{aligned}&m,c,k>0\ \Longrightarrow\\&\text{both roots have negative real part}\end{aligned}

Read the sum and the product of the roots off the coefficients: the sum is minus c over m, negative, and the product is k over m, positive. A complex pair has twice its real part as its sum, so the real part is negative. Two real roots with a positive product share a sign, and their negative sum makes it negative. Either way every term of the solution decays, so the motion dies away.

Proof steps

r_{1}+r_{2}=-\frac cm<0,\qquad r_{1}r_{2}=\frac km>0

The sum and product of the roots, read off the coefficients.

r_{1,2}=\alpha\pm i\beta\ \Longrightarrow\ 2\alpha=r_{1}+r_{2}<0

A complex pair sums to twice its real part, which is therefore negative.

r_{1},r_{2}\ \text{real}\ \Longrightarrow\ r_{1},r_{2}<0

Real roots with a positive product share a sign, and a negative sum makes it negative.

x(t)\to0\ \text{as}\ t\to\infty

Every term of the solution decays, so the motion dies away.

Applications

Practice

One Discriminant, Three Regimes

Compare $c^{2}$ with $4mk$ : smaller oscillates, equal is critical, larger creeps.

c^{2}-4mk<0,\ =0,\ >0

Try it

Which kind of motion does $x''+2x'+10x=0$ describe?