Damping Determines How Oscillations Fade

The unforced mass–spring–damper equation is

$mx''+cx'+kx=0.$

Its characteristic discriminant $c^2-4mk$ determines the response.

Three regimes

If $c^2<4mk$, the system is underdamped and oscillates with a decaying envelope. If $c^2=4mk$, it is critically damped and returns without oscillation as quickly as possible. If $c^2>4mk$, it is overdamped and returns without oscillation more slowly.

Initial conditions

Initial displacement and velocity determine the particular combination of the two fundamental solutions, but the damping regime is determined by the system coefficients.

Normalize the equation

Dividing by $m$ gives

$x''+2\zeta\omega_n x'+\omega_n^2x=0,$

where $\omega_n=\sqrt{k/m}$ is the undamped natural frequency and

$\zeta=\frac{c}{2\sqrt{mk}}$

is the dimensionless damping ratio. The three regimes become $0<\zeta<1$, $\zeta=1$, and $\zeta>1$. Nondimensionalization reveals that systems with very different masses and springs can share the same response shape.

Energy interpretation

For $E=\tfrac12mx'^2+\tfrac12kx^2$,

$\frac{dE}{dt}=x'(mx''+kx)=-cx'^2\le0.$

Damping removes energy at a rate proportional to velocity squared. This proves monotone energy loss even when position oscillates.

Check your understanding

Can an overdamped system cross equilibrium once?

Show the reasoning

Yes. “Non-oscillatory” means it does not repeatedly cross. Particular initial conditions can carry it across once before the two decaying exponential modes bring it back.