Real oscillations die out. Add a drag force like the one in Linear Drag and Terminal Velocity to the spring of The Equation of Simple Harmonic Motion, and the cosine gets wrapped in a shrinking exponential, the shape you graphed in Damped Oscillation.
Why it works
With drag $-b\dot{x}$: $m\ddot{x} + b\dot{x} + kx = 0$. Try $x = e^{rt}$: $mr^2 + br + k = 0$, so $$r = -\dfrac{b}{2m} \pm \sqrt{\left(\dfrac{b}{2m}\right)^2 - \dfrac{k}{m}}.$$ If $b^2 < 4mk$ the square root is imaginary, $\pm i\omega_d$, and the real solutions are $e^{-bt/2m}\cos(\omega_dt + \phi)$: oscillation inside a decaying envelope. If $b^2 = 4mk$ there is one repeated root (critical damping); if $b^2 > 4mk$, two negative real roots (overdamped): no oscillation at all.
Key idea
Underdamped ($b < 2\sqrt{km}$): $$x = A_0e^{-bt/2m}\cos(\omega_dt + \phi), \qquad \omega_d = \sqrt{\dfrac{k}{m} - \left(\dfrac{b}{2m}\right)^2}.$$ The amplitude falls by half every $\dfrac{2m\ln 2}{b}$, and the energy as $e^{-bt/m}$. Critical damping, $b_c = 2\sqrt{km}$, returns to equilibrium fastest without overshooting; more damping (overdamped) returns more slowly.
Worked example: How fast it fades
A $1\ \text{kg}$ mass on a spring ($k = 100\ \text{N/m}$) has $b = 2\ \text{N·s/m}$. How long until the amplitude halves, and what is $\omega_d$?
- Envelope $e^{-bt/2m} = e^{-t}$; half at $t = \ln 2 = 0.693\ \text{s}$.
- $\omega_d = \sqrt{100 - 1^2} = 9.95\ \text{rad/s}$, barely below $\omega_0 = 10\ \text{rad/s}$.
Answer: $0.693\ \text{s}$; $\omega_d = 9.95\ \text{rad/s}$.
Worked example: Designing a door closer
What damping makes the same system critically damped?
- Critical: $b^2 = 4mk$, so $b_c = 2\sqrt{(1)(100)} = 20\ \text{N·s/m}$.
- With this damping, a pushed door swings shut quickly without banging past the frame.
Answer: $b_c = 20\ \text{N·s/m}$.
Common mistake
Thinking a critically damped system swings once before it stops. Released from rest, it never crosses equilibrium: critical damping is the boundary between oscillating and not. Any overshoot means the system is (slightly) underdamped. Check the roots of $mr^2 + br + k = 0$: when they are real, there is no cosine in the solution, so nothing to swing.
| Damping | Roots of $mr^2 + br + k = 0$ | Motion after release from rest |
|---|---|---|
| $b < 2\sqrt{km}$ (under) | complex: $-\tfrac{b}{2m} \pm i\omega_d$ | oscillates inside a decaying envelope |
| $b = 2\sqrt{km}$ (critical) | one repeated real root | fastest return, no overshoot |
| $b > 2\sqrt{km}$ (over) | two negative real roots | slow return, no overshoot |
Car shock absorbers are tuned near critical damping, so a bump is absorbed without bouncing. A guitar string is very lightly damped, which is why it rings for seconds.