Damped Oscillations §

The characteristic equation is $r^2 + 2\gamma r + \omega_0^2 = 0$ with roots $r = -\gamma \pm \sqrt{\gamma^2 - \omega_0^2}$. Three regimes result depending on the sign of the discriminant $\gamma^2 - \omega_0^2$.

Physical origin of damping. For a mass moving through a fluid at low speed, the drag force is approximately linear: $F_d = -c\dot{x}$. In an RLC circuit, the resistor provides the analogous energy loss. In a pendulum, air drag and pivot friction combine to give an effective $c$.

Electrical analog: the RLC circuit. The same equation governs a series RLC circuit:

The mapping is exact: $m \leftrightarrow L$, $c \leftrightarrow R$, $k \leftrightarrow 1/C$, $x \leftrightarrow q$. Under this map $\omega_0 = 1/\sqrt{LC}$, $\gamma = R/(2L)$, and the three regimes correspond to:

• Underdamped RLC ($R < 2\sqrt{L/C}$): the circuit rings (oscillatory charge decay).
• Critically damped RLC ($R = 2\sqrt{L/C}$): fastest settling without oscillation.
• Overdamped RLC ($R > 2\sqrt{L/C}$): slow exponential discharge.

This is the basis of radio tuning, band-pass filters, and pulse-shaping circuits. See Driven oscillations for the driven (forced) RLC case.

Authoritative source for this section: OpenStax University Physics Volume 2, §16.5: RLC Series Circuits

Why ζ is the natural dimensionless parameter. The equation of motion has three parameters ($m$, $c$, $k$), but only two independent ratios can be formed from them (since the equation is homogeneous in $x$). The natural frequency $\omega_0 = \sqrt{k/m}$ sets the time scale, and the damping rate $\gamma = c/(2m)$ sets the decay rate. Their ratio $\zeta = \gamma/\omega_0$ is the single dimensionless number that determines the qualitative behavior: it is the ratio of the actual damping to the critical damping $c_{\text{crit}} = 2\sqrt{mk}$.

• If $\zeta < 1$: damping is weaker than critical → the system oscillates.
• If $\zeta = 1$: damping is exactly critical → fastest non-oscillatory return.
• If $\zeta > 1$: damping exceeds critical → the system creeps back.

Concrete example. A car suspension has $m = 300$ kg (one wheel), $k = 30{,}000$ N/m, and $c = 2000$ N·s/m.

• $\omega_0 = \sqrt{30000/300} = 10$ rad/s.
• $\gamma = 2000/(2 \times 300) = 3.33$ s⁻¹.
• $\zeta = 3.33/10 = 0.33 < 1$: underdamped. The wheel bounces a few times after hitting a bump.

The critical damping would be $c_{\text{crit}} = 2\sqrt{300 \times 30000} = 2 \times 3000 = 6000$ N·s/m. The actual damper provides only about one-third of critical damping, which is typical for car suspensions (they are deliberately underdamped for ride comfort).

Authoritative source for this section: OpenStax University Physics Volume 1, §15.5: Damped Oscillations

The full solution is $x(t) = A\,e^{-\gamma t}\cos(\omega_d\, t + \phi)$.

Phase-space picture. In the $(x, \dot{x})$ plane, the trajectory is a spiral converging to the origin. This contrasts with the undamped Harmonic oscillator, whose phase portrait is a family of closed orbits.

Numerical example. Let $m = 1$ kg, $k = 25$ N/m, $c = 2$ N·s/m. Then $\omega_0 = 5$, $\gamma = 1$, $\zeta = 0.2$, $\omega_d = \sqrt{24} \approx 4.90$ rad/s. With $x(0) = 1$, $\dot{x}(0) = 0$:

• $x(0) = 1.000$
• $x(0.5) = e^{-0.5}\cos(2.45) = 0.607 \times (-0.770) = -0.467$
• $x(1.0) = e^{-1}\cos(4.90) = 0.368 \times (-0.183) = -0.067$
• $x(1.5) = e^{-1.5}\cos(7.35) = 0.223 \times 0.558 = 0.124$
• $x(2.0) = e^{-2}\cos(9.80) = 0.135 \times (-0.999) = -0.135$

The amplitude envelope $e^{-t}$ drops to $e^{-1} \approx 0.37$ after one time constant ($t = 1/\gamma = 1$ s).

Authoritative source for this section: OpenStax University Physics Volume 1, §15.5: Damped Oscillations

Why the repeated root gives $t\,e^{-\gamma t}$. At $\zeta = 1$ the discriminant vanishes: $r = -\gamma$ is a double root. The general solution of a second-order ODE with a repeated root $r$ is $x(t) = (A + Bt)\,e^{rt}$. The factor $t$ appears because the two linearly independent solutions of $(D - r)^2 x = 0$ are $e^{rt}$ and $t\,e^{rt}$ (the second is obtained by differentiating the first with respect to $r$).

Numerical example. Let $m = 1$, $k = 25$, $c = 10$ (so $\gamma = 5 = \omega_0$, $\zeta = 1$). With $x(0) = 1$, $\dot{x}(0) = 0$:

• $A = 1$, $B = \gamma = 5$ (from $x'(0) = B - \gamma A = 0$).
• $x(t) = (1 + 5t)\,e^{-5t}$.
• $x(0.1) = 1.5 \times e^{-0.5} = 1.5 \times 0.607 = 0.910$
• $x(0.5) = 3.5 \times e^{-2.5} = 3.5 \times 0.0821 = 0.287$
• $x(1.0) = 6.0 \times e^{-5} = 6.0 \times 0.00674 = 0.0404$
• $x(2.0) = 11.0 \times e^{-10} \approx 0.0005$

The system is essentially at equilibrium by $t \approx 1/\gamma = 0.2$ s.

Design target. Critical damping is the design goal for door closers, galvanometers, and many measurement instruments. Any deviation — slightly underdamped — introduces a small overshoot; slightly overdamped — slows the return.

Authoritative source for this section: Landau & Lifshitz, Mechanics, §21: Free oscillations in one dimension

Two time constants. The two roots give two decay rates: $|r_1| = \gamma - \sqrt{\gamma^2 - \omega_0^2}$ (slow) and $|r_2| = \gamma + \sqrt{\gamma^2 - \omega_0^2}$ (fast). The fast mode dies quickly, leaving the slow mode to govern the long-time behavior.

Large-ζ limit. For $\gamma \gg \omega_0$ (i.e. $\zeta \gg 1$), expand the slow root:

The slow-mode time constant is therefore:

Numerical example. Let $m = 1$, $k = 1$, $c = 10$ (so $\gamma = 5$, $\omega_0 = 1$, $\zeta = 5$).

• $r_1 = -5 + \sqrt{25 - 1} = -5 + 4.899 = -0.101$
• $r_2 = -5 - 4.899 = -9.899$
• Slow time constant: $\tau_1 = 1/0.101 \approx 9.9$ s.
• Fast time constant: $\tau_2 = 1/9.899 \approx 0.101$ s.
• Check: $c/k = 10/1 = 10$ s ≈ $\tau_1$ (the approximation $\tau \approx c/k$ is excellent for $\zeta = 5$).

The system creeps back to equilibrium over roughly 10 s without any oscillation. This regime is common in heavily viscous systems (e.g., a mass moving through honey).

Authoritative source for this section: Landau & Lifshitz, Mechanics, §21: Free oscillations in one dimension

The energy half-life is $t_{1/2} = \ln 2 / (2\gamma)$. The number of oscillations before the energy drops by a factor $e$ is $N = Q/(2\pi)$. A high-$Q$ system (e.g., a tuning fork, $Q \sim 10^3$) rings for hundreds of cycles.

The logarithmic decrement is $\delta = \gamma T_d = 2\pi\gamma/\omega_d \approx \pi/Q$ for $Q \gg 1$. It is measured experimentally as the natural log of the ratio of successive peak amplitudes.

Connection to resonance. The same quality factor $Q = \omega_0/(2\gamma)$ that quantifies how slowly the free oscillation decays also quantifies the sharpness of the resonance peak in a driven system: a high-$Q$ oscillator has a narrow resonance with a large peak amplitude. See Driven oscillations for the full resonance analysis.

Authoritative source for this section: OpenStax University Physics Volume 1, §15.5: Damped Oscillations

Problem: A 2 kg mass on a spring ($k = 50$ N/m) has a damping coefficient $c = 4$ N·s/m. Determine the regime, the damped frequency, and the time for the amplitude to fall to 1% of its initial value.