Damped oscillations

An ideal spring-mass system oscillates forever with fixed amplitude — a convenient fiction. Every real oscillator sits inside a medium that resists its motion, so friction or fluid drag steadily drains its mechanical energy and the amplitude shrinks with time. This is damped oscillation, and PH11001 asks you to go beyond describing that shrinkage qualitatively: you set up the linear differential equation for a mass-spring system with velocity-proportional (viscous) damping, solve it, and classify the result into three distinct regimes depending on how strong the damping is relative to the natural frequency. This article carries the Pass-1 equation and worked numbers forward and adds the full derivation, a second worked pass through the same underdamped example, and the quiz/FAQ scaffolding you need to be exam-ready. Only linear viscous damping is in scope; nonlinear drag (velocity-squared, dry friction) is not examined here.

The core insight

A mass-spring system with viscous damping obeys x¨+2γx˙+ω02x=0\ddot x+2\gamma\dot x+\omega_0^2x=0, and comparing the damping rate γ\gamma to the natural frequency ω0\omega_0 splits its motion into underdamped (oscillates with decaying amplitude), critically damped (fastest non-oscillating return), or overdamped (slow, non-oscillating creep).

The core idea

A mass mm on a spring of constant kk, moving through a medium that exerts a resistive force proportional to velocity, −bx˙-b\dot x, obeys Newton's second law in the form

mx¨+bx˙+kx=0.m\ddot x + b\dot x + kx = 0.

where: mm is the mass, bb is the damping coefficient (force per unit velocity), kk is the spring constant, and x(t)x(t) is displacement from equilibrium. Dividing by mm and defining the damping rate γ=b/2m\gamma = b/2m and the natural frequency ω0=k/m\omega_0 = \sqrt{k/m} gives the standard form used throughout this topic:

x¨+2γx˙+ω02x=0.\ddot x + 2\gamma\dot x + \omega_0^2 x = 0.

where: γ\gamma has units of s−1\text{s}^{-1} and sets how fast energy leaks out, while ω0\omega_0 is the frequency the same mass and spring would show with no damping at all. The sign of γ2−ω02\gamma^2-\omega_0^2 decides which of three regimes the system falls into.

  • Underdamped (γ<ω0\gamma<\omega_0): the motion oscillates inside a decaying envelope, x(t)=Ae−γtcos⁡(ω1t+ϕ)x(t)=Ae^{-\gamma t}\cos(\omega_1 t+\phi) with ω1=ω02−γ2\omega_1=\sqrt{\omega_0^2-\gamma^2} — a frequency always slightly below ω0\omega_0.
  • Critically damped (γ=ω0\gamma=\omega_0): x(t)=(A+Bt)e−γtx(t)=(A+Bt)e^{-\gamma t} — the fastest possible return to equilibrium with no oscillation.
  • Overdamped (γ>ω0\gamma>\omega_0): x(t)=C1eλ1t+C2eλ2tx(t)=C_1e^{\lambda_1 t}+C_2e^{\lambda_2 t} with two distinct negative real roots — a slow, non-oscillating creep back to equilibrium.

Because energy E∝A2E\propto A^2 for an oscillator and the underdamped amplitude decays as e−γte^{-\gamma t}, the energy decays twice as fast:

E(t)=E0 e−2γt.E(t) = E_0\,e^{-2\gamma t}.

The quality factor Q=ω0/2γQ=\omega_0/2\gamma measures roughly how many radians of oscillation occur before the energy falls by a factor 1/e1/e. High QQ means light damping and many visible cycles (a tuning fork, Q∼103Q\sim10^3–10410^4); low QQ means heavy damping (a car's shock absorber, tuned near critical damping so the car does not bounce after a bump).

Figure
Three displacement versus time curves showing underdamped oscillation decaying inside an exponential envelope, critically damped motion returning fastest without crossing zero, and overdamped motion creeping back slowly
Three damping regimes, same starting displacement time t x(t) underdamped critically damped overdamped equilibrium

The characteristic equation and the three regimes

Derivation AFrom Newton's law to underdamped, critical and overdamped motion

The goal is to start from the damped oscillator equation x¨+2γx˙+ω02x=0\ddot x+2\gamma\dot x+\omega_0^2x=0 and show, using nothing beyond the algebra of a quadratic equation, exactly why the sign of γ2−ω02\gamma^2-\omega_0^2 is what separates oscillating decay from non-oscillating decay, and why the boundary case γ=ω0\gamma=\omega_0 gives the fastest possible non-oscillating return.

Step 1. Try an exponential trial solution x(t)=eλtx(t)=e^{\lambda t}, since the equation is linear with constant coefficients and exponentials reproduce themselves under differentiation. Substituting x˙=λeλt\dot x=\lambda e^{\lambda t} and x¨=λ2eλt\ddot x=\lambda^2e^{\lambda t} into x¨+2γx˙+ω02x=0\ddot x+2\gamma\dot x+\omega_0^2x=0 and dividing through by the common factor eλte^{\lambda t} (never zero) leaves the characteristic equation λ2+2γλ+ω02=0\lambda^2+2\gamma\lambda+\omega_0^2=0.

Step 2. Solve this quadratic for λ\lambda using the standard formula: λ=−2γ±4γ2−4ω022=−γ±γ2−ω02\lambda=\dfrac{-2\gamma\pm\sqrt{4\gamma^2-4\omega_0^2}}{2}=-\gamma\pm\sqrt{\gamma^2-\omega_0^2}. Everything hinges on the sign of the quantity under the square root, γ2−ω02\gamma^2-\omega_0^2.

Step 3. If γ<ω0\gamma<\omega_0, the discriminant γ2−ω02\gamma^2-\omega_0^2 is negative, so the square root is imaginary: γ2−ω02=iω02−γ2=iω1\sqrt{\gamma^2-\omega_0^2}=i\sqrt{\omega_0^2-\gamma^2}=i\omega_1. The two roots are the complex conjugate pair λ=−γ±iω1\lambda=-\gamma\pm i\omega_1. A real solution built from these roots is x(t)=e−γt(Acos⁡ω1t+Bsin⁡ω1t)x(t)=e^{-\gamma t}(A\cos\omega_1 t+B\sin\omega_1 t), which can be rewritten as Ae−γtcos⁡(ω1t+ϕ)Ae^{-\gamma t}\cos(\omega_1 t+\phi) — an oscillation at frequency ω1=ω02−γ2\omega_1=\sqrt{\omega_0^2-\gamma^2} riding inside an exponentially decaying envelope. This is the underdamped case.

Step 4. If γ>ω0\gamma>\omega_0, the discriminant is positive, so both roots λ1,2=−γ±γ2−ω02\lambda_{1,2}=-\gamma\pm\sqrt{\gamma^2-\omega_0^2} are real and negative (since γ2−ω02<γ\sqrt{\gamma^2-\omega_0^2}<\gamma always). The general solution x(t)=C1eλ1t+C2eλ2tx(t)=C_1e^{\lambda_1t}+C_2e^{\lambda_2t} is a sum of two decaying exponentials with no oscillation — the overdamped case, which creeps back to zero, dominated at long times by the slower-decaying root (the one closer to zero).

Step 5. If γ=ω0\gamma=\omega_0 exactly, the discriminant vanishes and the quadratic has a repeated root λ=−γ\lambda=-\gamma. A single exponential e−γte^{-\gamma t} is only one independent solution of a second-order equation, so the second independent solution is t e−γtt\,e^{-\gamma t} (checked directly by substitution), giving the general solution x(t)=(A+Bt)e−γtx(t)=(A+Bt)e^{-\gamma t}. This critically damped case sits exactly on the boundary between oscillating and non-oscillating behaviour, and among all three regimes it returns to equilibrium fastest without ever oscillating — an overdamped system, despite the name suggesting "more damping," is actually slower because its dominant decay rate γ−γ2−ω02\gamma-\sqrt{\gamma^2-\omega_0^2} is smaller than γ\gamma.

Worked example

A 0.50 kg0.50\ \text{kg} mass on a spring of constant k=200 N/mk=200\ \text{N/m} oscillates in a fluid producing a damping force −bx˙-b\dot x with b=4.0 kg/sb=4.0\ \text{kg/s}. Find the regime, the oscillation frequency ω1\omega_1, the time for the amplitude to fall to 1/e1/e of its initial value, the quality factor QQ, and the time for the oscillator's energy to fall to half its initial value.

Step 1. Compute the two governing constants. ω0=k/m=200/0.50=400=20 rad/s\omega_0=\sqrt{k/m}=\sqrt{200/0.50}=\sqrt{400}=20\ \text{rad/s}, and γ=b/2m=4.0/(2×0.50)=4.0 s−1\gamma=b/2m=4.0/(2\times0.50)=4.0\ \text{s}^{-1}.

Step 2. Classify. Since γ=4.0 s−1<ω0=20 rad/s\gamma=4.0\ \text{s}^{-1}<\omega_0=20\ \text{rad/s}, the system is underdamped.

Step 3. Find ω1\omega_1 and the 1/e1/e amplitude time. ω1=ω02−γ2=400−16=384≈19.6 rad/s\omega_1=\sqrt{\omega_0^2-\gamma^2}=\sqrt{400-16}=\sqrt{384}\approx19.6\ \text{rad/s}. The amplitude envelope Ae−γtAe^{-\gamma t} falls to 1/e1/e when γt=1\gamma t=1, so t=1/γ=1/4.0=0.25 st=1/\gamma=1/4.0=0.25\ \text{s}.

Step 4. Find QQ. Q=ω0/2γ=20/(2×4.0)=2.5Q=\omega_0/2\gamma=20/(2\times4.0)=2.5 — a fairly low quality factor, consistent with visibly damped motion over just a few cycles.

Step 5. Extend to the energy half-life. Since E(t)=E0e−2γtE(t)=E_0e^{-2\gamma t}, energy halves when e−2γt=12e^{-2\gamma t}=\tfrac12, i.e. 2γt=ln⁡22\gamma t=\ln2, so t1/2=ln⁡2/(2γ)=0.693/8.0≈0.0866 st_{1/2}=\ln2/(2\gamma)=0.693/8.0\approx0.0866\ \text{s}. The motion is underdamped, with ω1≈19.6 rad/s\omega_1\approx19.6\ \text{rad/s}, amplitude 1/e1/e-time 0.25 s0.25\ \text{s}, Q=2.5Q=2.5, and the energy halves in about 0.087 s0.087\ \text{s} — roughly a third of the amplitude decay time, exactly as expected since energy decays twice as fast as amplitude.

Common traps

  • Writing ω1=ω0\omega_1=\omega_0 for the underdamped case — the true oscillation frequency ω1=ω02−γ2\omega_1=\sqrt{\omega_0^2-\gamma^2} is always slightly less than ω0\omega_0.
  • Mixing amplitude decay e−γte^{-\gamma t} with energy decay e−2γte^{-2\gamma t} — forgetting the factor of 2 is the single most common slip when finding a "time to halve."
  • Thinking overdamped means "returns fastest because it is damped the most" — critical damping (γ=ω0\gamma=\omega_0) is the fastest non-oscillating return; overdamping (γ>ω0\gamma>\omega_0) is actually slower.
  • Treating Q=ω0/2γQ=\omega_0/2\gamma as meaningful for overdamped or critically damped systems — it is a useful cycles-to-decay measure only for underdamped, reasonably high-QQ oscillators.

Ask the tutor: if you would like to work through an overdamped numerical example or see the full initial-value algebra for the underdamped case, just ask.

Why does critical damping return to equilibrium faster than overdamping, when overdamping sounds like "more damping"?

The name is misleading. In the overdamped case the motion is a sum of two decaying exponentials with rates γ±γ2−ω02\gamma\pm\sqrt{\gamma^2-\omega_0^2}, and the slower of the two, γ−γ2−ω02\gamma-\sqrt{\gamma^2-\omega_0^2}, is always smaller than γ\gamma itself and controls the long-time behaviour. Critical damping sits exactly at the boundary where this slow rate is maximised at γ=ω0\gamma=\omega_0, so it decays faster than any overdamped case. Overdamping means the system is so sluggish (high viscosity relative to the restoring force) that it takes longer to creep home, not shorter.

Does the mass ever cross the equilibrium position in the critically damped or overdamped case?

Generally no, or at most once, depending on the initial conditions. Both solutions, (A+Bt)e−γt(A+Bt)e^{-\gamma t} for critical damping and C1eλ1t+C2eλ2tC_1e^{\lambda_1t}+C_2e^{\lambda_2t} for overdamping, are built from exponentials with no built-in oscillation, so unless the initial velocity is large and pointed strongly toward equilibrium there is no repeated crossing. In the worked critical-damping case released from rest, the mass moves monotonically back toward zero without ever overshooting it — that is precisely what "no oscillation" means here.

How is Q related to how many oscillations I actually see before the motion dies out?

Roughly, QQ tells you the number of radians of oscillation that occur while the energy falls by a factor 1/e1/e; dividing by 2π2\pi gives an approximate number of full cycles. In the worked example, Q=2.5Q=2.5 means the oscillator only completes roughly 2.5/(2π)≈0.42.5/(2\pi)\approx0.4 of a cycle before its energy drops to about 37%37\% of its starting value — matching the fact that with only ω1≈19.6 rad/s\omega_1\approx19.6\ \text{rad/s} against a fairly fast decay rate γ=4.0 s−1\gamma=4.0\ \text{s}^{-1}, you would see the oscillation die out within a cycle or two on a graph, not ring on indefinitely like a high-QQ tuning fork.

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Where this leads