Damped oscillations

A loss term has to reverse when the motion reverses, and velocity is the only quantity in the problem that does. Writing the retarding force as −b dx/dt with b > 0 guarantees that the power it delivers, −b(dx/dt)², is negative whatever the mass is doing. The choice is physically honest in two common cases: a small sphere in a viscous fluid at low Reynolds number obeys Stokes drag F = 6πηrv, and a conductor moving through a magnetic field feels an eddy-current force proportional to its speed. It is wrong for a body moving fast through air, where drag goes as v², and wrong for dry sliding friction, whose magnitude does not depend on speed at all. The payoff for the linear choice is that m d²x/dt² + b dx/dt + kx = 0 keeps constant coefficients, so superposition still holds and a single exponential trial solves it. In SI units b is measured in N s m⁻¹, which is the same as kg s⁻¹.

Substitute x = e^(st) into m d²x/dt² + b dx/dt + kx = 0. Every term carries the same factor e^(st), which is never zero, so it divides out and leaves ms² + bs + k = 0 — an algebraic equation for a number s of dimension T⁻¹. Its roots are s = [−b ± √(b² − 4mk)]/(2m). Divide through by 2m and name the two combinations that appear: the damping rate γ = b/(2m) in s⁻¹, and the undamped angular frequency ω₀ = √(k/m) in rad s⁻¹. Then s = −γ ± √(γ² − ω₀²). Both roots have negative real part whenever b > 0, which is the algebra saying the motion must die. Everything else in this topic is the sign of the discriminant γ² − ω₀², and it is tidier to track the dimensionless ratio ζ = γ/ω₀ = b/(2√(mk)): ζ < 1 rings, ζ = 1 is critical, ζ > 1 creeps.

For ζ < 1 the discriminant is negative and s = −γ ± iω′ with ω′ = √(ω₀² − γ²). Combining the two complex solutions into a real one gives x(t) = A₀e^(−γt)cos(ω′t + φ), where A₀ and φ are fixed by x(0) and v(0). This is not periodic in the strict sense — no later time reproduces the state exactly — but the zero crossings stay evenly spaced by π/ω′, and each maximum is a constant factor e^(−γT′) smaller than the last, with T′ = 2π/ω′. That constant ratio is what makes a graph of ln(amplitude) against time a straight line of slope −γ, and it is the standard way to measure b from data. Take ω₀ = 12.65 rad s⁻¹ and γ = 1.20 s⁻¹: ω′ = √(160 − 1.44) = 12.59 rad s⁻¹, a period stretched by only 0.45%, while the amplitude falls to 55% of its value every single cycle. Decay and frequency shift differ so wildly in size because γ enters ω′ squared.

For ζ > 1 both roots are real and negative, and x(t) = Ae^{(−(γ−β)t)} + Be^{(−(γ+β)t)} with β = √(γ² − ω₀²). The fast term dies quickly and the slow one sets the settling time. Because (γ − β)(γ + β) = ω₀², the slow rate is γ − β = ω₀²/(γ + β), and that decreases as b grows: an overdamped system becomes more sluggish, not crisper, the harder you damp it. At ζ = 1 exactly the two roots merge at −ω₀, and a repeated root forces a second solution te^(−ω₀t), so x(t) = (A + Bt)e^(−ω₀t). This is the quickest return the system can make without overshooting, which is why it is the design target for car dampers, door closers and analogue meter needles — most are set a little under critical, so the response stays sharp and one small overshoot buys a faster settle. For m = 0.50 kg and k = 200 N m⁻¹, critical damping needs b_c = 2√(mk) = 20.0 N s m⁻¹.

Stored energy goes as the square of amplitude, E = ½kA², so an envelope decaying as e^(−γt) gives E(t) = E₀e^(−2γt). The energy time constant is τ_E = 1/(2γ) = m/b, half the amplitude time constant. The quality factor packages the same information in cycles rather than seconds: Q = ω₀τ_E = ω₀/(2γ) = √(mk)/b, a dimensionless number. Since one cycle takes T₀ = 2π/ω₀, the energy drops to 1/e of its value in Q/2π cycles, and for Q ≫ 1 the fraction lost per cycle is 1 − e^(−2γT₀) ≈ 2π/Q. A system with Q = 137 loses 2π/137 = 4.6% per cycle, against an exact 4.5%, and rings for about 22 cycles before its energy is down by a factor of e. The range in nature is enormous: a car suspension has Q ≈ 1, a guitar string Q ≈ 10³, a quartz timing crystal Q ≈ 10⁵.

Slide the same mass on a dry surface instead. Kinetic friction has constant magnitude μₖ mg and always opposes motion, so over a half cycle from amplitude A₁ to A₂ it removes μₖ mg(A₁ + A₂) of energy. Setting that equal to ½kA₁² − ½kA₂² = ½k(A₁ + A₂)(A₁ − A₂) and cancelling the common factor gives A₁ − A₂ = 2μₖ mg/k — the same drop every half cycle however large the swing. The amplitude therefore falls along a straight line, not an exponential, by 4μₖ mg/k per full cycle. For m = 0.40 kg, k = 25 N m⁻¹ and μₖ = 0.15 that is 4.7 cm lost per half cycle. Two further differences: the period stays exactly 2π√(m/k) = 0.79 s, because no velocity-dependent term ever entered the equation, and the motion stops dead the moment a turning point lands inside |x| ≤ μₛ mg/k, leaving a permanent offset rather than creeping to zero.