Damped Oscillations

Keep the spring force −kx and add a drag that opposes the velocity, −bv. The constant b is a force per unit speed, N per (m s⁻¹), which reduces to kg s⁻¹. Newton's second law gives m d²x/dt² = −kx − b dx/dt. Divide by m and only two combinations survive: k/m = ω₀², the square of the natural angular frequency, and b/m = 2γ, where γ = b/2m is a decay rate in s⁻¹. Both are rates, so they can be compared directly, and that comparison is the whole topic. Take m = 2.0 kg, k = 200 N m⁻¹ and b = 4.0 kg s⁻¹: ω₀ = √(200/2.0) = 10 rad s⁻¹ and γ = 4.0/(2 × 2.0) = 1.0 s⁻¹. The spring acts ten times faster than the drag, so expect the block to oscillate with the drag as a slow correction. The 2 in the definition is there so that γ itself, not γ/2, turns out to be the rate at which the amplitude decays.

A linear equation with constant coefficients is solved by guessing x = e^(rt), because differentiating only multiplies by r. Substituting gives (mr² + br + k)e^(rt) = 0, so r² + 2γr + ω₀² = 0, with roots r = −γ ± √(γ² − ω₀²). Everything hangs on the sign under the root. If γ is less than ω₀ it is negative, the roots are the complex pair −γ ± iω′ with ω′ = √(ω₀² − γ²), and their real combinations are e^(−γt) cos ω′t and e^(−γt) sin ω′t: a decaying oscillation. If γ equals ω₀ the roots merge into −ω₀ and the second solution is t e^(−ω₀t). If γ exceeds ω₀ both roots are real and negative: two plain decays. If complex exponentials are new to you, skip the guess and substitute Ae^(−γt) cos(ω′t + φ) directly. The sine terms cancel identically, and the cosine terms cancel only when ω′² = ω₀² − γ².

With γ = 1.0 s⁻¹ and ω₀ = 10 rad s⁻¹ the damped angular frequency is ω′ = √(100 − 1.0) = 9.95 rad s⁻¹, so the period is 2π/9.95 = 0.6315 s against 0.6283 s undamped: half a percent longer. Meanwhile the envelope Ae^(−γt) shrinks by the factor e^(γT′) = e^(0.6315) = 1.88 every cycle. That contrast is the thing to remember. Damping strong enough to nearly halve the amplitude each cycle barely touches the timing, because γ enters ω′ only through γ²/ω₀² = 1%. The period is also the same from the first swing to the last: a decaying amplitude does not mean a slowing oscillation. The amplitude halves in ln 2/γ = 0.69 s whatever its starting value, which is the signature of exponential decay. For a block released from rest at x₀, every turning point lies exactly on ±x₀e^(−γt), at times that are whole multiples of π/ω′.

Raise b until γ = ω₀ and ω′ reaches zero: b_c = 2mω₀ = 2√(km) = 2√(200 × 2.0) = 40 kg s⁻¹. Released from rest at x₀, the critically damped block follows x = x₀(1 + ω₀t)e^(−ω₀t) and never crosses equilibrium; after 0.50 s it is down to 6.0 × e^(−5.0) = 4.0% of x₀. Now go past it to b = 50 kg s⁻¹. Then γ = 12.5 s⁻¹, √(γ² − ω₀²) = √(156.25 − 100) = 7.5 s⁻¹, and the two decay rates are r₁ = 5.0 s⁻¹ and r₂ = 20 s⁻¹. Their product is always ω₀² = 100 s⁻², so pushing one rate up drags the other down. The fast term is gone within a fraction of a second and the slow one, e^(−5.0t), governs the return: at 0.50 s the block is still at 10.9% of x₀, more than twice as far out as with critical damping. For very heavy damping r₁ ≈ ω₀²/2γ = k/b. The mass has dropped out, and the spring is simply dragging the block through the fluid.

Write E = ½mv² + ½kx² and differentiate: dE/dt = mv(dv/dt) + kxv = v(m dv/dt + kx). The equation of motion says the bracket equals −bv, so dE/dt = −bv² exactly, in every regime. This is just force times velocity for the drag, and it is never positive. It also says when the loss happens: fastest as the block sweeps through equilibrium, where v is largest, and not at all at the turning points, so E(t) descends like a staircase rather than a smooth slope. Averaged over a cycle of light damping, E follows the square of the amplitude: E ≈ ½kA²e^(−2γt) = E₀e^(−bt/m). Energy therefore decays twice as fast as amplitude, with time constant m/b. For the block above, released at x₀ = 8.0 cm, E₀ = ½ × 200 × 0.080² = 0.64 J. One full cycle later it turns round at 1/1.88 = 53% of x₀ with all of its energy potential again, so E = 0.532² × 0.64 J = 0.18 J, just 28% of the start: the amplitude ratio gets squared.

An experiment hands you peaks, not b. Successive maxima on the same side are one period T′ apart, so their ratio is e^(γT′), and its logarithm δ = γT′ is called the logarithmic decrement. Use peaks n cycles apart to beat down reading error: γ = ln(A₀/Aₙ)/(nT′). A plot of ln(peak height) against time should be a straight line of slope −γ; if it curves, the drag is not linear and the model is wrong for your apparatus. The quality factor Q = ω₀/2γ = √(km)/b packages the same information as a pure number: 2π times the stored energy divided by the energy lost per cycle. A lightly damped oscillator loses the fraction 1 − e^(−2π/Q) ≈ 2π/Q of its energy each cycle, and its amplitude takes Q/π cycles to fall by the factor e. With b lowered to 1.0 kg s⁻¹ in the running example, γ = 0.25 s⁻¹ and Q = 20: about 6.4 cycles for each factor of e, and 27% of the energy gone every cycle.