Damped free vibration and logarithmic decrement

Why it matters

A car that keeps bouncing after a speed breaker has worn dampers: its suspension is too lightly damped. The damping ratio decides whether a disturbed system rings for many cycles, settles in one or two, or creeps back slowly, and the logarithmic decrement is how you measure it from a recorded decay curve. Shock absorbers, engine mounts and instrument needles are all tuned with these ideas.

Key ideas

Viscous damping. A damping force proportional to velocity and opposing it, F_d = c·ẋ, where c is the damping coefficient (N·s/m). It models an oil-filled shock absorber well and makes the equation linear: m·ẍ + c·ẋ + k·x = 0.

Critical damping and damping ratio. The critical damping coefficient c_c = 2√(k·m) = 2·m·ω_n is the smallest damping for which the system returns to equilibrium without oscillating. The damping ratio ζ = c / c_c classifies the motion:

• ζ < 1, underdamped: oscillation at ω_d = ω_n√(1 − ζ²) inside an exponentially decaying envelope e^(−ζ·ω_n·t). Almost all mechanical structures, and car suspensions (ζ about 0.2 to 0.4), are here.
• ζ = 1, critically damped: no oscillation and the fastest return to equilibrium without overshoot. Used for instrument movements and some door closers.
• ζ > 1, overdamped: no oscillation, and the return is slower the larger ζ is.

Damped natural frequency. ω_d = ω_n√(1 − ζ²) is always lower than ω_n, but the difference is tiny for light damping: at ζ = 0.1 it is 0.5%, at ζ = 0.2 it is 2%.

Logarithmic decrement. In underdamped free vibration, the ratio of any two successive peaks on the same side is constant, x_i / x_(i+1) = e^(ζ·ω_n·T_d). Its natural logarithm, δ, is the logarithmic decrement. It can be measured from n cycles to reduce reading error, and it gives ζ directly. For small damping δ ≈ 2πζ.

Energy dissipated. In one cycle of amplitude X at frequency ω, a viscous damper dissipates π·c·ω·X². The fractional energy loss per cycle is about 2δ for light damping.

Coulomb (dry friction) damping. A constant friction force F opposes motion. The frequency stays ω_n, but the amplitude falls linearly, by 2F/k every half cycle (4F/k per cycle), and motion stops once the amplitude at a turning point is within ±F/k of equilibrium, because the spring force can no longer overcome friction. Leaf springs with inter-leaf friction behave partly like this.

Other damping. Structural (hysteretic) damping comes from internal friction in materials; it is small for steel, larger for rubber and cast iron. Real systems are usually modelled with an equivalent viscous ζ.

Formulas

m·ẍ + c·ẋ + k·x = 0

• c: viscous damping coefficient (N·s/m); m, k as before.

c_c = 2·√(k·m) = 2·m·ω_n and ζ = c / c_c

• c_c: critical damping coefficient (N·s/m); ζ: damping ratio (dimensionless).

ω_d = ω_n·√(1 − ζ²) and T_d = 2π / ω_d

• ω_d: damped natural frequency (rad/s); T_d: damped period (s). Underdamped only (ζ < 1).

x(t) = X·e^(−ζ·ω_n·t)·sin(ω_d·t + φ)

• Underdamped response; X, φ from initial conditions.

δ = ln(x_i / x_(i+1)) = (1/n)·ln(x_0 / x_n) = 2π·ζ / √(1 − ζ²)

• δ: logarithmic decrement (dimensionless); x_0, x_n: peak amplitudes n cycles apart (same side of equilibrium).

ζ = δ / √(4π² + δ²) (≈ δ / 2π for small δ)

ΔX_cycle = 4·F / k (Coulomb damping)

• F: friction force (N); amplitude falls linearly by this amount each full cycle; motion stops when the amplitude is less than F/k.

Worked examples

Example 1 (standard). A mass of 10 kg is supported by a spring of 4000 N/m and a damper of 80 N·s/m. Find ω_n, ζ, ω_d, the logarithmic decrement and the ratio of successive amplitudes.

1. ω_n = √(4000/10) = 20.0 rad/s.
2. c_c = 2√(4000 × 10) = 2 × 200 = 400 N·s/m; ζ = 80 / 400 = 0.20.
3. ω_d = 20 × √(1 − 0.04) = 20 × 0.9798 = 19.60 rad/s; T_d = 2π / 19.60 = 0.321 s.
4. δ = 2π × 0.2 / √0.96 = 1.2566 / 0.9798 = 1.283.
5. x_i / x_(i+1) = e^1.283 = 3.61: each peak is 28% of the one before.
6. Cycles to fall to 10% of the starting amplitude: n = ln 10 / δ = 2.303 / 1.283 = 1.8, so about two cycles.

Example 2 (GATE level: measuring damping). A quarter-car rig (sprung mass 250 kg, spring 25 kN/m) is released and the recorded peaks fall from 12 mm to 3 mm in 5 cycles. Find the damping ratio and the damping coefficient. Is the damper healthy if the design value is ζ = 0.25?

1. δ = (1/5)·ln(12/3) = (1/5) × 1.3863 = 0.2773.
2. ζ = δ / √(4π² + δ²) = 0.2773 / √(39.478 + 0.0769) = 0.2773 / 6.289 = 0.0441.
3. c = 2ζ√(k·m) = 2 × 0.0441 × √(25 000 × 250) = 0.0882 × 2500 = 220 N·s/m.
4. Design: c = 2 × 0.25 × 2500 = 1250 N·s/m. The measured value is less than a fifth of it: the damper has failed, which is why the body keeps bouncing.

Example 3 (Coulomb damping). A 5 kg block on a spring of 2000 N/m slides on a surface with μ = 0.1 and is released from 20 mm. Friction F = 0.1 × 5 × 9.81 = 4.905 N, so each half-cycle reduces the amplitude by 2F/k = 4.905 mm. Turning-point amplitudes: 20 → 15.10 → 10.19 → 5.29 → 0.38 mm. Since 0.38 mm < F/k = 2.45 mm, the block stops after two full cycles, 0.38 mm from the unstretched position.

Common mistakes

• Using c_c = 2√(k/m). It is 2√(k·m) (units N·s/m).
• Forgetting the 1/n when amplitudes are n cycles apart, or using amplitudes on opposite sides of equilibrium (half a cycle apart).
• Using log base 10 instead of the natural logarithm.
• Saying a larger ζ always returns the system faster. Beyond ζ = 1 the return gets slower.
• Applying the exponential decay or the log decrement to Coulomb damping, where the decay is linear.
• Writing ω_d in Hz with an ω symbol; keep rad/s and Hz separate.

For GATE ME

Expect ζ, c_c and ω_d from m, k and c; log decrement from amplitude readings and the resulting ζ; the number of cycles for a given decay; classification as under-, critically or over-damped; and amplitude loss under Coulomb friction. Practise the δ–ζ conversion both ways and keep four significant figures through the logarithms.

Quick check

1. m = 2 kg, k = 800 N/m. Critical damping coefficient?
2. ζ = 0.3, ω_n = 15 rad/s. ω_d?
3. Successive amplitudes 10 mm and 8 mm. δ and ζ?
4. A system with ζ = 1.5 is displaced and released. Does it oscillate?
5. How does the amplitude decay under Coulomb damping?

Answers: 1. 2√1600 = 80 N·s/m. 2. 15√0.91 = 14.31 rad/s. 3. δ = ln 1.25 = 0.223, ζ = 0.0355. 4. No, it is overdamped. 5. Linearly, by 4F/k per cycle.