Damped free vibration and logarithmic decrement

Why it matters

Real vibrations die away because energy is dissipated in oil, air, joints and materials. How fast they die is set by damping, and damping controls the peak response at resonance, the settling time of a servo axis or a measuring instrument, and the ride of a vehicle on its shock absorbers. The logarithmic decrement is the standard way to measure damping from a simple tap test.

Key ideas

Viscous damping. A dashpot force proportional to velocity and opposing it: F_d = c·ẋ, with c the damping coefficient (N·s/m). It is the usual model because it gives a linear equation and represents oil dampers and many lightly damped structures well.

Equation of motion. m·ẍ + c·ẋ + k·x = 0. Trying x = e^(s·t) gives m·s² + c·s + k = 0, and the nature of the roots decides the motion.

Critical damping and damping ratio. The critical damping coefficient is the value of c that makes the two roots equal: c_c = 2√(k·m) = 2·m·ω_n. The damping ratio ζ = c / c_c classifies the motion:

• Underdamped (ζ < 1): oscillation at the damped natural frequency ω_d = ω_n·√(1 − ζ²) inside an exponentially decaying envelope e^(−ζ·ω_n·t). Almost all structures and machines (ζ ≈ 0.01–0.1).
• Critically damped (ζ = 1): returns to equilibrium fastest without oscillating (at most one overshoot through zero for some initial velocities). Used for gun recoil mechanisms and door closers.
• Overdamped (ζ > 1): creeps back without oscillation, more slowly than critical damping. Measuring instruments and many servo loops are tuned to ζ ≈ 0.6–0.7, a compromise between speed and overshoot; car shock absorbers are typically around 0.2–0.4.

Natural frequency is unchanged. Damping does not change ω_n = √(k/m); it changes the frequency of the free oscillation to ω_d, which is slightly lower. For ζ = 0.1, ω_d is only 0.5% below ω_n.

Logarithmic decrement. For an underdamped system the ratio of any two successive peaks on the same side is constant, x_n / x_(n+1) = e^(ζ·ω_n·T_d). Its natural logarithm is the logarithmic decrement δ, linked directly to ζ. Measuring the decay of a free vibration (tap test, pluck test) therefore gives ζ without knowing m, k or c. Over N cycles the ratio is the N-th power, so use many cycles for accuracy.

Coulomb (dry friction) damping. A constant friction force F opposes motion. The frequency stays ω_n, but the amplitude falls linearly by 4F/k per cycle, and motion stops when the spring force can no longer overcome friction (amplitude below F/k). The straight-line envelope distinguishes it from the exponential envelope of viscous damping.

Quality factor. For light damping Q = 1/(2ζ): the resonance amplification of a forced system (next topic), and a measure of how sharp the resonance is.

Formulas

m·ẍ + c·ẋ + k·x = 0 c_c = 2√(k·m) = 2·m·ω_n (N·s/m) ; ζ = c / c_c (dimensionless) ω_n = √(k/m) ; ω_d = ω_n·√(1 − ζ²) ; T_d = 2π / ω_d (rad/s, s)

Underdamped response: x(t) = X·e^(−ζ·ω_n·t)·sin(ω_d·t + φ)

δ = ln(x_n / x_(n+1)) = 2π·ζ / √(1 − ζ²) ; for small ζ, δ ≈ 2π·ζ δ = (1/N)·ln(x₀ / x_N) (amplitude measured N cycles apart) ζ = δ / √(4π² + δ²)

Coulomb damping: amplitude loss per cycle = 4F/k ; per half cycle = 2F/k

Q ≈ 1 / (2ζ)

Worked examples

Example 1 (standard): classify and find the decay. Given: m = 10 kg, k = 4000 N/m, c = 60 N·s/m.

1. ω_n = √(4000 / 10) = 20 rad/s
2. c_c = 2·m·ω_n = 2 × 10 × 20 = 400 N·s/m → ζ = 60 / 400 = 0.15 (underdamped).
3. ω_d = 20 × √(1 − 0.0225) = 19.77 rad/s
4. δ = 2π × 0.15 / √(1 − 0.0225) = 0.953; each peak is e^0.953 = 2.59 times the next.
5. After 3 cycles the amplitude is e^(−3 × 0.953) = 0.057 of the start, i.e. 5.7%. Answer: ζ = 0.15, ω_d ≈ 19.8 rad/s, δ ≈ 0.953; about 5.7% of the amplitude remains after 3 cycles.

Example 2 (GATE level): identify a system from a tap test. Given: a 2 kg mass on a spring and damper is tapped; the amplitude falls from 12 mm to 3 mm in 5 cycles, and the measured period of oscillation is 0.2 s. Find ζ, k and c.

1. δ = (1/5)·ln(12/3) = (1/5) × 1.386 = 0.2773
2. ζ = δ / √(4π² + δ²) = 0.2773 / √(39.48 + 0.077) = 0.2773 / 6.289 = 0.0441
3. Measured period is damped: ω_d = 2π / 0.2 = 31.42 rad/s → ω_n = 31.42 / √(1 − 0.0441²) = 31.45 rad/s
4. k = m·ω_n² = 2 × 31.45² = 1978 N/m
5. c = 2·ζ·m·ω_n = 2 × 0.0441 × 2 × 31.45 = 5.55 N·s/m Answer: ζ ≈ 0.044, k ≈ 1.98 kN/m, c ≈ 5.5 N·s/m.

Common mistakes

• Saying damping "reduces the natural frequency"; it is the damped frequency ω_d that is lower.
• Taking the ratio of a positive peak to the next negative peak for δ; use successive peaks on the same side (one full cycle apart).
• Forgetting the 1/N when amplitudes are N cycles apart, or counting N wrongly (from the 1st to the 4th peak is N = 3).
• Using δ = 2πζ for heavy damping; use the exact form.
• Treating the measured period as undamped when converting to k.
• Expecting an exponential envelope from dry friction; Coulomb damping decays linearly.

For GATE ME

Common: damping ratio from given m, k and c; logarithmic decrement from amplitude ratios over several cycles; ζ from δ; damped natural frequency; identifying under-, critically- and over-damped responses; the amplitude drop per cycle with Coulomb friction. Practise the δ–ζ conversion both ways.

Quick check

1. m = 1 kg, k = 100 N/m. What c gives critical damping?
2. δ = 0.5. Find ζ.
3. Successive peaks are 10 mm and 8 mm. Find δ.
4. With dry friction F = 5 N and k = 1000 N/m, how much amplitude is lost per cycle?
5. Is ω_d greater or less than ω_n?

Answers: 1. 20 N·s/m. 2. 0.5/√(39.48 + 0.25) = 0.0793. 3. ln 1.25 = 0.223. 4. 4F/k = 20 mm. 5. Less.