MechanicalVibrationshardCourseworkPE / Advanced

Damping Ratio from Amplitude Decay

A damped single-degree-of-freedom system is displaced and released. Its free-vibration amplitude decays from $10.0\ \text{mm}$ to $4.0\ \text{mm}$ over $3$ complete cycles.

Find the damping ratio $\zeta$ .

Given

x0	10 mm	Initial amplitude
xn	4 mm	Amplitude after n cycles
n	3 -	Number of cycles

Hint 1

The logarithmic decrement δ measures decay per cycle — but the data spans three cycles.

Hint 2

δ = (1/n)·ln(x₀/xₙ).

Hint 3

Then ζ = δ/√(4π² + δ²). For light damping this is very close to δ/2π.

Worked solution — try the problem first

Step 1 — Logarithmic decrement over $n$ cycles.

$\delta = \frac{1}{n}\ln\!\left(\frac{x_0}{x_n}\right) = \frac{1}{3}\ln\!\left(\frac{10.0}{4.0}\right) = \frac{1}{3}\ln(2.5)$

$= \frac{0.91629}{3} = 0.30543$

Dividing by $n$ is essential — $\delta$ is defined per cycle, and the decay here spans three.

Step 2 — Convert to damping ratio.

$\zeta = \frac{\delta}{\sqrt{4\pi^2 + \delta^2}} = \frac{0.30543}{\sqrt{39.478 + 0.0933}} = \frac{0.30543}{\sqrt{39.5715}} = \frac{0.30543}{6.2906}$

$\boxed{\zeta = 0.0486}$

Light-damping check. For $\zeta \ll 1$ the approximation $\zeta \approx \delta/(2\pi)$ holds:

$\frac{0.30543}{6.2832} = 0.0486 \ \checkmark$

The two agree to three figures here because $\delta^2 = 0.093$ is negligible beside $4\pi^2 = 39.5$ . At around 5% of critical damping, this system is lightly damped and will ring for many cycles.

Concepts:Logarithmic decrementDamping ratioFree vibration decay

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