Calculator
Example data table
| Method | Inputs | Key outputs |
|---|---|---|
| m, c, k | m = 2 kg, k = 200 N/m, c = 10 N·s/m | ζ = 0.25, ωn = 10 rad/s, ωd ≈ 9.682 rad/s |
| Overshoot & peak time | Mp = 16.3%, Tp = 0.9 s | ζ ≈ 0.5, ωn ≈ 4.03 rad/s |
| Log decrement & damped period | δ = 0.5, Td = 0.65 s | ζ ≈ 0.079, ωn ≈ 9.70 rad/s |
Formula used
1) Mass–spring–damper
ωn = √(k/m), ζ = c / (2√(km)), ωd = ωn√(1 − ζ²)
2) Log decrement
ζ = δ / √((2π)² + δ²), ωd = 2π/Td, ωn = ωd/√(1 − ζ²)
3) Overshoot relations (second-order underdamped)
ζ = −ln(Mp/100) / √(π² + ln²(Mp/100)), ωd = π/Tp
4) Settling-time approximation
Ts(2%) ≈ 4/(ζωn) and Ts(5%) ≈ 3/(ζωn)
How to use this calculator
- Select the input method that matches your data.
- Enter values using consistent units shown beside fields.
- Click Calculate to view results above the form.
- Use CSV or PDF buttons to export computed outputs.
- If ζ ≥ 1, treat the response as non-oscillatory.
Damping Ratio and Natural Frequency Explained
1) What natural frequency means
Natural frequency describes how fast a system would oscillate if it had no damping. For a classic mass–spring model, the calculator uses ωn = √(k/m). Doubling stiffness k increases ωn by √2, while doubling mass m reduces ωn by √2. You can convert to cycles per second with fn = ωn/(2π), and the natural period is Tn = 2π/ωn.
2) What damping ratio means
Damping ratio ζ is dimensionless and compares actual damping to critical damping. With measured m, c, and k, ζ = c / (2√(km)). A value of ζ = 0 means no damping, while ζ = 1 is the boundary between oscillatory and non‑oscillatory response. The decay rate is σ = ζ·ωn; for ζ = 0.25 and ωn = 10 rad/s, σ = 2.5 s−1 and τ = 1/σ ≈ 0.4 s.
3) Typical ζ ranges and behavior
Many mechanical assemblies operate around ζ = 0.02–0.20. Control systems often target ζ ≈ 0.5–0.8 to balance speed and overshoot. When ζ < 1 the response is underdamped; at ζ = 1 it is critically damped; and when ζ > 1 it becomes overdamped.
4) How ωd differs from ωn
Damped frequency is ωd = ωn√(1 − ζ²). For ζ = 0.1, ωd ≈ 0.995 ωn; for ζ = 0.5, ωd ≈ 0.866 ωn. If ζ ≥ 1, ωd is not defined as a sinusoidal frequency, which is why the calculator hides it.
5) Overshoot, peak time, and settling time
For a standard second‑order step response, percent overshoot relates to ζ through ζ = −ln(Mp/100)/√(π² + ln²(Mp/100)). Peak time uses ωd = π/Tp. Settling time commonly uses Ts(2%) ≈ 4/(ζωn) or Ts(5%) ≈ 3/(ζωn).
6) Practical measurement tips
When using log decrement, measure two successive peak amplitudes x0 and x1 and compute δ = ln(x0/x1), then measure the damped period Td between peaks. For noisy data, average δ across 5–10 cycles.
FAQs
1) What damping ratio typically limits overshoot?
For a classic second‑order response, ζ ≈ 0.7 gives fast settling with very small overshoot. Lower ζ (0.2–0.5) responds faster but overshoots more, while higher ζ reduces overshoot but can feel sluggish.
2) Why is damped frequency missing when ζ ≥ 1?
When ζ is one or greater, the system does not oscillate sinusoidally. Instead of a ringing frequency, the response is a sum of decaying exponentials. In that case ωd is not meaningful as an oscillation rate.
3) Can I enter frequency in Hz instead of rad/s?
Yes. If you have fn in Hz, convert with ωn = 2πfn. The calculator also reports both ωn and fn, plus periods Tn and Td when applicable.
4) How do I calculate log decrement from measurements?
Record two successive peak amplitudes x0 and x1 from the same waveform, then compute δ = ln(x0/x1). Use the time between peaks as Td. Averaging several cycles improves stability.
5) Are the settling‑time formulas exact?
They are widely used approximations for a dominant second‑order system. Real systems may have higher‑order dynamics, nonlinear damping, or actuator limits. Treat Ts(2%) ≈ 4/(ζωn) and Ts(5%) ≈ 3/(ζωn) as estimates.
6) What does it mean if ζ comes out greater than one?
ζ > 1 indicates an overdamped system that returns without oscillation. Check units for c, k, and m, then interpret results using decay rate σ and time constant τ. Overdamped systems can be stable but slower to respond.