How to use it
- Enter the mass m in kilograms.
- Enter the spring stiffness k in N/m.
- Enter the damping coefficient b in kg/s (N·s/m) — the force per unit of speed from the damper.
- Optionally change the number of cycles to see how much the amplitude decays.
Formulas
The motion obeys m·x″ + b·x′ + k·x = 0.
- Natural angular frequency ω₀ = √(k ÷ m), natural frequency f₀ = ω₀ ÷ 2π
- Damping ratio ζ = b ÷ (2√(k·m)); critical damping b_c = 2√(k·m)
- Damped angular frequency ω_d = ω₀ √(1 − ζ²) (only for ζ < 1), period T_d = 2π ÷ ω_d
- Quality factor Q = 1 ÷ (2ζ)
- Logarithmic decrement δ = 2πζ ÷ √(1 − ζ²)
- Amplitude decays as e^(−b·t ÷ 2m): decay time 2m ÷ b, after N cycles e^(−N·δ) is left
Worked example
m = 1 kg, k = 100 N/m, b = 2 kg/s.
- ω₀ = √100 = 10 rad/s, f₀ = 1.5915 Hz
- ζ = 2 ÷ (2 × √100) = 0.1, so the system is underdamped; Q = 5
- ω_d = 10 × √0.99 = 9.9499 rad/s, f_d = 1.5836 Hz, T_d = 0.6315 s
- δ = 2π × 0.1 ÷ √0.99 = 0.6315; the amplitude falls to 1/e in 1 s
- after 10 cycles only e^(−6.315) = 0.181 % of the starting amplitude is left
Tips
- Critical damping (b = 2√(k·m), here 20 kg/s) is what door closers and instrument needles aim for: no overshoot and the fastest return.
- Measure ζ from a recording: take two peaks N cycles apart, δ = ln(A₀ ÷ A_N) ÷ N, then ζ = δ ÷ √(4π² + δ²).
- The model assumes viscous damping (force proportional to speed). Dry friction makes the amplitude fall linearly instead.