Understanding the damped oscillator and resonance calculator — concept and principle
A mass on a spring with a damper (a shock absorber, friction or air drag) is the basic model for vibration in cars, buildings, machines, sensors and loudspeakers. This calculator takes mass, stiffness and damping coefficient and returns the natural frequency, damping ratio, damped frequency, quality factor and the type of response.
How it is calculated
- Natural frequency: ωₙ = √(k ÷ m) in rad/s, shown as fₙ = ωₙ ÷ 2π in Hz. m is mass in kg, k is stiffness in N/m.
- Damping ratio: ζ = c ÷ (2√(km)), where c is the damping coefficient in N·s/m. The value 2√(km) is the critical damping coefficient.
- Damped frequency: f_d = fₙ √(1 − ζ²), defined only when ζ < 1.
- Quality factor: Q = 1 ÷ (2ζ).
- Type: underdamped (ζ < 1), critically damped (ζ = 1) or overdamped (ζ > 1).
The principle
The equation of motion is m x'' + c x' + k x = 0: inertia, a damping force proportional to velocity, and a spring force proportional to displacement. Its solutions decay as e^(−ζωₙt). If ζ < 1 the system still oscillates, slightly slower than its natural frequency, with an envelope that shrinks each cycle. At ζ = 1 it returns to rest as fast as possible without overshooting. Above 1 it creeps back slowly with no oscillation.
Q describes how sharp the resonance is when the system is driven by a periodic force: for light damping the steady amplitude at resonance is about Q times the static deflection.
Worked example
- Inputs: m = 1 kg, k = 100 N/m, c = 2 N·s/m.
- ωₙ = √(100 ÷ 1) = 10 rad/s, so fₙ = 10 ÷ 2π ≈ 1.591549431 Hz.
- Critical damping is 2√(100 × 1) = 20 N·s/m, so ζ = 2 ÷ 20 = 0.1.
- f_d = 1.591549431 × √(1 − 0.01) ≈ 1.583571689 Hz.
- Q = 1 ÷ (2 × 0.1) = 5, and since ζ < 1 the type is underdamped.
| Mass (kg) | Natural frequency (Hz) |
|---|---|
| 0.25 | 3.183098862 |
| 0.5 | 2.25079079 |
| 0.75 | 1.837762985 |
| 1 | 1.591549431 |
| 1.25 | 1.423525087 |
| 1.5 | 1.299494669 |
| 2 | 1.125395395 |
How to read the result
With ζ = 0.1 each cycle the amplitude falls to about 53% of the previous one (logarithmic decrement 2πζ ÷ √(1 − ζ²) ≈ 0.63), and the envelope time constant is 1 ÷ (ζωₙ) = 1 s. Car suspensions are usually tuned well below critical damping; instruments and door closers often aim near ζ ≈ 0.6–1. Steel structures can have ζ of only a few percent, giving high Q and strong resonance. Avoid driving a lightly damped system near fₙ.
Common mistakes
- Using weight in newtons instead of mass in kg.
- Confusing ωₙ (rad/s) with fₙ (Hz); they differ by 2π.
- Using Q = 1 ÷ (2ζ) for heavy damping, where it loses its simple meaning.
Frequently asked questions
Why is the damped frequency lower than the natural frequency?
Damping slows each swing slightly, so f_d = fₙ√(1 − ζ²). For ζ = 0.1 the drop is only 0.5%.
At what frequency is a forced oscillation largest?
The displacement amplitude peaks at fₙ√(1 − 2ζ²), slightly below fₙ, and only if ζ < 1/√2.
How do I estimate ζ from a measurement?
Record successive peaks x₁ and x₂ one cycle apart, compute δ = ln(x₁ ÷ x₂), then ζ = δ ÷ √(4π² + δ²).
Sources: S. S. Rao, Mechanical Vibrations, Pearson · Halliday, Resnick and Walker, Fundamentals of Physics, chapter on oscillations