Library / Mechanical / MEC-27
Mechanical
Forced vibration amplitude
Steady-state vibration amplitude of a mass–spring–damper under a sinusoidal force.
- Ref
- MEC-27
- Rev
- v1
- Field
- Mechanical
- Result
- X [mm]
For the linear mass–spring–damper m ẍ + c ẋ + k x = F₀ sin ωt, with X the steady-state displacement amplitude. At r = 1 the amplification is 1/(2ζ) for positive damping. Peak amplification occurs just below r = 1 for light damping.
Shaking a spring-mounted mass at its natural frequency makes it swing far more than a steady push of the same size would.
Calculator
- Static deflection F₀/k0.5 mm
- Amplification factor10
What drives the result
Each variable is moved across its typical range while the others stay at the values above. The longest bar is the key driver.
- With ζ = 0.05 a 10 N force at resonance gives an amplitude of 5 mm, ten times the 0.5 mm static deflection.
- Well above resonance (r large) the amplitude falls and the mass is effectively isolated.
Chart data as tables
Ranked variables
| Variable | Typical range | Lowest result | Highest result | Swing (mm) | Local step | Change in result |
|---|---|---|---|---|---|---|
| k Stiffness | 0.1 to 1,000 kN/m | 0.1 | 1,000 | 999.9 | +10% | −9.09 % |
| F₀ Force amplitude | 0.1 to 1,000 N | 0.05 | 500 | 499.9 | +10% | +10 % |
| ζ Damping ratio | 0.01 to 1 | 0.25 | 25 | 24.75 | +10% | −9.09 % |
| r Frequency ratio | 1 to 5 | 0.02083 | 5 | 4.979 | +10% | −57.8 % |
Sweep of Frequency ratio
| Frequency ratio | Amplitude (mm) |
|---|---|
| 1 | 5 |
| 1.5 | 0.3972 |
| 2 | 0.1663 |
| 2.5 | 0.09513 |
| 3 | 0.06246 |
| 3.5 | 0.04442 |
| 4 | 0.03332 |
| 4.5 | 0.02597 |
| 5 | 0.02083 |
Sweep one variable
Variables and how each one contributes
Amplitude
What the equation gives you. Everything else feeds it.
Force amplitude
Amplitude of the applied force. Directly proportional.
Stiffness
Spring stiffness. Inversely proportional.
Damping ratio
Fraction of critical damping. Dominant near resonance: amplitude at r = 1 is 1/(2ζ) × static deflection.
Frequency ratio
Forcing frequency divided by natural frequency. Peaks near r = 1.
Worked example with the values above
- Static deflection: F₀/k = 10 / 20,000 = 0.5 mm.
- Amplification: 1/√[(1 − r²)² + (2ζr)²] = 10.
- X = 5 mm.
Assumptions and limits
- Linear system with viscous damping.
- Steady-state sinusoidal force.
Common mistakes
- Forgetting that the amplification is 1/(2ζ) only at r = 1.
- Confusing force excitation with base excitation.
Statistics and uncertainty
Near resonance the amplitude is very sensitive to damping and to the frequency ratio; ±5% in ζ is ±5% in X.
Where it is used
- Vibration of mounted equipment.
- Avoiding resonance in rotating machinery.
Maths you need
- Natural frequency: ωn = √(k/m)
- Damping ratio: ζ