Forced vibration, resonance and isolation

Separate transient motion from the repeating response

For the same linear mass-spring-damper model, a sinusoidal force of amplitude F0 and angular frequency ω produces a transient plus a steady sinusoidal response. The transient depends on initial conditions and decays when damping is positive. The steady displacement amplitude X depends on frequency, damping and stiffness. A short record before settling may not equal that steady response.

Define frequency ratio r=ω/ωn and damping ratio ζ as in the previous lesson. For a force applied directly to the mass with a fixed base, the displacement magnification relative to static deflection F0/k is D=X/(F0/k)=1/√[(1−r²)²+(2ζr)²]. This dimensionless ratio is an amplitude comparison, not a force rating or an efficiency.

A response peak is sensitive to model details

The left plot shows why a small drive can produce large displacement near r=1 when damping is low. At exactly r=1, the formula gives D=1/(2ζ). For displacement response, the actual maximum is slightly below r=1 for these damping values; the formula, not an exact-frequency slogan, defines the peak.

More damping reduces that near-resonance peak in this model. Real systems may have changing stiffness, friction, clearances or several modes. Unbounded steady amplitude in an undamped resonant idealisation signals a missing limiting mechanism; it does not predict that actual motion becomes physically infinite. Excessive motion or stress can invalidate the assumed linear range first.

Moving-base isolation has a different numerator

If the base itself moves with displacement amplitude Y, the mass’s absolute displacement ratio is T=X/Y=√[1+(2ζr)²]D. This is the right plot. It describes a different input/output pair from direct-force D. Relative movement between mass and base is another quantity again and can matter for available travel.

For this ideal base-motion model, T<1 means less absolute motion than the base and occurs for r>√2. Damping can reduce amplification near resonance yet raise high-frequency transmission. Source reduction, source isolation and receiver isolation are distinct options. A practical decision must retain static support, travel, other modes and startup conditions; simply choosing a softer mount is not a complete solution.

Calculate a defined response rather than an unspecified reduction

Let k=100 N/m, F0=0.10 N and ζ=0.10. The reference static deflection is F0/k=0.001 m=1 mm. At r=1, D=5, so the steady directly forced amplitude is 5 mm. Raising ζ to 0.30 gives D≈1.67 at that same frequency. These are point comparisons, not the exact peak locations.

At r=3 and ζ=0.10, D≈0.12465. For a separate moving-base case, T=√1.36×0.12465≈0.14537. A 0.10 mm base-motion amplitude would then give about 0.0145 mm absolute mass-motion amplitude under that model. The direct-force magnification cannot simply be labelled the base transmission ratio.

Try the reasoning

1. Why can a brief startup trace differ from the plotted steady amplitude?

   Show the answer and reasoning

   It can still contain transient motion selected by the initial state.

2. Does more damping always reduce moving-base transmission?

   Show the answer and reasoning

   No. It helps near resonance but can increase transmission well above resonance in this model.

3. What must be named before claiming a vibration reduction ratio?

   Show the answer and reasoning

   The excitation, input/output quantities, frequency, amplitude convention and model conditions.

What to carry forward

Identify the excitation path and response quantity before interpreting resonance or isolation.