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Structural dynamics and vibration testing

Critically damped response to a force step

Displacement of a critically damped oscillator at rest after a constant force F is applied.

Ref
SDY-10
Rev
v1
Field
Structural dynamics and vibration testing
Result
x [mm]

For a critically damped oscillator at rest, a constant force F produces x = (F/k)[1 − (1 + ωₙt)e−ωₙt]. The displacement rises to the static deflection F/k without any overshoot, which is why critical damping is used where overshoot must be avoided. Output is in mm.

Displacement of a critically damped oscillator at rest after a constant force F is applied. It creeps up to the static deflection F/k with no overshoot.

Calculator

Solve for
mm
N
−5050
N/m
102,000
rad/s
1100
s
02
  • Static deflection, F/k50 mm
  • Fraction of the final value reached0.594
  • Time to reach 95 % of the final value, 4.74/ωₙ0.4744 s

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.

Displacement (mm) current value 29.7 mm
kStiffnessKey driver
2.97594
FConstant force
−148148
tTime after the force is applied
050
ωnUndamped angular frequency
0.87650
  • After 0.2 s the displacement has reached 59.4 % of its final value of 50 mm.
  • About 4.7/ωₙ = 0.474 s are needed to settle within 5 % of the final value.
Chart data as tables

Ranked variables

VariableTypical rangeLowest resultHighest resultSwing (mm)Local stepChange in result
k Stiffness10 to 2,000 N/m2.97594591+10%−9.09 %
F Constant force−50 to 50 N−148.5148.5297+2 N+20 %
t Time after the force is applied0 to 2 s05050+10%+8.66 %
ωn Undamped angular frequency1 to 100 rad/s0.87625049.12+10%+8.66 %

Sweep of Time after the force is applied

Time after the force is applied (s)Displacement (mm)
00
0.2535.64
0.547.98
0.7549.76
149.98
1.2550
1.550
1.7550
250

Sweep one variable

Displacement (mm)Hover, touch, or use the arrow keys on the chart
01020304050 00.511.52 Time after the force is applied (s) now 29.7

Variables and how each one contributes

x

Displacement

Result · mm

What the equation gives you. Everything else feeds it.

F

Constant force

N · typical −50 to 50

The constant force, in N. Negative values reverse the motion. Scales the response in direct proportion. Raising it by 10 % (to 11) moves the result by +10 %, from 29.7 to 32.67 mm.

k

Stiffness

N/m · typical 10 to 2,000

The stiffness, in N/m. The static deflection under the force is F/k. A stiffer system moves less. Raising it by 10 % (to 220) moves the result by −9.09 %, from 29.7 to 27 mm.

ωn

Undamped angular frequency

rad/s · typical 1 to 100

The undamped natural angular frequency, in rad/s. Sets how quickly the motion develops and decays. Raising it by 10 % (to 11) moves the result by +8.66 %, from 29.7 to 32.27 mm.

t

Time after the force is applied

s · typical 0 to 2

The time after the load is applied, in s. Moves the time at which the response is read. Raising it by 0.05 (to 0.25) moves the result by +20 %, from 29.7 to 35.64 mm.

Worked example with the values above

  1. Dimensionless time: ωₙt = 10 × 0.2 = 2.
  2. Approach factor: 1 − (1 + 2) e^(−2) = 0.59399.
  3. x = (10 / 200) × that = 29.7 mm.

Assumptions and limits

  • A single-degree-of-freedom linear oscillator that starts at rest, with an ideal impulse or step.
  • The force acts on a mass-spring system with no other loads, and the damping is viscous.

Common mistakes

  • Treating a real, finite-duration impulse as ideal when its length is not short compared with the period.
  • Mixing up the natural frequency in Hz with the angular frequency in rad/s.
  • Forgetting that a suddenly applied force doubles the static deflection in an undamped system.

Statistics and uncertainty

The response scales directly with the load and the inverse of the stiffness or mass, so errors in those carry through proportionally. Frequency errors matter most when the time or pulse length is near a multiple of the period, where the response changes quickly.

Where it is used

  • Predicting peak response of structures and instruments to impacts, shocks and suddenly applied loads.
  • Understanding shock response spectra and the effect of pulse length on residual vibration.

Maths you need

  • Natural frequency: ω = √(k/m) for a mass on a spring
  • Impulse: force integrated over time
  • Damping ratio: damping as a fraction of critical

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