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Structural dynamics and vibration testing

Critically damped response to an impulse

Displacement of a critically damped mass that is struck by an impulse J at t = 0 while at rest.

Ref
SDY-06
Rev
v1
Field
Structural dynamics and vibration testing
Result
x [mm]

An impulse J gives a stationary mass m an initial velocity J/m. With critical damping (ζ = 1) the displacement is (J/m) t e−ωₙt: it rises, peaks at t = 1/ωₙ at the value J/(emωₙ), and then decays without any oscillation. Output is in mm.

Displacement of a critically damped mass that is struck by an impulse J at t = 0 while at rest. It rises, peaks at t = 1/ωₙ and then decays without oscillating.

Calculator

Solve for
mm
N·s
0.120
kg
0.550
rad/s
1100
s
01
  • Time of the peak, 1/ωₙ0.1 s
  • Peak displacement, J/(e m ωₙ)18.39 mm
  • Initial velocity, J/m0.5 m/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 18.39 mm
JImpulseKey driver
0.92184
mMass
1.47147
ωnUndamped angular frequency
0.0022745.2
tTime after the impulse
0
  • 0.1 s after a 2 N·s impulse the mass has moved 18.39 mm; its peak travel would be 18.39 mm at 0.1 s.
  • The peak scales with J/(mωₙ): a stiffer or heavier system moves much less for the same impulse.
Chart data as tables

Ranked variables

VariableTypical rangeLowest resultHighest resultSwing (mm)Local stepChange in result
J Impulse0.1 to 20 N·s0.9197183.9183+10%+10 %
m Mass0.5 to 50 kg1.472147.2145.7+10%−9.09 %
ωn Undamped angular frequency1 to 100 rad/s0.0022745.2445.24+10%−9.52 %
t Time after the impulse0 to 1 s018.3918.39+10%−0.468 %

Sweep of Time after the impulse

Time after the impulse (s)Displacement (mm)
00
0.12517.91
0.2510.26
0.3754.41
0.51.684
0.6250.6033
0.750.2074
0.8750.06933
10.0227

Sweep one variable

Displacement (mm)Hover, touch, or use the arrow keys on the chart
051015 00.20.40.60.81 Time after the impulse (s) now 18.39

Variables and how each one contributes

x

Displacement

Result · mm

What the equation gives you. Everything else feeds it.

J

Impulse

N·s · typical 0.1 to 20

The impulse, in N·s: the force integrated over the very short time that it acts. Scales the response in direct proportion. Raising it by 10 % (to 2.2) moves the result by +10 %, from 18.39 to 20.23 mm.

m

Mass

kg · typical 0.5 to 50

The mass, in kg. More mass gives a smaller response for the same impulse. Raising it by 10 % (to 4.4) moves the result by −9.09 %, from 18.39 to 16.72 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 −9.52 %, from 18.39 to 16.64 mm.

t

Time after the impulse

s · typical 0 to 1

The time after the load is applied, in s. Moves the time at which the response is read. Raising it by 0.05 (to 0.15) moves the result by −9.02 %, from 18.39 to 16.73 mm.

Worked example with the values above

  1. Initial velocity: J/m = 2 / 4 = 0.5 m/s.
  2. Time factor: t e^(−ωₙt) = 0.1 × e^(−1) = 0.036788 s.
  3. x = 0.5 × 0.036788 = 18.394 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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