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Library / Structural dynamics and vibration testing / SDY-07

Structural dynamics and vibration testing

Peak displacement after an underdamped impulse

Size of the first positive peak of an underdamped oscillator after a positive impulse J from rest.

Ref
SDY-07
Rev
v1
Field
Structural dynamics and vibration testing
Result
xp [mm]

After a positive impulse J, an underdamped oscillator first swings to xₚ = [J/(mωₙ)] exp[−ζ arccos ζ/√(1−ζ²)]. With no damping this is J/(mωₙ); damping lowers the peak and moves it earlier. The formula gives the first positive peak. Output is in mm.

Size of the first positive peak of an underdamped oscillator after a positive impulse J from rest. With no damping the peak is J/(mωₙ); damping lowers it.

Calculator

Solve for
mm
N·s
0.120
kg
0.550
rad/s
1100
00.95
  • Undamped peak, J/(m ωₙ)50 mm
  • Reduction by damping0.2439
  • 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.

First peak displacement (mm) current value 37.81 mm
JImpulseKey driver
1.89378
ωnUndamped angular frequency
3.78378
mMass
3.02302
ζDamping ratio
1950
  • A damping ratio of 0.2 cuts the first peak from 50 mm to 37.81 mm.
  • Adding damping lowers the peak less than you might expect at small ζ, since it acts over a whole quarter-cycle.
Chart data as tables

Ranked variables

VariableTypical rangeLowest resultHighest resultSwing (mm)Local stepChange in result
J Impulse0.1 to 20 N·s1.89378.1376.2+10%+10 %
ωn Undamped angular frequency1 to 100 rad/s3.781378.1374.3+10%−9.09 %
m Mass0.5 to 50 kg3.025302.5299.4+10%−9.09 %
ζ Damping ratio0 to 0.95 19.035030.97+10%−2.44 %

Sweep of Damping ratio

Damping ratioFirst peak displacement (mm)
050
0.118742.03
0.237536.11
0.356231.56
0.47527.97
0.593825.08
0.712522.7
0.831220.71
0.9519.03

Sweep one variable

First peak displacement (mm)Hover, touch, or use the arrow keys on the chart
20304050 00.20.40.60.8 Damping ratio now 37.81

Variables and how each one contributes

xp

First peak 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 37.81 to 41.59 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 37.81 to 34.37 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 −9.09 %, from 37.81 to 34.37 mm.

ζ

Damping ratio

· typical 0 to 0.95

The viscous damping ratio, from 0 (undamped) to just below 1. More damping lowers the first peak. Raising it by 0.05 (to 0.25) moves the result by −5.9 %, from 37.81 to 35.58 mm.

Worked example with the values above

  1. Undamped peak: J/(mωₙ) = 2 / (4 × 10) = 0.05 m.
  2. Damping factor: exp(−ζ arccos ζ / √(1 − ζ²)) = 0.75613.
  3. xₚ = 37.807 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

Related formulas

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