Library / Mechanical / MEC-30

Mechanical

Free underdamped response

Displacement of a damped oscillator released with no applied force.

Ref: MEC-30
Rev: v1
Field: Mechanical
Result: x [mm]

x = e^{- ζ ω_{n} t} (A \cos ω_{d} t + B \sin ω_{d} t)

For 0 ≤ ζ < 1 with no applied force. ω_d = ω_n√(1 − ζ²). Initial conditions set A = x(0) and B = [ẋ(0) + ζω_nA]/ω_d. The calculator solves for x within the first half damped cycle.

Released from a displacement, a lightly damped system swings back and forth with a shrinking amplitude.

Calculator

Solve for

xDisplacementComputed

x₀Initial displacement

−5050

ẋ₀Initial velocity

mm/s

−500500

ζDamping ratio

0.010.5

ω_nNatural frequency

rad/s

5100

tTime

160

Damped frequency ωd49.94 rad/s
Envelope e^(−ζ ωn t)0.9048

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 −3.332 mm

x₀Initial displacementKey driver

−16.716.7

tTime

−8.459.99

ω_nNatural frequency

−8.549.8

ẋ₀Initial velocity

−11.64.91

ζDamping ratio

−3.991.51

The amplitude decays exponentially at rate ζω_n; the oscillation frequency is ω_d, slightly below ω_n.
At ζ = 0.05 the damped frequency differs from ω_n by only 0.1%.

Chart data as tables

Ranked variables

Variable	Typical range	Lowest result	Highest result	Swing (mm)	Local step	Change in result
x₀ Initial displacement	−50 to 50 mm	−16.66	16.66	33.32	+10%	−0.333 mm
t Time	1 to 60 ms	−8.454	9.988	18.44	+10%	−1.56 mm
ω_n Natural frequency	5 to 100 rad/s	−8.54	9.802	18.34	+10%	−1.56 mm
ẋ₀ Initial velocity	−500 to 500 mm/s	−11.58	4.915	16.49	+10%	0 mm
ζ Damping ratio	0.01 to 0.5	−3.989	1.506	5.495	+10%	+0.0785 mm

Sweep of Time

Time (ms)	Displacement (mm)
1	9.988
8.375	9.148
15.75	7.131
23.13	4.246
30.5	0.9052
37.88	−2.434
45.25	−5.332
52.63	−7.42
60	−8.454

Sweep one variable

Vary

Displacement (mm)Hover, touch, or use the arrow keys on the chart

Variables and how each one contributes

Displacement

Result · mm

What the equation gives you. Everything else feeds it.

x₀

Initial displacement

mm · typical −50 to 50

Displacement at t = 0. Directly proportional.

ẋ₀

Initial velocity

mm/s · typical −500 to 500

Velocity at t = 0. Adds an extra sinusoidal term.

Damping ratio

· typical 0.01 to 0.5

Fraction of critical damping. Larger damping shrinks the response faster.

ω_n

Natural frequency

rad/s · typical 5 to 100

Undamped natural frequency. A higher frequency completes more cycles.

Time

ms · typical 1 to 60

Time since release. Response oscillates and decays.

Worked example with the values above

ω_d = 50√(1 − 0.05²) = 49.94 rad/s; A = 10; B = 0.5006.
Envelope: e^−ζω_nt = 0.9048; angle ω_dt = 1.997 rad.
x = −3.332 mm.

Assumptions and limits

Linear viscous damping.
No applied force after release.

Common mistakes

Using ω_n in place of ω_d inside the cosine.
Using the formula for ζ ≥ 1.

Statistics and uncertainty

The phase of the response is sensitive to ω_n at later times; an error of 1% in ω_n shifts the phase by 1% of the total angle.

Where it is used

Tap-test and ring-down analysis.
Settling time of mounts and suspension.

Maths you need

Natural frequency: ωn = √(k/m)
Exponential decay: e^(−at)

Related formulas

Practise this formula