Module 4 · Lesson 4.1

Free vibration and the natural frequency

Where ω_n = √(k/m) comes from, what critical damping means, and why the three regimes are one solution rather than three.

Why this matters

The natural frequency is the most useful single number in structural dynamics. It is the yardstick against which every loading is judged, it is the first output of any modal analysis, and it is the number you check before believing anything else the software says.

It comes out of the simplest problem there is: displace the structure, release it, and see what it does on its own.

By the end of this lesson you should be able to

Derive the natural frequency from the homogeneous equation of motion
Derive critical damping and say what it is the boundary between
Write the response in each of the three regimes
Explain why damping barely affects the frequency

What you should already know

The equation of motion (Module 3)
The idea that a trial solution can be substituted into a differential equation

From first principles

Natural frequency and critical damping

We want to show: Solve m ẍ + c ẋ + kx = 0 and show that the character of the answer changes at c = 2√(km), which defines critical damping.

With no forcing, the only thing the structure can do is exchange energy between the spring and the mass. If nothing dissipates it, that exchange goes on for ever at a rate set by how stiff the spring is and how heavy the mass is. Add dissipation and the exchange runs down. Add enough dissipation and the exchange never gets going at all — the mass simply creeps back. The value of damping at which the oscillation stops existing is what we are looking for.

Try it

Free vibration in every damping regime

Displace the mass, release it, and see what damping does to what follows.

Natural period T_ns

Damping ratio ζ

Initial displacement x₀mm

Initial velocity x′₀mm/s

Compare all four regimes

ζ = 0 (undamped)
ζ = 0.05
ζ = 1 (critical)
ζ = 2 (overdamped)

Displacement against Time. ζ = 0 (undamped) reaches a peak magnitude of 30 mm. ζ = 0.05 reaches a peak magnitude of 30 mm. ζ = 1 (critical) reaches a peak magnitude of 30 mm. ζ = 2 (overdamped) reaches a peak magnitude of 30 mm.

Show the numbers behind this plot

t = 0.00 s

Animation time — drag to examine any instant

ζ = 0 (undamped)
ζ = 0.05
ζ = 1 (critical)
ζ = 2 (overdamped)

Phase-plane view: displacement on the horizontal axis against velocity on the vertical. An undamped response traces a closed ellipse for ever; a damped one spirals in to the origin; an overdamped one runs straight in without circling.

ω_n: 6.28 rad/s
T_n: 1 s
Regime: underdamped
ω_d: 6.28 rad/s
T_d: 1 s
Cycles to half amplitude: 2.2

At ζ = 0.05 the damped period is only 0.13% longer than the undamped one. Damping has an enormous effect on how long the vibration lasts and almost none on how fast it vibrates — which is why the distinction between ω_n and ω_d is nearly always ignorable in practice, and why it must not be forgotten in the derivation.

What this shows: The three damping regimes are not three phenomena. They are one solution whose character changes as ζ passes through 1, and at ζ = 1 the oscillation does not slow down — it stops existing.

Note

Critical damping is a boundary, not a target

'Critically damped' sounds like an optimum, and for a car suspension or a door closer it is close to one — you want the fastest return to equilibrium without overshoot.

For a structure it is neither achievable nor desirable. Buildings have ζ between about 0.01 and 0.07. Reaching ζ = 1 would need a damping coefficient between fifteen and a hundred times larger than anything a real structure has, and it would make the structure stiff against every movement including thermal expansion and settlement.

The value of c_cr is entirely as a REFERENCE. Saying a building has 5% damping is meaningful across every size and type of structure. Saying it has 18.7 kN·s/m is meaningful only for that one building.

One genuine subtlety: beyond critical damping the return to equilibrium gets SLOWER, not faster. An overdamped system has so much resistance that it creeps back. Critical damping is the fastest non-oscillatory return, and that is what 'critical' names.

Worked example

Free vibration of a pedestrian bridge after a jump test

Given

A footbridge with a measured fundamental frequency of 2.3 Hz
Modal mass 18 000 kg
A test in which a group jumps once and then stops, leaving the bridge at 9 mm amplitude
Damping measured at 0.8% — a bare steel bridge with no surfacing

Find

How long the bridge keeps moving perceptibly after the test.

Assumptions

The response is dominated by the first mode
Damping is constant over the amplitude range — a poor assumption at the tail, and the answer is a lower bound because of it
Perception threshold taken as 0.5 mm amplitude at this frequency

Predict first

Two identical structures differ only in damping: one has ζ = 0.02, the other ζ = 0.10 — five times as much. How do their natural frequencies compare?

Practice

A system has ω_n = 25 rad/s and ζ = 0.08. What is the damped circular frequency, in rad/s?

Practice

A structure with 3% damping is displaced and released. After how many cycles has the amplitude fallen to 10% of its starting value?

Practice

A mass of 6 000 kg on a spring of 2.4 MN/m has a damping coefficient of 12 000 N·s/m. What is the damping ratio?

Summary

ω_n = √(k/m), and the square root is why stiffening is an expensive lever
c_cr = 2√(km) = 2mω_n is a reference value, not a target
ζ = c/c_cr makes damping comparable across structures of any size
ω_d = ω_n√(1 − ζ²) — a shift of 0.13% at 5% damping
The three regimes are one solution as the roots move from complex to real
Beyond critical damping the return gets slower, not faster
Damping controls duration and resonant response, not frequency

Progress is kept in this browser only.

This is educational material. It uses simplified examples to teach principles, and must not be relied on for real design or safety-critical decisions. Module overview and checkpoint