The resonance that ran out of time

The resonant amplification of a lightly damped oscillator is $1/2\zeta$, and for a structure at two per cent damping that is twenty-five. For a machine shaft at half a per cent it is a hundred.

Those numbers are steady-state answers. They are what the response settles down to after the forcing has been applied at exactly the resonant frequency for long enough — and “long enough” contains the same small damping in the denominator: the response builds with a time constant $1/\zeta\omega$, so the more spectacular the resonance, the longer it takes to develop.

Almost nothing sits at a resonance. A machine runs up to speed and passes through; a turbine coasts down through the same point; a train crosses a bridge at a speed that matches for a few seconds; a crane hoist accelerates. Each of them sweeps, and how much amplification it collects depends on how long it spent inside the peak.

Which free body produced the number #

The oscillator, driven by a force whose frequency is a function of time — $\ddot{u} + 2\zeta\omega_0\dot{u} + \omega_0^2 u = (F/m)\sin\phi(t)$ with $\dot\phi = \omega_i + \beta t$.

There is no closed form worth quoting, and the integration is done numerically. What can be got in closed form is the parameter the answer depends on, and it comes from comparing two times.

The time available is how long the sweep spends inside the resonance. The half-power bandwidth of a resonance is $2\zeta\omega_0$ wide, so at a sweep rate $\beta$ the time inside it is $2\zeta\omega_0/\beta$.

The time needed is the build-up time constant, $1/\zeta\omega_0$.

Their ratio is $2\zeta^2\omega_0^2/\beta$, so the whole problem depends on the single dimensionless group

$$\eta = \frac{\beta}{\zeta^2\omega_0^2}$$

and the fraction of the steady-state peak reached is a function of $\eta$ and of nothing else. At $\eta = 0.05$ it is 99.8%; at 0.5 it is 95%; at 3.2 it is 75%; at 25.6 it is 41%.

Damping, and the case where it stops mattering #

Look at where $\zeta$ appears and something odd shows up: it is in the steady-state peak as $1/2\zeta$, and it is in the parameter as $\zeta^2$. The two do not cancel, and at a fast enough sweep the second wins.

Hold the absolute sweep rate fixed and vary the damping. At a leisurely rate, the peaks reached are 96.8, 49.8, 25.0 and 10.0 for damping ratios of 0.5, 1, 2 and 5 per cent — which is $1/2\zeta$ almost exactly, a factor of 9.7 across the set. Damping is the whole answer.

At a rate forty times faster, the same four give 11.9, 11.1, 9.6 and 6.7. A factor of 1.8, on steady-state values that are twenty times apart.

The reason is straightforward once stated: at a fast sweep nothing has had time to reach a level where damping would have limited it. The response is bounded by the clock rather than by the dashpot, and the amplitude is set by the impulse the resonance received rather than by the balance between input and dissipation.

Which has a practical edge. Adding damping to a machine that passes quickly through a resonance buys very little, and money spent on damping there is better spent on getting through faster — which is the opposite of the advice for a machine that operates near a resonance, where damping is everything.

The peak arrives late, and then it beats #

Two features come out of the integration that are not guessable from the steady-state picture, and both are visible on an instrument.

The peak arrives after the frequency has passed resonance. The response is still growing when the driving frequency leaves the peak, and it goes on growing for a while afterwards because the energy already in the oscillator has not yet been dissipated. The lag grows with the sweep rate: 1.2% of the natural frequency at the rate drawn, 8% at thirteen times it, 20% at fifty.

So an operator watching a vibration meter during a run-up sees the maximum reading while the machine is already above its critical speed, and the natural conclusion — that the critical speed is where the peak was — is wrong by up to a fifth.

And the response beats afterwards. Past the resonance the oscillator is ringing at its own frequency while being driven at a different one, so the two superpose and the envelope pulses at the difference. The beating decays as the free response damps out, over the same $1/\zeta\omega_0$, which for a lightly damped machine is many cycles.

Both features are diagnostic. A trace that peaks after the nominal critical speed and then beats is a transient resonance; one that peaks at the critical speed and stays there is a steady one.

What is actually growing, and what limits it #

It helps to have a physical picture of the build-up, because the “time constant” language hides a simple energy statement.

At resonance the force is in phase with the velocity, so it does work on every part of every cycle — nothing is given back. The energy in the oscillator therefore grows steadily, and since energy goes as the square of the amplitude, the amplitude grows in proportion to the number of cycles rather than exponentially. The damping dissipates energy at a rate proportional to the square of the amplitude, so it starts negligible and catches up: the steady state is where the two rates are equal, and that happens when the amplitude has reached $1/2\zeta$ times the static one.

So the build-up is a race between an input growing linearly and a dissipation growing quadratically. A lightly damped system takes longer to reach its limit because its dissipation is so weak that it has to grow very large before it catches the input — which is the same $\zeta$ appearing in both places, and the reason the two effects cannot be separated.

That picture also explains what a sweep does. Off resonance the force no longer keeps step with the velocity, so part of each cycle takes energy back out. The band over which the two stay close enough in step for net input is the half-power band, and outside it the oscillator is simply ringing down.

Run-up and run-down are not the same #

The asymmetry is worth stating because it decides which event a machine is designed for.

A motor running up is being driven through its critical speed by a drive with power to spare, so the sweep rate is high and the resonance is passed quickly. A machine running down is coasting, decelerating on friction and windage alone, so the sweep rate is low — and at the resonance itself the drive is no longer there to push it through.

Lower $\beta$ means lower $\eta$, which means a larger fraction of the steady-state peak. The run-down is the dangerous pass, and it is the one that happens on a power failure, when the machine is unattended and nobody is watching the meter.

It is also why a machine that has to stop in an emergency is fitted with a brake rather than left to coast: braking through the critical speed converts a slow sweep into a fast one, and the peak falls with it. That is the same lever the machine that shakes the building identifies as a snubber, reached from the other direction — change the time available rather than the dissipation.

There is a worse case still, and it is why large machines have barring gear. A rotor that stops in its resonance — a drive tripping out at the wrong moment, or a system whose deceleration stalls near the critical speed because the resonance itself is absorbing power — has left the transient regime entirely and is at $1/2\zeta$.

The same argument, in structures #

The pattern occurs wherever a forcing frequency changes, and three cases in this collection are versions of it.

A train crossing a bridge. The excitation frequency is the axle spacing over the speed, and it sweeps as the train accelerates or decelerates on the span. The train that arrives in time with itself is the resonant case; whether the bridge collects the resonance depends on how many axles pass while the match holds, which is the same time-in-band argument with the number of cycles counted directly.

A machine on isolators. Every start and stop takes the machine through its mount’s resonance, and the machine that shakes the building makes the design compromise: damping that helps during the run-up spoils the isolation at running speed. The transient argument says how much damping is worth adding — very little if the run-up is quick — and points instead at a snubber, which is present only at large amplitude.

A structure in an earthquake. The forcing is broadband rather than swept, but the accounting is identical: a structure sees a few strong cycles at any one frequency, so the response is a build-up problem and not a steady-state one. That is precisely why a response spectrum is computed by integration rather than from an amplification factor, and the spectrum is not a load is what the resulting object is.

Counting cycles instead of computing #

There is a shortcut that gets most of the answer without integrating anything, and it is worth having because it can be done on the back of a drawing.

The half-power bandwidth of a resonance is $2\zeta$ of the frequency, so at a sweep rate $\beta$ the number of cycles spent inside the band is roughly

$$N = \frac{2\zeta\,\omega_0}{\beta}\cdot\frac{\omega_0}{2\pi} = \frac{\zeta\,\omega_0^2}{\pi\beta}$$

and the response of a resonating oscillator grows by about $\pi/2$ of the static deflection per cycle in the early stages, so the peak reached is roughly $N$ times that until it approaches $1/2\zeta$.

Setting the two equal recovers the parameter: the sweep collects the full resonance when $N \gtrsim 1/\zeta$, which is $\eta \lesssim 1$. So the rule of thumb is that a sweep is slow if it spends more than $1/\zeta$ cycles inside the half-power band, and fast if it spends fewer — fifty cycles at 2% damping, two hundred at half a per cent.

That number is worth carrying because it is checkable in the field. A machine running up through 1,500 rpm at 300 rpm per second spends about 0.2 s in a 2%-wide band at 25 Hz, which is five cycles. Against a requirement of fifty, that is a fast sweep, and it explains immediately why the measured peak is a fifth of what the steady-state calculation predicted.

Where the model stops #

The system was linear. A large response opens up non-linearity — a bearing clearance, a mount going solid, a joint slipping — and a non-linear resonance is not symmetric: its peak leans, so a sweep upward and a sweep downward through the same range give different answers and can jump between branches. That is the standard behaviour of a Duffing oscillator and it is common in real machinery.

One mode was considered. A structure has several — a structure has more than one period — and a real shaft has several critical speeds, and a sweep through a range may pass two of them close together — in which case the response from the first has not decayed when the second arrives, and they add.

The amplitude of the forcing was constant. For an out-of-balance rotor it is not: the force is $me\omega^2$, so it grows as the square of the speed and the excitation is much larger at the second critical speed than at the first. The sweep therefore hits a larger force with less time available, and the two effects partly cancel.

Fatigue was not counted. A pass through a resonance delivers a burst of large-amplitude cycles, and a machine that starts and stops daily delivers that burst every day. The stress range matters more than the peak because the load that never came near failing anything is what accumulates — and a transient resonance is a fatigue loading rather than a strength one.

And the sweep was linear in frequency. A real run-up is not, and the rate near the critical speed is what matters — which is a local property of the drive’s torque curve rather than a global rate.

The generalisation #

The idea to carry away is that an amplification is a limit, and reaching a limit takes time that a structure may not be given.

Almost every dynamic amplification in this collection is a steady-state quantity: the resonant $1/2\zeta$; a transmissibility; a gust factor; a lock-in amplitude. Each is what a system settles to under a forcing that persists, and each therefore comes with an implied number of cycles. The right question to ask of any of them is how many cycles the real excitation supplies, and the answer is often far fewer than the amplification needs.

The corollary runs the other way too and is the reason the impact factor of two exists. Twice the deflection, for the same load is a limit reached in half a cycle, so it is collected by essentially any suddenly applied load however brief the event. A dynamic amplification that needs one cycle is always available; one that needs a hundred is often not.

So the useful classification of dynamic effects is not by mechanism but by how many cycles they take to develop: impact needs half, a build-up to a fifth of resonance needs a few, full resonance needs $1/\zeta$ of them, and self-excitation needs as many as it takes. Knowing which of those a structure is exposed to decides whether damping is the answer, whether speed is the answer, or whether the effect will simply not have time to happen.

And there is a companion habit about instruments. A measurement of a transient is a measurement of the transient and not of the system, so a peak read off a run-up trace does not give the resonant amplification, does not sit at the resonant frequency, and cannot be compared with a steady-state calculation without the sweep rate. A great many arguments about whether a structure “matches its model” are arguments in which one side has measured a sweep and the other has computed a steady state. The only thing that stops it sets out how damping is actually measured, and every one of the methods there is careful about exactly this.