Driving, resonance and the frequency response

Every solution in the previous lesson decayed to nothing, because nothing was pushing, and a system that is being pushed periodically behaves in a way that depends violently on how fast.

The equation is now driven:

m x'' + c x' + k x = F_{0} \cos ω t

Its general solution, by the structure theorem, is a particular solution plus the complementary function. The complementary function is the free motion of the previous lesson, which decays whenever there is any damping at all, so after a few time constants nothing is left of it. What survives is the particular solution, and that is what a driven system settles into regardless of how it started. It is called the steady state, and finding it is the business of this lesson.

Undetermined coefficients

For a constant coefficient equation with a well-behaved forcing there is a direct method: guess a form with unknown coefficients, substitute, and match. It works because the derivatives of an exponential, a sinusoid or a polynomial stay within the same family.

The guess mirrors the forcing. For $g = e^{a t}$ try $A e^{a t}$ . For a polynomial of degree $n$ try a general polynomial of degree $n$ , all coefficients unknown, including those the forcing lacks. For $\cos ω t$ or $\sin ω t$ , try $A \cos ω t + B \sin ω t$ , both terms, because differentiating once produces the other. Products multiply the guesses together.

There is one exception, and it is the interesting case. If the guess already solves the homogeneous equation, substituting it gives zero and cannot match the forcing. Multiply the guess by $t$ , and again by $t$ if it still solves the homogeneous equation. That rule is where resonance comes from.

Example. Find a particular solution of $y'' + 4 y = 5 \cos 3 t$ .

The homogeneous solutions are $\cos 2 t$ and $\sin 2 t$ , and $\cos 3 t$ is not among them, so try $y_{p} = A \cos 3 t + B \sin 3 t$ . Then $y_{p}'' = - 9 A \cos 3 t - 9 B \sin 3 t$ , and the left side is $- 5 A \cos 3 t - 5 B \sin 3 t$ . Matching gives $- 5 A = 5$ and $- 5 B = 0$ , so $A = - 1$ and $B = 0$ , and $y_{p} = - \cos 3 t$ . The minus sign is not an accident: the drive is above the natural frequency, and the response is in antiphase with it, which the general formula below explains.

Now you. Find a particular solution of $y'' - y' - 6 y = 4 e^{t}$ .

Answer

The homogeneous roots are $3$ and $- 2$ , so $e^{t}$ is not a homogeneous solution and the guess $A e^{t}$ is safe. Substituting gives $A (1 - 1 - 6) e^{t} = - 6 A e^{t} = 4 e^{t}$ , so $A = - 2 / 3$ and $y_{p} = - {\frac{23}{e}}^{t}$ .

The steady state of a driven oscillator

Now do the general case. Substituting $x_{p} = A_{1} \cos ω t + A_{2} \sin ω t$ into $m x'' + c x' + k x = F_{0} \cos ω t$ and matching the cosine and sine terms gives two linear equations, and the tidy way to present the answer is as a single sinusoid $x_{p} = A \cos (ω t - φ)$ with

A = \frac{F_{0} / k}{\sqrt{{(1 - ρ^{2})}^{2} + {(2 ζ ρ)}^{2}}}, \tan φ = \frac{2 ζ ρ}{1 - ρ^{2}}

where $ρ = ω / ω_{0}$ is the drive frequency measured in units of the natural one. The quantity $F_{0} / k$ is the static deflection, what the force would produce if applied steadily, and the fraction multiplying it is the dimensionless amplification factor.

Three regimes are visible in that formula without any plotting.

Drive slowly, $ρ ≪ 1$ : the amplification is $1$ and $φ \approx 0$ . The mass follows the force exactly, as if the spring alone were present. Stiffness controls the response.

Drive quickly, $ρ ≫ 1$ : the amplification falls as $1 / ρ^{2}$ , so $A \to F_{0} / m ω^{2}$ , and $φ \to 180$ degrees. The mass barely moves and does the opposite of what the force asks. Inertia controls the response.

Drive near $ρ = 1$ : the first bracket vanishes and only the damping term is left, so $A \approx F_{0} / (k \cdot 2 ζ) = Q F_{0} / k$ . The response is $Q$ times the static deflection, and the phase passes through exactly $90$ degrees, whatever the damping. That phase quadrature is the sharpest experimental signature of resonance, because the amplitude peak is broad and the phase crossing is not.

Where the peak actually is

Maximising $A$ means minimising $(1 - ρ^{2})^{2} + (2 ζ ρ)^{2}$ . Differentiating with respect to $ρ^{2}$ and setting the result to zero gives $ρ_{r} = \sqrt{1 - 2 ζ^{2}}$ , with maximum amplification

\frac{A_{\max}}{F_{0} / k} = \frac{1}{2 ζ \sqrt{1 - ζ^{2}}}

So the resonant peak sits slightly below the natural frequency, and below the damped frequency $ω_{d}$ too. For light damping the shift is negligible: at $ζ = 0.05$ the peak is at $ρ = 0.9975$ and the amplification is $10.01$ , against $1 / 2 ζ = 10$ . At $ζ = 0.3$ it is at $ρ = 0.906$ with amplification $1.75$ . And for $ζ > 1 / \sqrt{2} = 0.707$ there is no peak at all: the response falls monotonically from the static value, which is why heavily damped instruments have no preferred frequency.

Example. A $500$ kg machine sits on mounts with total stiffness $2 \times 10^{5}$ N m⁻¹ and damping ratio $ζ = 0.05$ . A rotating imbalance applies a force of amplitude $1000$ N at $3.0$ Hz. Find the steady state amplitude and the phase lag.

First $ω_{0} = \sqrt{2 \times 10^{5} / 500} = 20$ rad s⁻¹, which is $3.18$ Hz, so the machine is running just below its own resonance: $ω = 2 π (3.0) = 18.85$ rad s⁻¹ and $ρ = 0.9425$ . The static deflection is $1000 / 2 \times 10^{5} = 5.0$ mm. The bracket is $(1 - 0.8883)^{2} + (2 \times 0.05 \times 0.9425)^{2} = 0.01248 + 0.00888 = 0.02136$ , whose square root is $0.1462$ , so the amplification is $6.84$ and the amplitude is $34$ mm. The phase lag is $\arctan (0.0943 / 0.1117) = 40$ degrees. Thirty-four millimetres of shake from a five millimetre static deflection is a machine that will destroy its mounts, and the fix is to move the operating speed or the mount stiffness, not to add damping, which at $ζ = 0.05$ would have to be increased manyfold to help.

Now you. The same machine runs at $1.5$ Hz instead. Find $ρ$ , the amplification, and the amplitude.

Answer

$ω = 9.42$ rad s⁻¹, so $ρ = 0.471$ and $ρ^{2} = 0.222$ . The bracket is $(0.778)^{2} + (0.0471)^{2} = 0.6076$ , whose square root is $0.7795$ , so the amplification is $1.28$ and the amplitude is $6.4$ mm. Halving the speed has removed nearly all of the trouble.

Resonance without damping

Set $c = 0$ and drive exactly at the natural frequency. The formula above divides by zero, which is the algebra reporting that the guess has failed: $\cos ω_{0} t$ now solves the homogeneous equation. Following the rule, multiply by $t$ and try $x_{p} = A t \sin ω_{0} t$ . Then

x_{p}'' = 2 A ω_{0} \cos ω_{0} t - A ω_{0}^{2} t \sin ω_{0} t

so $x_{p}'' + ω_{0}^{2} x_{p} = 2 A ω_{0} \cos ω_{0} t$ , and matching $F_{0} / m$ gives

x_{p} = \frac{F_{0}}{2 m ω_{0}} t \sin ω_{0} t

an oscillation whose amplitude grows linearly and without bound. That is undamped resonance, and it is the reason the word carries the connotation it does. Real systems always have some damping, so the growth stops at $Q$ times the static deflection, but the approach is worth seeing because it shows what damping is holding back.

Drive slightly off the natural frequency with no damping and something different happens. Starting from rest, the solution is the difference of two cosines at $ω$ and $ω_{0}$ , and the product formula turns it into

x = \frac{2 F_{0}}{m (ω_{0}^{2} - ω^{2})} \sin \frac{(ω_{0} - ω) t}{2} \sin \frac{(ω_{0} + ω) t}{2}

a fast oscillation at the average frequency inside a slow envelope at half the difference. These are beats, the wobble heard when two instruments are nearly in tune, and the energy is being handed to the oscillator and taken back again. As $ω \to ω_{0}$ the envelope's period lengthens and the amplitude prefactor grows, and in the limit the taking back never happens, which is resonance again.

What resonance does and does not explain

The Tacoma Narrows bridge, which twisted itself apart on 7 November 1940 in a wind of about $19$ m s⁻¹, is the illustration in every textbook, and the usual caption is wrong. Wind that steady contains no periodic force at the bridge's $0.2$ Hz torsional frequency. What happened was aeroelastic flutter: the deck's own twisting changed the airflow so as to feed energy back into the twisting, in phase with the velocity. In the equation that is a negative $c$ , not a forcing term, and the growth is exponential rather than linear. The distinction matters practically, because flutter is not cured by adding damping in the usual amounts or by avoiding a frequency; it is cured by changing the shape of the deck so the feedback loses its sign.

The London Millennium Bridge, closed two days after opening in June 2000, is a second cautionary case. Pedestrians on a slightly swaying deck adjust their gait to stay balanced, which pushes sideways in time with the sway. Again the input is created by the response, and once the crowd exceeded a critical size the lateral mode grew. The cure was a retrofit of several dozen viscous dampers and tuned mass absorbers, which is to say: raise $ζ$ until the feedback cannot beat it.

Straightforward resonance is real, though, and older than either. Bridges have collapsed under troops marching in step, at Broughton near Manchester in 1831 and at Angers in 1850, where $226$ soldiers died. Marching in step supplies a genuine periodic force at around $2$ Hz, close to the frequency of a light suspension span, which is why troops are ordered to break step on a bridge to this day.

The same formula underlies the useful side. Radio tuning selects one station because a circuit with $Q$ of a hundred amplifies its own resonant frequency by that factor and neighbours by far less. Magnetic resonance imaging drives nuclear spins at their precession frequency. A microwave oven does not, contrary to the popular account, drive a resonance of the water molecule; it heats by dielectric loss over a broad band, which is why it works on a wide range of foods rather than only at one frequency.

Vibration isolation

One more reading of the same curve settles a practical question: how do you keep a vibrating floor from shaking an instrument? Mount the instrument on springs, and the fraction of the floor's motion that reaches it, the transmissibility, is

T = \frac{\sqrt{1 + {(2 ζ ρ)}^{2}}}{\sqrt{{(1 - ρ^{2})}^{2} + {(2 ζ ρ)}^{2}}}

which equals one at $ρ = \sqrt{2}$ regardless of damping, and falls below one only above it. So isolation requires the mounts to be soft enough that the natural frequency is well below the disturbance, by a factor of at least $\sqrt{2}$ and in practice three or more. That is why a sensitive balance sits on a slack, heavy table rather than a stiff one, and why the isolation stages of a gravitational wave detector are a stack of pendulums with periods of seconds.

Example. An instrument is mounted with $ρ = 2$ and $ζ = 0.1$ . What fraction of the floor's motion reaches it?

The numerator is $\sqrt{1 + (0.4)^{2}} = 1.077$ . The denominator is $\sqrt{(1 - 4)^{2} + (0.4)^{2}} = \sqrt{9.16} = 3.027$ . So $T = 0.356$ : about a third gets through.

Now you. The same mounts are used at $ρ = 4$ . What is the transmissibility?

Answer

The numerator is $\sqrt{1 + (0.8)^{2}} = 1.281$ and the denominator is $\sqrt{225 + 0.64} = 15.02$ , so $T = 0.085$ . Doubling the frequency ratio has cut the transmitted motion fourfold, roughly as $1 / ρ^{2}$ .

Note that damping hurts here. At high $ρ$ the numerator grows with $ζ$ while the denominator does not, so a stiff damper carries vibration across the mount. Damping is wanted only near resonance, which is the tension every real isolator has to resolve, usually by making the damping frequency dependent.

What is still missing

All of this assumed the forcing was a single sinusoid. Real disturbances are not: a hammer blow, a switch closing, a road surface, an earthquake record. Undetermined coefficients has nothing to say about any of them, since none has a guessable form.

Two things rescue the situation, and the next lesson takes both. There is a method, variation of parameters, that produces a particular solution for any forcing at all in the form of an integral. And there is a structural fact: because the system is linear, its response to a complicated input is the sum of its responses to simple pieces, so the frequency response computed here turns out to answer far more than the question it was asked.