Free oscillation and damping

The three root cases of a constant coefficient quadratic are not an algebraic curiosity: they are the difference between a car that rides well, one that bounces down the road, and one that wallows.

This lesson attaches physics to the algebra of the previous one. The equation is the same throughout, and the whole of its behaviour is controlled by two numbers that can be read off the coefficients before anything is solved.

Building the equation

Hooke's law says a spring stretched by $x$ from its natural length pulls back with force $- k x$ , the sign saying that the force opposes the displacement, and the constant $k$ measured in newtons per metre. Resistance to motion, from a dashpot or from air, is modelled at low speeds as proportional to velocity and opposing it, giving $- c x'$ . Newton's second law then assembles

m x'' = - k x - c x' that is m x'' + c x' + k x = 0

Divide by $m$ and give the two resulting numbers names:

x'' + 2 ζ ω_{0} x' + ω_{0}^{2} x = 0, ω_{0} = \sqrt{\frac{k}{m}}, ζ = \frac{c}{2 \sqrt{m k}}

The natural frequency $ω_{0}$ is in radians per second and the damping ratio $ζ$ is dimensionless. Every free linear oscillator in existence is described by that pair of numbers, and nothing else about the system matters to its motion.

The characteristic equation is $r^{2} + 2 ζ ω_{0} r + ω_{0}^{2} = 0$ , whose roots are

r = ω_{0} (- ζ \pm \sqrt{ζ^{2} - 1})

so the sign of $ζ - 1$ decides everything.

No damping at all

With $ζ = 0$ the roots are $\pm i ω_{0}$ and the solution is $x = A \cos ω_{0} t + B \sin ω_{0} t$ , or equivalently $x = R \cos (ω_{0} t - δ)$ with $R = \sqrt{A^{2} + B^{2}}$ . This is simple harmonic motion: a pure sinusoid of period $T = 2 π / ω_{0}$ , running forever.

Two features deserve attention. The frequency depends on the spring and the mass, not on how hard the system was started: a large oscillation and a small one take the same time, which is isochronism and is the property that made the pendulum clock possible. And energy is conserved, sloshing between kinetic $\frac{12}{m} (x')^{2}$ and potential $\frac{12}{k} x^{2}$ with a constant sum $\frac{12}{k} R^{2}$ , which you can verify by differentiating that sum and using the equation.

Example. A $0.50$ kg mass hangs on a spring of stiffness $200$ N m⁻¹. Find the natural frequency in radians per second and in hertz, and the period.

$ω_{0} = \sqrt{200 / 0.50} = \sqrt{400} = 20$ rad s⁻¹. In hertz that is $20 / 2 π = 3.18$ Hz, and the period is $2 π / 20 = 0.314$ s. Hanging the mass also stretches the spring by $m g / k = 0.5 \times 9.81 / 200 = 24.5$ mm at rest, but that only shifts the equilibrium: measuring $x$ from the new rest position gives back the same equation, with gravity absorbed.

Now you. A $2.0$ kg mass sits on a spring of stiffness $50$ N m⁻¹. Find $ω_{0}$ , the frequency in hertz, and the period.

Answer

$ω_{0} = \sqrt{50 / 2} = 5$ rad s⁻¹, which is $5 / 2 π = 0.796$ Hz, and the period is $2 π / 5 = 1.26$ s.

The three regimes

Turn the damping on. Since the roots depend only on $ζ$ and $ω_{0}$ , the classification is complete and has three cases.

Underdamped, $ζ < 1$ . The roots are complex, $r = - ζ ω_{0} \pm i ω_{d}$ with $ω_{d} = ω_{0} \sqrt{1 - ζ^{2}}$ , and

x = R e^{- ζ ω_{0} t} \cos (ω_{d} t - δ)

an oscillation inside a decaying envelope. The damped frequency $ω_{d}$ is always below the natural one, though only slightly for light damping: at $ζ = 0.1$ it is lower by half a per cent. The envelope has time constant $1 / ζ ω_{0}$ .

Critically damped, $ζ = 1$ . The root $- ω_{0}$ is repeated, so $x = (c_{1} + c_{2} t) e^{- ω_{0} t}$ . There is no oscillation, and this is the fastest possible return to equilibrium without overshoot, since any smaller $ζ$ overshoots and any larger one is slower.

Overdamped, $ζ > 1$ . Two negative real roots, and $x = c_{1} e^{r_{1} t} + c_{2} e^{r_{2} t}$ . The system creeps back, dominated eventually by the slower root, the one nearer zero. Increasing the damping further makes it slower still, which is the counterintuitive fact of the subject: too much damping does not stop motion faster, it drags it out.

Example. A $2.0$ kg mass on a spring with $k = 20$ N m⁻¹ has a dashpot with $c = 6.0$ N s m⁻¹. Classify the motion, and find the damped period.

$ω_{0} = \sqrt{20 / 2} = 3.162$ rad s⁻¹ and $ζ = 6 / (2 \sqrt{2 \times 20}) = 6 / 12.65 = 0.474$ , so the system is underdamped. Then $ω_{d} = 3.162 \sqrt{1 - 0.225} = 2.784$ rad s⁻¹, giving a damped period of $2 π / 2.784 = 2.26$ s, against $1.99$ s undamped. The envelope decays with time constant $1 / (ζ ω_{0}) = 1 / 1.5 = 0.667$ s, so the oscillation is essentially over in a few seconds.

Now you. A $1.0$ kg mass on a spring with $k = 16$ N m⁻¹ has $c = 10$ N s m⁻¹. Classify the motion and find the two roots.

Answer

$ω_{0} = 4$ rad s⁻¹ and $ζ = 10 / (2 \sqrt{16}) = 1.25 > 1$ , so it is overdamped. The roots solve $r^{2} + 10 r + 16 = 0$ , giving $r = - 2$ and $r = - 8$ . The return is governed by the slow root, so the decay time constant is $0.5$ s.

Which regime a designer wants

The choice of $ζ$ is a design decision, and different machines want different answers.

A car suspension is a spring and damper carrying about a quarter of the car's mass. Take $m = 300$ kg and $k = 30 000$ N m⁻¹, giving $ω_{0} = 10$ rad s⁻¹, or $1.59$ Hz, which is close to the frequency of a walking human and is chosen deliberately: ride comfort is poor at much higher frequencies and motion sickness sets in below about $0.5$ Hz. Critical damping would need $c = 2 \sqrt{m k} = 6000$ N s m⁻¹. Real dampers are set nearer $ζ = 0.3$ , so $c = 1800$ N s m⁻¹, because a critically damped suspension transmits too much of a sharp bump into the cabin. The price is a single small overshoot, which is what the body of a car does after a speed bump.

A door closer is deliberately overdamped: an overshoot would slam the door. A moving-coil galvanometer, and the analogue meters that descend from it, is critically damped so the needle reaches its reading in the least time without swinging past it, and instrument makers specify the coil circuit resistance that achieves it. A bell or a tuning fork is as lightly damped as the material allows, since the whole point is that the oscillation persists.

Measuring the damping

None of these numbers is usually known in advance. What can be measured is a decaying trace, and two consecutive peaks are enough.

Successive maxima are one damped period apart, so their ratio is the envelope's decay over that period:

\frac{x_{n}}{x_{n + 1}} = e^{ζ ω_{0} T_{d}}, T_{d} = \frac{2 π}{ω_{d}}

Taking the logarithm defines the logarithmic decrement $δ = \ln (x_{n} / x_{n + 1}) = 2 π ζ / \sqrt{1 - ζ^{2}}$ , which inverts to

ζ = \frac{δ}{\sqrt{4 π^{2} + δ^{2}}}

Example. A trace shows successive peaks of $12.0$ mm and $9.2$ mm. Find the logarithmic decrement, the damping ratio, and the quality factor $Q = 1 / 2 ζ$ .

$δ = \ln (12.0 / 9.2) = 0.2657$ . Then $ζ = 0.2657 / \sqrt{39.48 + 0.071} = 0.0423$ , light damping, and $Q = 1 / (2 \times 0.0423) = 11.8$ . Since $ζ$ is small, the correction from $ω_{d}$ to $ω_{0}$ is under a tenth of a per cent and can be ignored.

Now you. Successive peaks are $8.0$ mm and $5.0$ mm. Find $δ$ , $ζ$ and $Q$ .

Answer

$δ = \ln (8 / 5) = 0.470$ , so $ζ = 0.470 / \sqrt{39.48 + 0.221} = 0.0746$ and $Q = 6.7$ .

The quality factor $Q$ is the standard way to quote light damping, and it has a physical reading. Energy goes as amplitude squared, so it decays as $e^{- 2 ζ ω_{0} t}$ , and $Q$ is $2 π$ times the energy stored divided by the energy lost per cycle. It also counts oscillations: the amplitude falls by a factor of $e$ in about $Q / π$ cycles. A car suspension has $Q$ near $1.7$ , a guitar string a few thousand, a quartz watch crystal $10^{5}$ , and the mirror suspensions of the LIGO gravitational wave detectors exceed $10^{7}$ , which is why they ring for hours.

The same equation in a circuit

An inductor, resistor and capacitor in series obey Kirchhoff's voltage law: $L \frac{d i}{d t} + R i + \frac{q}{C} = 0$ , and since $i = q'$ this is

L q'' + R q' + \frac{q}{C} = 0

Term by term this is the mechanical equation, with inductance playing the part of mass, resistance the part of the dashpot, and the reciprocal capacitance the part of the spring constant. So

ω_{0} = \frac{1}{\sqrt{L C}}, ζ = \frac{R}{2} \sqrt{\frac{C}{L}}

Take $L = 10$ mH and $C = 100$ nF. Then $ω_{0} = 1 / \sqrt{10^{- 2} \times 10^{- 7}} = 31 623$ rad s⁻¹, or $5.03$ kHz. Critical damping needs $R = 2 \sqrt{L / C} = 632$ Ω, and a real coil with $50$ Ω of resistance gives $ζ = 0.079$ and $Q = 6.3$ : a ringing circuit, which is what a tuned radio stage is for and what a designer of a digital signal path works to avoid.

The correspondence is exact rather than an analogy, and it is why oscillation is studied once rather than once per discipline. The same three regimes appear in a thermostat's temperature swing, in the response of a servo motor, and in the pitch of a ship in a swell.

Where the model stops

Two assumptions are doing real work here, and both fail eventually.

Viscous damping, force proportional to velocity, is a good model for a fluid dashpot at low speed and for a resistor exactly. It is a poor model for dry friction, whose magnitude is roughly constant and whose direction flips with the velocity. That difference is qualitative rather than a matter of accuracy: a viscously damped oscillator approaches equilibrium exponentially and never quite reaches it, while a dry-friction oscillator loses a fixed amount of amplitude per cycle, so the peaks fall on a straight line and the motion stops dead in finite time, generally not at the equilibrium position.

Linearity is the other. Hooke's law is the first term of a Taylor expansion of a real restoring force, and stretching a spring far enough introduces $x^{3}$ terms that make the frequency depend on amplitude. A pendulum has $\sin θ$ in place of $θ$ and is only harmonic for small swings, which the final lesson quantifies.

So far, though, nothing has pushed on the system. Every solution here decays to nothing, and a machine that only ever rings down is not doing any work. The next lesson adds a driving force and finds that the response depends violently on the frequency at which it is applied.