Part A · the equation of motion · spine

The damped driven oscillator, which is most of linear physics

One equation, solved completely — and with it the habit that runs to the end of the course: find the response to a single kick, and you have the response to everything.

needs 01, 02~2.5 h slides + 5 h problemsproblems: ./mr problems 06

Unit 01 said that a solution exists and is unique; unit 02 said what a solution looks like when there is no formula. This unit is the other half of that bargain: the one family of equations where a formula always exists, where it can be written down in closed form for an arbitrary forcing, and where almost every linear phenomenon in physics is already visible.

How to read this deck

One instrument runs through it: a moving-coil galvanometer, a coil on a torsion fibre between magnet poles. It has a knob — the resistance you hang across its terminals — and that knob is exactly the damping constant $\gamma$. Section 2 turns the knob through all three regimes, section 3 drives the instrument with an alternating current, and section 4 kicks it with a charge pulse and then reads the same number off it twice, by two unrelated routes.

Words with a dotted underline are glossary terms; click one, or press G for the index. N opens the notes, where the long proofs sit under a Full proof heading. A solid thin underline names a piece of the running example and shows it on hover.

New mathematics here: the complex numbers and the complex exponential, used purely as algebra — no complex analysis until unit 20. Everything else is linear algebra over a function space, single-variable calculus, and unit 01's existence and uniqueness theorem doing structural work.

06 · overview

One equation, four sections

1 · why this equation

Expand any potential about a minimum and the leading term is quadratic. Add friction and a drive and you have $Lx := \ddot x + \gamma\dot x + \omega_0^2 x = f(t)$ — the local model of everything. Then complex numbers, which make it easy.

2 · the free equation

The solutions of $Lx = 0$, written $\ker L$, form a two-dimensional vector space — a theorem, from unit 01, not folklore. The characteristic polynomial, the Wronskian, and the three damping regimes the galvanometer's knob sweeps.

3 · driving it

Steady state, amplitude and phase lag; where the peak really is; the width that is exactly $\gamma$; and a galvanometer tuned to a frequency, gaining a factor $Q = \omega_0/\gamma = 20$.

4 · one response to everything

The impulse response $G$, built from a jump in the derivative; convolution; and one charge pulse measured by one swing.

the running example

A moving-coil galvanometer: $I\ddot\theta + b\dot\theta + \kappa\theta = \alpha\, i(t)$, with $\omega_0 = 8\ \mathrm{rad\,s^{-1}}$ and a damping you set by turning a dial. Every result below is checked on it, in numbers.

The order is chosen so that nothing is asserted twice. Section 2 establishes that the solution space is a plane, which is what licenses every "and that is the general solution" later. Section 3 solves one particular drive by an ansatz, which is fast but works only for that drive. Section 4 solves every drive at once, and then reproduces section 3's answer as a special case — the same number by two unrelated routes, which is the strongest check a piece of physics can get.

What this unit is the prerequisite for: unit 07 (many coupled oscillators, where $\ker L$ has dimension $2n$ and the spectral theorem picks its basis), unit 08 (the orbit equation is this equation with an extra term), unit 20 (which replaces the jump condition by a distribution and turns causality into a statement about poles), unit 21 (perturbation series built on this solution), and unit 24, whose retarded Green's function for the wave operator is this construction done in four dimensions.

Problem sheet: ./mr problems 06. Five hours. P7, P10 and P12 are the ones the checkpoint is built from.

section 1

Why this equation, and the numbers that make it easy

Every minimum is a spring. Then the field $\mathbb{C}$, introduced for one reason: so that every quadratic factorises.

06 · why this equation

Every minimum is a spring

Let $V$ be $C^3$ near $x_\ast$ with $V'(x_\ast) = 0$ and $V''(x_\ast) > 0$. Unit 00's third case gives $m\ddot x = -V'(x)$. Put $\xi = x - x_\ast$ and use Taylor with remainder:

$$m\ddot\xi = -V'(x_\ast + \xi) = -V''(x_\ast)\,\xi - \tfrac12 V'''(\eta)\,\xi^2 ,$$

small oscillations

Dropping the $O(\xi^2)$ term leaves the harmonic oscillator

$$\ddot\xi + \omega_0^2\xi = 0, \qquad \omega_0 = \sqrt{V''(x_\ast)/m}.$$

The approximation is uniform on $|\xi| \le \varepsilon$ with error bounded by $\tfrac12\varepsilon^2\max|V'''|$, so it is a statement about a neighbourhood, not a hope.

A lopsided well, $V = (1 - e^{-\xi})^2$, and its quadratic Taylor polynomial $\xi^2$. Inside the shaded band $|\xi| \le \varepsilon$ the two cannot be told apart; outside it they part, because the true well is steep on one side and soft on the other — that asymmetry is the dropped $V'''$ term.

What to look for in the figure. The parabola is not a fit; it is fixed entirely by two numbers at one point, $V(x_\ast)$ and $V''(x_\ast)$ (the slope there is zero). That it hugs the true curve over a whole band is Taylor's theorem with remainder, and the width of the band over which the hug is good to a given tolerance is what $\max|V'''|$ controls. The well drawn is a Morse potential, the standard model of a chemical bond: a molecule vibrates harmonically for small stretches and dissociates — the plateau — for large ones.

Where the damping term comes from. Nowhere, honestly: $-b\dot x$ is not derived from anything in this course. It is a phenomenological force law, admitted exactly as unit 00 admitted force laws as input. It is a good model when the true dissipation is linear in the velocity — a coil moving in a magnetic field with a resistor across it, as in the running example, where the back-e.m.f. is exactly proportional to $\dot\theta$ and Ohm's law is exactly linear. It is a bad model for dry friction, which is $-\mu N\operatorname{sgn}(\dot x)$ and not linear at all, and for turbulent drag, which is quadratic (unit 00's second case).

Taylor with remainder in Lagrange form is single-variable calculus and the learner has it: $V'(x_\ast+\xi) = V'(x_\ast) + V''(x_\ast)\xi + \tfrac12 V'''(\eta)\xi^2$ for some $\eta$ between $x_\ast$ and $x_\ast+\xi$. The multivariable version, which is what unit 16 needs for a rigid body's small oscillations, waits for unit 03's derivative as a linear map.

06 · why this equation

Three checks, and the hilltop

checks

Dimensions: $[V''/m] = (\mathrm{M\,L^2T^{-2}}/\mathrm{L^2})/\mathrm{M} = \mathrm{T^{-2}}$, so $\omega_0$ is a frequency.

Special case: $V = \tfrac12 k x^2$ gives $\omega_0 = \sqrt{k/m}$ with no approximation at all — the expansion terminates.

Against unit 00: the pendulum has $V = mg\ell(1-\cos\theta)$, $V''(0) = mg\ell$, inertia $m\ell^2$, so $\omega_0 = \sqrt{g/\ell}$, and unit 00's exact period $4\sqrt{\ell/g}\,K(\sin\tfrac{\theta_0}{2})$ tends to $2\pi\sqrt{\ell/g}$ as $\theta_0 \to 0$. ✓

The expansion is why the rest of the unit is worth the time: it says the harmonic oscillator is not an example, it is the generic behaviour of matter near equilibrium.

The pendulum's potential has two equilibria. At the bottom the matching parabola opens upwards — a spring, with $\omega_0 = \sqrt{g/\ell}$. At the top it opens downwards: $V'' < 0$, the same expansion gives growing exponentials, and the equilibrium is unit 02's saddle.

Why $V''(x_\ast) > 0$ and not $\ge 0$. If $V''(x_\ast) = 0$ the quadratic term is absent and the leading behaviour is quartic or worse; the motion is still periodic but its period depends on amplitude, and none of this unit applies. Unit 02 classified such a point as a non-hyperbolic equilibrium, and unit 21 computes the amplitude-dependent period as a perturbation series. If $V''(x_\ast) < 0$ the same expansion gives $\ddot\xi = +|\omega_0^2|\xi$, hyperbolic functions, and a saddle — unit 02's upright pendulum, the hilltop in the figure.

Reading the figure against the checks. The upward parabola at $\theta = 0$ is $\tfrac12\theta^2$ in units of $mg\ell$, so $V''(0) = mg\ell$; divided by the inertia $m\ell^2$ that is $g/\ell$, the third check. The two curves agree to $1\%$ out to about $20^\circ$, since $1-\cos\theta = \tfrac12\theta^2(1 - \theta^2/12 + \dots)$ — which is the same $20^\circ$, roughly, at which unit 00 found the period itself $1\%$ off the small-angle value.

06 · the running example

The galvanometer, and the knob that is $\gamma$

A coil on a torsion fibre between magnet poles. Three torques act: the fibre's restoring torque $-\kappa\theta$ (exactly linear, no approximation); the driving torque $\alpha\,i(t)$; and a torque from the current the coil's own motion induces.

where the damping comes from

Motion at rate $\dot\theta$ generates a back-e.m.f. $\alpha\dot\theta$, hence a current $-\alpha\dot\theta/(R+r)$, hence a torque $-\alpha^2\dot\theta/(R+r)$. So $b = \alpha^2/(R+r)$ and

$$\ddot\theta + \gamma\dot\theta + \omega_0^2\theta = \frac{\alpha}{I}\,i(t),$$ $$\gamma = \frac{\alpha^2}{I(R+r)},\qquad \omega_0^2 = \frac{\kappa}{I}.$$

With the numbers on the card: $\omega_0 = 8\ \mathrm{rad\,s^{-1}}$ and $\gamma = 400/(R+r)\ \mathrm{s^{-1}}$. Turning $R$ from $0$ to $\infty$ sweeps $\gamma$ from $80$ down to $0$.

Seen from above: the coil (axis out of the page) turned through $\theta$ between the pole pieces. The drive pushes it one way; the fibre and the induced current pull it back. Right: the same coil in its circuit — the resistor $R$ across its terminals is the knob. Schematic.

Reading the figure. The fibre is seen end-on, as the dot the coil turns about; its torque $-\kappa\theta$ grows with the angle drawn. The induced torque $-b\dot\theta$ is drawn opposing the drive, but strictly it opposes the velocity, whichever way the coil happens to be moving. The current that produces it flows round the loop on the right — through the coil's own resistance $r$ and the external $R$ — which is why $R$ sets its size, and why opening the circuit ($R\to\infty$) removes it altogether.

The back-e.m.f. is the same constant $\alpha$. That the coil constant appearing in "torque per unit current" also appears in "e.m.f. per unit angular velocity" is not a coincidence; it is forced by energy conservation, and unit 24 will derive it from Faraday's law. Here it is taken as given, with the physical content noted: the same number governs both directions of the energy exchange, which is why $b = \alpha^2/(R+r)$ carries $\alpha$ squared.

A third torque we are ignoring: air damping on the coil and the mirror. It is real, it is roughly linear in $\dot\theta$, and it simply adds to $b$; taking it as zero means the instrument can never be made to ring for ever, only for a long time. At $R = 995\ \Omega$ the electromagnetic damping alone gives a decay time of $5$ s, and the air contribution for a coil this small is an order of magnitude weaker, so ignoring it changes the fourth figure, not the picture.

06 · the running example

The knob, checked

$\gamma = 400/(R+r)$ against the total resistance, on a log scale. Every setting used in this unit is a dot; above the dashed line $\gamma = 2\omega_0$ the instrument is overdamped, below it, it rings.

checks

Dimensions: $[\alpha^2/(I\,\Omega)]$ reduces to $\mathrm{s^{-1}}$, using $\mathrm{V = \Omega\,A}$ and $\mathrm{J = V\,A\,s}$. ✓

Limit $R \to \infty$: open circuit, no induced current, $\gamma \to 0$ — the coil swings freely, and it does.

Limit $R \to 0$: short circuit, maximal damping. Shorting the terminals to stop a galvanometer swinging is laboratory practice, and this is why it works.

two dials, one equation

$\omega_0$ is fixed by the hardware; $\gamma$ is set by a resistor you choose. That is unusual and it is why this instrument makes a good running example: the parameter the whole unit turns on is, here, literally a knob.

What to look for in the figure. The knob is very non-linear. The whole overdamped range is squeezed into the first $20\ \Omega$ of external resistance, while the ringing regime takes everything from $20\ \Omega$ to infinity; on a log scale the curve is a hyperbola, $\gamma \propto (R+r)^{-1}$, and the coil's own $5\ \Omega$ is why it stops at $80$ instead of rising for ever. The four dots are the four settings the deck returns to: shorted, critical, a moderate $Q = 1.6$, and the $Q = 20$ setting of the tuned and ballistic examples.

The numbers, chosen to be clean. $\alpha^2 = 4\times10^{-6}$ and $I = 10^{-8}$ give $\gamma = 400/(R+r)$ exactly. Critical damping needs $\gamma = 2\omega_0 = 16$, so $R + r = 25\ \Omega$ and, with $r = 5\ \Omega$, an external resistance of exactly $20\ \Omega$. Real galvanometers quote precisely this number on their datasheets, under the name external critical damping resistance (CDRX), and it is the one number a user must match.

06 · the complex numbers

$\mathbb{C}$, and the one property we want

complex numbers

$\mathbb{C}$ is the set $\mathbb{R}^2$ with the usual addition and the multiplication $(a,b)\cdot(c,d) = (ac - bd,\ ad + bc)$.

Writing $1 = (1,0)$ and $i = (0,1)$, every element is $a + bi$ uniquely, and $i^2 = -1$. The complex conjugate is $\overline{a+bi} = a - bi$, the modulus is $|z| = \sqrt{z\bar z} = \sqrt{a^2+b^2}$, and for $z \ne 0$ the argument $\arg z$ is the angle $\varphi$ with $z = |z|(\cos\varphi + i\sin\varphi)$.

$\mathbb{C}$ is a field

Associativity, commutativity and distributivity are four short computations. For $z \ne 0$, $\;z^{-1} = \bar z/|z|^2\;$ works, since $z\bar z = |z|^2 \ne 0$. $\square$

why we want it

The whole of section 2 turns on factorising $\lambda^2 + \gamma\lambda + \omega_0^2$. Over $\mathbb{R}$ that fails when $\gamma^2 < 4\omega_0^2$ — the interesting case, the one where the thing oscillates. Over $\mathbb{C}$ every quadratic factorises, so one calculation covers all three regimes instead of three.

algebra, not analysis

We use $\mathbb{C}$ as a field of coefficients and as a place where $e^{i\theta}$ lives — not holomorphy, contour integration or residues, which are unit 20.

A complex-valued function of a real variable $t$ is a pair of real functions; derivative and integral are componentwise, and every rule of single-variable calculus transfers term by term.

The argument, more carefully. For $z \ne 0$, $z/|z|$ lies on the unit circle, so $z = |z|(\cos\varphi + i\sin\varphi)$ for a $\varphi$ unique modulo $2\pi$; that $\varphi$ is $\arg z$. We will always name a specific branch when it matters — in section 3 the phase lag is pinned to $(0,\pi)$ by taking $\arg$ with positive imaginary part.

A cleaner construction, for the algebraically minded. $\mathbb{C} = \mathbb{R}[X]/(X^2+1)$. Since $X^2+1$ is irreducible over $\mathbb{R}$, the ideal it generates is maximal and the quotient is a field; the class of $X$ is $i$. This makes the field axioms free, and it makes the point of the construction explicit: we are adjoining a root of the one polynomial we could not factor. The $\mathbb{R}^2$ definition is used above because the learner can check it in four lines and because the identification of $\mathbb{C}$ with the plane is what makes $|z|$ obviously a length.

What we are not claiming. That every polynomial over $\mathbb{C}$ splits — the fundamental theorem of algebra — is true, and unit 20 proves it by Liouville's theorem. We need only the quadratic case, which is the quadratic formula plus the existence of $\sqrt{w}$ for complex $w$, both elementary. Nothing here depends on the general theorem.

Multiplication is a rotation-and-scaling. The map $z \mapsto wz$ is $\mathbb{R}$-linear on $\mathbb{R}^2$ with matrix $\left(\begin{smallmatrix}c&-d\\d&c\end{smallmatrix}\right)$ for $w = c+di$ — which the learner has already met, in unit 02's real canonical form for a matrix with complex eigenvalues. That slide and this one are the same fact seen from two sides, and it is worth going back to look.

06 · the complex numbers

The complex exponential

complex exponential

For $z \in \mathbb{C}$, $\;\exp z = \sum_{n\ge0} z^n/n!$. The series converges absolutely for every $z$, since $\sum |z|^n/n!$ converges; so $\exp$ is defined on all of $\mathbb{C}$.

the two properties we need

(a) $\exp(z+w) = \exp z\,\exp w$.

(b) Euler's formula: $\;\exp(i\vartheta) = \cos\vartheta + i\sin\vartheta$ for real $\vartheta$; hence $\;\exp(a+bi) = e^a(\cos b + i\sin b)$ and $|\exp(a+bi)| = e^a$.

both proved in the notes (N), from real analysis only

the lemma that does all the work

For fixed $\lambda = a+bi$ and real $t$, $\,t \mapsto e^{\lambda t}$ is differentiable and

$$\frac{\mathrm{d}}{\mathrm{d}t}e^{\lambda t} = \lambda e^{\lambda t}, \qquad \int_{t_1}^{t_2}\! e^{\lambda t}\,\mathrm{d}t = \frac{e^{\lambda t_2} - e^{\lambda t_1}}{\lambda}$$

proof

$e^{\lambda t} = e^{at}\cos bt + i\,e^{at}\sin bt$. Differentiate each real part by the product rule: $(a\cos bt - b\sin bt)e^{at} + i(a\sin bt + b\cos bt)e^{at}$, which is $(a+bi)e^{at}(\cos bt + i \sin bt)$. The integral ($\lambda \ne 0$) then follows from the fundamental theorem, componentwise. $\square$

Full proof of (a)

Both series converge absolutely, so their Cauchy product converges to the product of the sums (Mertens' theorem, cited: Rudin, Principles of Mathematical Analysis, Thm 3.50; absolute convergence of one factor suffices, and here both are absolute). The $n$-th term of the Cauchy product is $\sum_{k=0}^{n} \frac{z^k}{k!}\frac{w^{n-k}}{(n-k)!} = \frac{1}{n!}\sum_{k=0}^n \binom{n}{k}z^k w^{n-k} = \frac{(z+w)^n}{n!}$ by the binomial theorem, which holds in any commutative ring. Summing over $n$ gives $\exp(z+w)$. $\square$

Full proof of (b)

Put $z = i\vartheta$ with $\vartheta$ real. Then $i^{2m} = (-1)^m$ and $i^{2m+1} = (-1)^m i$, so splitting the absolutely convergent series into even and odd $n$ — legitimate precisely because it is absolutely convergent — gives $\exp(i\vartheta) = \sum_m \frac{(-1)^m\vartheta^{2m}}{(2m)!} + i\sum_m \frac{(-1)^m\vartheta^{2m+1}}{(2m+1)!}$. Those are the real Maclaurin series for $\cos\vartheta$ and $\sin\vartheta$, which converge to $\cos$ and $\sin$ for every real $\vartheta$ — a fact from single-variable analysis, provable from the Lagrange remainder $|R_N| \le |\vartheta|^{N+1}/(N+1)! \to 0$. The general case follows from (a): $\exp(a+bi) = \exp(a)\exp(bi)$. $\square$

Consequences used without further comment. $|e^{i\vartheta}| = 1$; $\overline{e^{z}} = e^{\bar z}$; $e^z \ne 0$ ever, since $e^z e^{-z} = 1$; $\cos\vartheta = \tfrac12(e^{i\vartheta} + e^{-i\vartheta})$ and $\sin\vartheta = \tfrac1{2i}(e^{i\vartheta} - e^{-i\vartheta})$, which is the identity that converts every trigonometric manipulation in this unit into algebra with exponentials.

Historically Euler published the formula in 1748 (Introductio in analysin infinitorum), having used it since the 1740s; Cotes had the logarithmic form in 1714. The modern reading — that $\vartheta \mapsto e^{i\vartheta}$ is a homomorphism from $(\mathbb{R},+)$ to the circle group — is exactly unit 02's one-parameter group property, and unit 03 will meet it again as $\mathrm{SO}(2) \subset \mathrm{SO}(3)$.

section 2

The free equation: a two-dimensional space

Not "the general solution has two constants" as folklore, but as a theorem with a proof, inherited from unit 01.

06 · the free equation

$L$ is a linear map between function spaces

the linear differential operator

Fix $\gamma \ge 0$ and $\omega_0 > 0$. Define $L : C^2(\mathbb{R},\mathbb{R}) \to C^0(\mathbb{R},\mathbb{R})$ by

$$Lx = \ddot x + \gamma\dot x + \omega_0^2 x .$$

$L$ is linear: $L(cx + y) = cLx + Ly$, because differentiation is.

So the equation $Lx = f$ is a linear equation in a vector space, and every theorem of linear algebra applies, word for word. $Lx = 0$ is the homogeneous equation, $Lx = f$ with $f \ne 0$ the inhomogeneous one.

the structure theorem, for free

If $x_p$ satisfies $Lx_p = f$, then the full solution set is the coset

$$\{x : Lx = f\} = x_p + \ker L .$$

proof

If $Lx = f$ then $L(x - x_p) = f - f = 0$, so $x = x_p + h$ with $h \in \ker L$. Conversely $L(x_p+h) = f + 0 = f$. $\square$

This is the superposition principle, and it is the reason the rest of the unit splits cleanly: find $\ker L$ once, find one $x_p$ per drive.

The spaces are infinite-dimensional; that does not matter. $C^2(\mathbb{R},\mathbb{R})$ and $C^0(\mathbb{R},\mathbb{R})$ are enormous, and $L$ is surjective onto the second (which the next section effectively proves, by producing a solution for every continuous $f$). The only finite object in sight is $\ker L$, and the next slide shows it has dimension exactly two. All the linear algebra we use — kernel, coset, basis, dimension — is valid over any field and in any dimension, so nothing needs to be re-proved for function spaces.

Why "operator" rather than "matrix". There is no matrix here because there is no finite basis; but $L$ is still a linear map, and the habit of asking "what is its kernel, what is its image, what is a preimage of this particular element" is exactly the right habit. Unit 07 makes the analogy literal — $n$ coupled oscillators give a genuine matrix — and unit 19 completes it, showing that for a Sturm–Liouville operator the eigenfunctions form an orthonormal basis, so that $L$ really does become diagonal.

Superposition is a physical claim, not only an algebraic one. It says two drives applied together produce the sum of their separate responses. Nature obeys it only to the extent that the equation is linear: a real spring stiffens at large extension, a real galvanometer's fibre yields, and then superposition fails. Every statement in this unit inherits the caveat of the first slide — it is a statement about a neighbourhood of equilibrium.

06 · the free equation

$\dim \ker L = 2$, and unit 01 is the reason

theorem

$\ker L$ is a real vector space of dimension exactly $2$. Precisely: for each $t_0$, the evaluation map

$$\mathrm{ev} : \ker L \to \mathbb{R}^2, \qquad \mathrm{ev}(x) = \big(x(t_0),\ \dot x(t_0)\big)$$

is an isomorphism of vector spaces.

proof

Linear: evaluation and differentiation are.

Injective: if $x \in \ker L$ has $x(t_0) = \dot x(t_0) = 0$, then $y = (x,\dot x)$ solves unit 01's system $\dot y = Ay$, $A = \left(\begin{smallmatrix}0&1\\-\omega_0^2&-\gamma\end{smallmatrix}\right)$, with $y(t_0) = 0$, as does $y \equiv 0$. $Ay$ is Lipschitz with constant $\|A\|$, so Picard–Lindelöf gives $x \equiv 0$.

Surjective: $\|Ay\| \le \|A\|\|y\|$ is linear growth, so unit 01's global existence theorem produces a solution with $y(t_0) = (x_0,v_0)$ on all of $\mathbb{R}$; its first component is in $\ker L$ and evaluates to $(x_0,v_0)$. $\square$

this is not a formality

"Second order, so two constants" is folklore, and it is false without a hypothesis: unit 01's dome, $\ddot r = k\sqrt r$, is second order with a one-parameter family of solutions through one state. Linearity rules that out — nothing else does.

what it licenses

Any two independent solutions you can guess and verify span every solution there is: section 2 becomes a check rather than a search.

The reduction to a first-order system is unit 01's phase-space reduction theorem, applied to $m = 1$, $F(x,\dot x) = -\gamma\dot x - \omega_0^2 x$: it says $x$ is a $C^2$ solution of $Lx = 0$ if and only if $(x,\dot x)$ is a $C^1$ solution of $\dot y = Ay$. Both directions are needed above: injectivity uses $\Rightarrow$, surjectivity uses $\Leftarrow$.

A Lipschitz constant for a linear field, by hand. $\|Ay - Az\| = \|A(y-z)\| \le \|A\|\|y-z\|$ where $\|A\|$ is the operator norm — no calculus needed, so unit 01's ban on "$C^1$ implies locally Lipschitz" (which waits for unit 03) is respected. Concretely $\|A\| \le \sqrt{1 + \omega_0^4 + \gamma^2}$ by the Frobenius bound, which is all that is required.

Over $\mathbb{C}$ as well. Everything above holds verbatim for $L : C^2(\mathbb{R},\mathbb{C}) \to C^0(\mathbb{R},\mathbb{C})$, giving $\dim_{\mathbb{C}}\ker L = 2$: split into real and imaginary parts and apply the real theorem twice. We use both versions — the complex one to find a basis, the real one to say what the instrument does.

The generalisation, for later. For an $n$-th order linear equation with continuous coefficients the identical argument gives $\dim\ker = n$; for a system of $n$ second-order equations (unit 07, unit 16's rigid body) it gives $2n$. The proof never changes: linearity gives a Lipschitz bound, Picard–Lindelöf gives a bijection with the space of initial data, and the dimension of the solution space is the dimension of the state space. That is the one sentence to remember.

06 · the free equation

The characteristic polynomial

Try $x = e^{\lambda t}$ with $\lambda \in \mathbb{C}$. By the lemma,

$$L\,e^{\lambda t} = \big(\lambda^2 + \gamma\lambda + \omega_0^2\big)\,e^{\lambda t} = p(\lambda)\,e^{\lambda t},$$

the characteristic polynomial

$p(\lambda) = \lambda^2 + \gamma\lambda + \omega_0^2$, with roots

$$\lambda_\pm = -\frac{\gamma}{2} \pm \sqrt{\frac{\gamma^2}{4} - \omega_0^2}\,.$$

$e^{\lambda t} \in \ker L \iff p(\lambda) = 0$, because $e^{\lambda t}$ never vanishes.

Where the two roots sit in $\mathbb{C}$ decides everything that follows; the next slide draws them.

the identity for a repeated root

For any $\lambda$, $\;L\big(t\,e^{\lambda t}\big) = p(\lambda)\,t\,e^{\lambda t} + p'(\lambda)\,e^{\lambda t}$.

proof

$\frac{\mathrm{d}}{\mathrm{d}t}(te^{\lambda t}) = (1+\lambda t)e^{\lambda t}$ and $\frac{\mathrm{d}^2}{\mathrm{d}t^2}(te^{\lambda t}) = (2\lambda + \lambda^2 t)e^{\lambda t}$, so $L(te^{\lambda t}) = \big[(\lambda^2+\gamma\lambda+\omega_0^2)t + (2\lambda+\gamma)\big]e^{\lambda t}$, and $p'(\lambda) = 2\lambda+\gamma$. $\square$

So when $p$ has a double root, $p(\lambda) = p'(\lambda) = 0$ and $t e^{\lambda t}$ is a second solution. It will be needed exactly once, and it is the whole of critical damping.

The same identity, used the other way round at $p(\lambda) \ne 0$, gives the resonant solution in section 3. One computation, two jobs.

Why the exponential and not something else. Because $L$ has constant coefficients, it commutes with time translation: $L(x(\cdot + s)) = (Lx)(\cdot + s)$. The functions that are eigenvectors of time translation — $x(t+s) = c(s)x(t)$ — are exactly the exponentials, and any linear map commuting with translation must act on them by a scalar. So the ansatz is not a guess; it is the representation theory of $(\mathbb{R},+)$ in one line. Unit 20 makes this the Fourier transform, and unit 07 makes it the matrix exponential.

Real and imaginary parts. If $z \in \ker L$ is complex-valued and $L$ has real coefficients, then $Lz = 0$ gives $L(\operatorname{Re}z) + iL(\operatorname{Im}z) = 0$, and a complex number is zero exactly when both parts are, so $\operatorname{Re}z$ and $\operatorname{Im}z$ are real solutions. This is the bridge used twice below: compute in $\mathbb{C}$, answer in $\mathbb{R}$.

06 · the free equation

Where the roots go as the knob turns

Read the roots off the coefficients of $p$:

$$\lambda_+\lambda_- = \omega_0^2, \qquad \lambda_+ + \lambda_- = -\gamma .$$

$\gamma < 2\omega_0$: a conjugate pair, so $|\lambda_\pm|^2 = \lambda_+\lambda_- = \omega_0^2$ — both roots lie on the circle $|\lambda| = \omega_0$, at real part $-\gamma/2$.
$\gamma = 2\omega_0$: they meet at $-\omega_0$, the repeated root.
$\gamma > 2\omega_0$: two negative reals with product $\omega_0^2$; as one runs off to $-\infty$ the other is pushed towards $0$.

no free motion grows

Product $> 0$ and sum $\le 0$ put both roots in $\operatorname{Re}\lambda \le 0$, on the boundary only when $\gamma = 0$.

The roots $\lambda_\pm$ as $\gamma$ grows from $0$, with the knob's settings marked. The pair slides round the circle, collides at $-\omega_0$ and splits along the real axis; the root heading for $0$ is why heavy damping is slow.

What to look for in the figure. The real part of a root is a decay rate and the imaginary part a ringing frequency, so the whole of section 2 can be read off this one picture. Moving down the circle, the ringing slows ($\omega_\mathrm{d}$ is the height) and the decay quickens ($\gamma/2$ is the distance from the imaginary axis). How fast a general solution dies is set by the root nearest the imaginary axis, and that distance is greatest — $\omega_0$ — exactly at the collision: critical damping, seen geometrically. Past it, the slow root creeps back towards $0$, which is the "more damping is not faster" of the worked example.

On $\operatorname{Re}\lambda \le 0$. Written out: if the roots are real they are $-\gamma/2 \pm \sqrt{\gamma^2/4-\omega_0^2}$, and the square root is strictly less than $\gamma/2$ because $\omega_0^2 > 0$; if they are complex they are $-\gamma/2 \pm i\omega_\mathrm{d}$ with real part $-\gamma/2$. Equality holds only when $\gamma = 0$. Unit 02 said the same thing in its own language: $\tau = -\gamma \le 0$ and $\Delta = \omega_0^2 > 0$, so the origin sits in the stable half of the trace–determinant plane, and on its boundary exactly when the damping is switched off.

06 · the free equation

The Wronskian, and Abel's identity

the Wronskian

For $x_1, x_2 \in \ker L$, $\;W(t) = \det\begin{pmatrix}x_1 & x_2\\ \dot x_1 & \dot x_2\end{pmatrix} = x_1\dot x_2 - x_2\dot x_1$.

Abel's identity $$\dot W = -\gamma W \ \Longrightarrow\ W(t) = W(t_0)\,e^{-\gamma(t-t_0)} .$$

proof

$\dot W = \dot x_1\dot x_2 + x_1\ddot x_2 - \dot x_2\dot x_1 - x_2\ddot x_1 = x_1(-\gamma\dot x_2 - \omega_0^2x_2) - x_2(-\gamma\dot x_1-\omega_0^2x_1) = -\gamma W$. Then unit 00's separable case, or simply $(We^{\gamma t})^{\boldsymbol\cdot} = 0$. $\square$

the test for a fundamental system

For $x_1,x_2 \in \ker L$, the following are equivalent: (i) $W(t_0) \ne 0$ for some $t_0$; (ii) $W(t) \ne 0$ for every $t$; (iii) $\{x_1,x_2\}$ is a basis of $\ker L$.

proof in the notes (N)

Abel says $W$ is never zero or identically zero: there is no in-between. Independence is therefore checkable at one instant of your choosing, usually $t = 0$.

Full proof of the equivalence

(i) $\Rightarrow$ (ii): Abel's formula, since $e^{-\gamma(t-t_0)} \ne 0$. (ii) $\Rightarrow$ (iii): $W(t_0) \ne 0$ means the matrix $M = \left(\begin{smallmatrix}x_1(t_0)&x_2(t_0)\\ \dot x_1(t_0)&\dot x_2(t_0)\end{smallmatrix}\right)$ is invertible, so $\mathrm{ev}(c_1x_1+c_2x_2) = M(c_1,c_2)^{\!\top}$ ranges over all of $\mathbb{R}^2$; since $\mathrm{ev}$ is an isomorphism onto $\mathbb{R}^2$, the span is all of $\ker L$, and two spanning vectors in a two-dimensional space are a basis. (iii) $\Rightarrow$ (i): if $W(t_0) = 0$ for some $t_0$, pick $(c_1,c_2) \ne 0$ in $\ker M$; then $x = c_1x_1+c_2x_2$ has $\mathrm{ev}(x) = 0$, so $x \equiv 0$ by injectivity of $\mathrm{ev}$, and $x_1,x_2$ are dependent — not a basis. $\square$

Abel's identity in general. For $\ddot x + p(t)\dot x + q(t)x = 0$ the same two lines give $\dot W = -p(t)W$, so $W(t) = W(t_0)\exp\big(-\int_{t_0}^{t}p\big)$ — and note that $q$ never appears. Abel published it in 1827. It is the tool that makes reduction of order work: knowing one solution $x_1$, the Wronskian relation is a first-order equation for $x_2$, solvable by quadrature. Problem P3 uses exactly that route to produce $te^{\lambda t}$ without guessing it.

06 · the free equation

Abel's identity, drawn: phase-space area shrinks

Two solutions as state vectors $y_i = (x_i, \dot x_i/\omega_0)$, for $\gamma = 5$. They start as a unit square; the flow turns and shears them, and the area they span — $W/\omega_0$ — falls by exactly $e^{-\gamma t}$, never reaching zero. Dotted: the tips' paths.

The Wronskian is a determinant, so it is a signed area: the parallelogram spanned by the two states $(x_1,\dot x_1)$ and $(x_2,\dot x_2)$ in the phase plane.

Abel's identity says the flow multiplies every such area by the same factor $e^{-\gamma t}$, whatever the two solutions. A parallelogram of non-zero area keeps non-zero area; one of zero area — two parallel states — stays flat.

checks

Dimensions: $[W] = [x]^2/\mathrm{T}$, and $e^{-\gamma(t-t_0)}$ is dimensionless. ✓ Limit $\gamma = 0$: $W$ is constant — the undamped flow preserves phase-space area, which is unit 14's Liouville theorem arriving eight units early.

The numbers in the figure. With $\gamma = 5$, $\omega_0 = 8$, take $x_1$ released from $(1,0)$ and $x_2 = \omega_0 G$, released from $(0,\omega_0)$; in the scaled coordinates both start as unit vectors. At $t = 0.1$ s the computed determinant is $0.6065$ and $e^{-0.5} = 0.6065$; at $t = 0.2$ s it is $0.3679$ against $e^{-1} = 0.3679$. The shape changes — the square is rotated by roughly $\omega_\mathrm{d}t$ and sheared — but only the area is controlled by Abel, and it is controlled exactly.

What the Wronskian is. Take the two solutions as columns of a $2\times2$ matrix of initial data carried along by the flow. $W$ is the determinant of that matrix, that is, the signed area of the parallelogram the two states span in phase space. Abel's identity then reads: the flow of $\dot y = Ay$ multiplies areas by $e^{t\operatorname{tr}A}$, and $\operatorname{tr}A = -\gamma$. Unit 02 drew that contraction; unit 14 proves the general statement (Liouville) and shows that Hamiltonian flows are exactly the ones with $\operatorname{tr} = 0$.

06 · the free equation · example

Worked: the galvanometer's three regimes

worked example 1 — turning the knob

Step 1. $\gamma = 400/(R+r)$ with $r = 5\,\Omega$; $\omega_0 = 8\ \mathrm{s^{-1}}$; the discriminant of $p$ is $\gamma^2 - 4\omega_0^2 = \gamma^2 - 256$. Step 2. Compare $\gamma$ with $2\omega_0 = 16$. Step 3. Solve $p(\lambda) = 0$ and read the basis off.

$R$	$\gamma$	roots $\lambda_\pm$	regime	basis of $\ker L$	$t$ to settle within $1\%$
$995\ \Omega$	$0.4$	$-0.2 \pm 7.9975\,i$	underdamped	$e^{-0.2t}\cos 7.9975t,\ e^{-0.2t}\sin 7.9975t$	$22.8$ s
$75\ \Omega$	$5$	$-2.5 \pm 7.5993\,i$	underdamped	$e^{-2.5t}\cos 7.5993t,\ e^{-2.5t}\sin 7.5993t$	$1.78$ s
$20\ \Omega$	$16$	$-8$ (twice)	critically damped	$e^{-8t},\ t\,e^{-8t}$	$\mathbf{0.83}$ s
$0$	$80$	$-0.8082,\ -79.19$	overdamped	$e^{-0.8082t},\ e^{-79.19t}$	$5.7$ s

Step 4, the check. For each row, $\lambda_+\lambda_- = \omega_0^2 = 64$ and $\lambda_++\lambda_- = -\gamma$: $\;0.2^2 + 7.9975^2 = 64.00$ ✓; $\;0.8082\times79.19 = 64.00$ ✓; $\;0.8082 + 79.19 = 80$ ✓.

The settling times, computed. Released from rest at $\theta_0$, each regime gives a different $\theta(t)/\theta_0$, and "settles within $1\%$" means $|\theta/\theta_0| \le 0.01$ from then on. Underdamped: the last crossing of $0.01$ is at a maximum of the oscillation, giving $22.82$ s at $\gamma = 0.4$ and $1.78$ s at $\gamma = 5$; the envelope estimate $\ln(100)/(\gamma/2)$ gives $23.0$ s and $1.84$ s, close but not exact. Critical: solve $(1+8t)e^{-8t} = 0.01$, giving $t = 0.8298$ s. Overdamped: $\theta/\theta_0 = (\lambda_- e^{\lambda_+t} - \lambda_+e^{\lambda_-t})/(\lambda_--\lambda_+)$, whose slow term $1.0103\,e^{-0.8082t}$ reaches $0.01$ at $t = 5.711$ s.

The lesson in one line: more damping is not faster. Going from $\gamma = 16$ to $\gamma = 80$ makes the instrument seven times slower, because the slow root $\lambda_+ = -\gamma/2 + \sqrt{\gamma^2/4-\omega_0^2} \to -\omega_0^2/\gamma$ tends to zero as $\gamma$ grows. Problem P4 proves that critical damping is the optimum, and the critical-damping slide states it.

Why the underdamped basis is real. $\lambda_\pm = -\gamma/2 \pm i\omega_\mathrm{d}$ gives the complex basis $e^{\lambda_\pm t}$; the real and imaginary parts of $e^{\lambda_+t}$ are $e^{-\gamma t/2}\cos\omega_\mathrm{d}t$ and $e^{-\gamma t/2}\sin\omega_\mathrm{d}t$, both in $\ker L$ by the remark in the characteristic-polynomial slide's notes, and their Wronskian at $t=0$ is $\omega_\mathrm{d} \ne 0$, so they are a basis. This is the standard passage from the complex basis to the real one and it is worth doing once by hand.

06 · the free equation

Underdamped: ringing, and how fast it dies

The galvanometer released from rest at $\theta_0$, for four settings of the knob. At $\gamma = 0.4$ the swings stay inside the envelope $e^{-\gamma t/2}$ (dashed), and the maxima $\theta_1, \theta_2$, one period $T_\mathrm{d}$ apart, are in the ratio $e^{-\delta} = 0.855$.

damped natural frequency

For $\gamma < 2\omega_0$, $\;\lambda_\pm = -\tfrac\gamma2 \pm i\omega_\mathrm{d}$ with

$$\omega_\mathrm{d} = \sqrt{\omega_0^2 - \gamma^2/4}\ <\ \omega_0 .$$

logarithmic decrement

Successive maxima are a period $T_\mathrm{d} = 2\pi/\omega_\mathrm{d}$ apart and their ratio is constant:

$$\delta = \ln\frac{\theta_n}{\theta_{n+1}} = \frac{\gamma T_\mathrm{d}}{2} = \frac{\pi\gamma}{\omega_\mathrm{d}} .$$

At $\gamma = 0.4$: $T_\mathrm{d} = 0.78564$ s, $\delta = 0.15713$, each swing $85.46\%$ of the last. Measuring $\delta$ is how $\gamma$ is measured.

Why the ratio of successive maxima is constant. With $\theta = Ae^{-\gamma t/2}\cos(\omega_\mathrm{d}t - \phi)$, $\dot\theta = 0$ gives $\tan(\omega_\mathrm{d}t-\phi) = -\gamma/2\omega_\mathrm{d}$, whose solutions are spaced by exactly $\pi/\omega_\mathrm{d}$; consecutive maxima are spaced by $T_\mathrm{d} = 2\pi/\omega_\mathrm{d}$, and at each of them the cosine takes the same value. So the ratio is the envelope's ratio, $e^{-\gamma T_\mathrm{d}/2}$, independent of which pair you pick. That independence is the experimental test that the damping really is linear in the velocity: for dry friction the successive differences, not ratios, are constant, and amplitude falls linearly to zero in finite time.

Frequency pulling, in numbers. $\omega_\mathrm{d}/\omega_0 = \sqrt{1-1/4Q^2}$, so at $Q = 20$ the shift is $-0.031\%$ — $7.99750$ against $8$ — and at $Q = 1.6$ ($\gamma = 5$) it is $-5.0\%$. The rule of thumb is that damping changes the frequency only at second order in $1/Q$, which is why a lightly damped oscillator can be used as a clock even though it is losing energy.

Energy. With $E = \tfrac12 I\dot\theta^2 + \tfrac12\kappa\theta^2$, a direct computation gives $\dot E = -b\dot\theta^2 \le 0$ — exactly unit 02's gradient-like structure, with the same Lyapunov function. Averaged over a cycle at small $\gamma$, $E$ decays as $e^{-\gamma t}$: amplitude decays at $\gamma/2$, energy at $\gamma$, because energy is quadratic in amplitude. Keeping those two rates apart is the commonest slip in this subject.

06 · the free equation

Critical damping, and why it is genuinely special

$\gamma = 2\omega_0$: the degenerate case

$p$ has the double root $\lambda = -\omega_0$, and $\{e^{-\omega_0t},\ te^{-\omega_0t}\}$ is a basis of $\ker L$.

proof

$p(\lambda) = p'(\lambda) = 0$, so the characteristic-polynomial slide's identity gives $L(te^{\lambda t}) = 0$. Wronskian at $t = 0$: $W = 1\cdot 1 - 0\cdot(-\omega_0) = 1 \ne 0$, so they are independent, hence a basis. $\square$

The $t$ is not a fudge. In unit 02's language the matrix $A$ has a single eigenvector and a $2\times2$ Jordan block: the same degeneracy, in the same system, seen through a different door. Unit 07 makes the connection exact.

why instruments are built this way

Among all $\gamma \ge 0$, $\gamma = 2\omega_0$ is the unique value that returns to equilibrium fastest without crossing it:

$\gamma < 2\omega_0$ overshoots — at $\gamma = 0.4$ the needle swings $92.4\%$ of the way to the other side before coming back;
$\gamma > 2\omega_0$ has slow root $\lambda_+ \to -\omega_0^2/\gamma \to 0$, so the approach gets slower without limit;
at $\gamma = 2\omega_0$ the decay rate $\omega_0$ is the largest available, and $te^{-\omega_0 t}$ crosses zero only at $t=0$.

the extremal claim is problem P4

"Dead-beat" is the trade name. A pointer instrument that rings is unusable — the eye has to wait for the swings to die — and one that creeps is worse. So meters are shipped damped at or just below critical, and the datasheet quotes the external resistance that achieves it: $20\ \Omega$ here. A user who connects the instrument to a source of the wrong resistance changes $\gamma$ and gets an instrument with the wrong dynamics, which is a real and common laboratory error.

The overshoot figure. Released from rest, $\theta/\theta_0 = e^{-\gamma t/2}(\cos\omega_\mathrm{d}t + \tfrac{\gamma}{2\omega_\mathrm{d}}\sin\omega_\mathrm{d}t)$. At $\gamma = 0.4$ the first minimum is $-0.9244$ at $t = 0.3928$ s: the needle goes $92.4\%$ as far the other way. The general formula for the fractional overshoot is $\exp(-\pi\gamma/2\omega_\mathrm{d}) = e^{-\delta/2}$, tying it to the logarithmic decrement.

The critical case is not a limit in disguise. It is tempting to define $te^{\lambda t} = \lim_{\mu\to\lambda}(e^{\mu t}-e^{\lambda t})/(\mu-\lambda)$ and say the critical solution is the limit of the overdamped ones. That limit is correct and it is a good way to remember the formula, but on its own it proves nothing: a pointwise limit of solutions is not automatically a solution, and making it one needs an argument about the derivatives as well. The proof on the slide needs no limit — the identity for $L(te^{\lambda t})$ checks the solution outright — and problem P3 produces $te^{\lambda t}$ a second way, by reduction of order, again with no limit taken.

06 · the free equation

Dead-beat, measured: the 1% band

$|\theta/\theta_0|$ on a log scale, released from rest. Each dip to the bottom is a zero crossing — an overshoot. "Settled" means inside the shaded band for good; the dotted lines mark when each curve enters it for the last time.

The number: $0.83$ s to settle at critical damping, against $5.71$ s shorted, $1.78$ s at $\gamma = 5$ and $22.8$ s at $R = 995\ \Omega$.

fastest depends on the criterion

Critical damping is fastest without crossing zero. Allow one overshoot of exactly $1\%$ and $\gamma = 13.2$ does better: it dips through zero once, never leaves the band again, and is settled at $0.52$ s — $37\%$ sooner.

The log scale is what makes the curves comparable: the critical curve is the straightest dive, and the slow root of $\gamma = 80$ is the shallow line still far above the band.

Reading the figure. On a log scale an exponential $e^{-\mu t}$ is a straight line of slope $-\mu/\ln 10$, so a curve's steepness is its decay rate. The ringing curves have notches because $|\theta|$ passes through zero; between notches their tops follow the envelope, a straight line. The critical curve $(1+8t)e^{-8t}$ is the steepest curve with no notch at all, which is the extremal claim of problem P4 drawn.

A different criterion gives a different optimum. "Fastest without overshoot" is not the only thing one might want. If a small overshoot is tolerable and the figure of merit is the time to enter and stay within $1\%$, the best damping is the one whose single overshoot is exactly $1\%$: $\exp(-\pi\gamma/2\omega_\mathrm{d}) = 0.01$, that is $\gamma/2\omega_0 = \ln 100/\sqrt{\pi^2+(\ln 100)^2} = 0.826$, or $\gamma = 13.22$ here. It settles in $0.524$ s against critical damping's $0.830$ s — $37\%$ faster. Meters are in fact shipped slightly underdamped for exactly this reason. State the criterion before claiming an optimum.

06 · the free equation

$Q$: one number for how good the oscillator is

the quality factor $$Q = \frac{\omega_0}{\gamma}.$$

Underdamped means $Q > \tfrac12$; critical is $Q = \tfrac12$.

four readings, two exact

exact: the resonant gain over the static deflection is $Q$ (worked example 2);
exact: $\omega_0/(\text{half-power width}) = Q$ (the power slide);
to $O(Q^{-1})$: energy decays by $e^{-1}$ in $Q$ radians of oscillation;
to $O(Q^{-1})$: $Q = 2\pi\times$ (energy stored)/(energy lost per cycle).

the last one, checked honestly

One period later the whole state is exactly $e^{-\gamma T_\mathrm{d}/2}$ times what it was, so the energy is $e^{-\gamma T_\mathrm{d}}$ times, and the fraction lost per cycle is exactly $1 - e^{-\gamma T_\mathrm{d}}$. At $Q = 20$ that is $\mathbf{0.2697}$, while $2\pi/Q = 0.3142$. The approximation is $16.5\%$ out at $Q = 20$ — an error of order $1/Q$, exactly as advertised, and a reminder to say which definition is meant.

scale

Car suspension $Q \approx 1$. Our galvanometer, open circuit, $Q \to \infty$; at $R = 995\ \Omega$, $Q = 20$. A guitar string $\approx 10^3$; a quartz watch crystal $\approx 10^5$; LIGO's test-mass suspension $\approx 10^8$. Each factor of ten is an engineering era.

Where "$Q$ radians" comes from. Energy $\propto e^{-\gamma t}$, so the $e$-folding time is $1/\gamma$; in that time the phase advances by $\omega_\mathrm{d}/\gamma \approx \omega_0/\gamma = Q$ radians, or $Q/2\pi$ cycles. This is the reading to carry: $Q$ counts radians of oscillation per $e$-fold of energy. It makes $Q = 10^8$ legible — LIGO's pendulum rings for about $10^8/2\pi \approx 1.6\times10^7$ cycles, which at about $1$ Hz is half a year.

Why two definitions disagree. "Energy lost per cycle" compares a continuous exponential decay with a discrete count, so it is exact only in the limit of small loss: $1 - e^{-\gamma T_\mathrm{d}} = \gamma T_\mathrm{d} + O((\gamma T_\mathrm{d})^2)$ and $\gamma T_\mathrm{d} = 2\pi\gamma/\omega_\mathrm{d} \to 2\pi/Q$. The two exact readings are both about the driven response, and they are the ones to use when a number has to be right; both are proved in section 3.

The name. It came from telephone engineering early in the twentieth century — K. S. Johnson of Western Electric, rating the coils of tuned circuits, as E. I. Green recounts in "The story of $Q$" (1955). By that account the letter was chosen simply because it was free, and "quality factor" is a later reading of it. It has since escaped into every branch of physics, which is itself evidence for this unit's thesis: one equation, many subjects.

section 3

Driving it: resonance, and the phase that sweeps

One drive, solved by an ansatz that works because $\mathbb{C}$ makes it work. Then: where the peak really is, and how wide.

06 · driving

Transient plus steady state

Take $Lx = F_0\cos\omega t$ with $\gamma > 0$. The solution set is $x_p + \ker L$, and every element of $\ker L$ decays.

transient and steady state

The $\ker L$ part is the transient: it carries the initial conditions and dies. What survives is the steady state, the same for every initial condition — so after a few decay times the oscillator has forgotten how it started.

For the tuned galvanometer, "a few decay times" is $23$ s: the transient is not a technicality, it is most of a minute of watching.

The tuned galvanometer driven at $\omega_0$, starting from rest. The swing is steady state minus a decaying transient; the transient's share dies like $e^{-\gamma t/2}$, so the envelope (dashed) climbs to the steady amplitude and is within $1\%$ of it after $23$ s.

What to look for in the figure. Exactly at resonance the steady state and the transient oscillate at almost the same frequency ($8$ against $\omega_\mathrm{d} = 7.9975$), and starting from rest they begin in exact opposition, so the needle starts at zero and the envelope is $0.0625\,(1 - e^{-\gamma t/2})$ to within a fraction of a per cent. Off resonance the two frequencies differ and the build-up shows beats instead — the same decomposition, drawn less tidily.

"Forgets how it started" is a physical statement with teeth. It is why a resonance curve is a property of the system rather than of the experiment: whatever you do to a driven oscillator, after a while it does the same thing. It is also why $\gamma = 0$ is pathological — with no damping the transient never dies, the initial conditions are remembered for ever, and there is no steady state at all. The undamped-resonance slide later in this section is about exactly that case.

On the word "particular". $x_p$ is not distinguished mathematically: any element of the coset would do, and they differ by transients. What picks out the steady state is a physical condition — boundedness as $t\to\pm\infty$ — and with $\gamma > 0$ there is exactly one solution of $Lx = F_0\cos\omega t$ bounded on all of $\mathbb{R}$, namely the sinusoid found on the steady-state slide. Proof: two such differ by a bounded element of $\ker L$, and the only bounded element is $0$ since both roots have strictly negative real part and hence every non-zero solution blows up as $t\to-\infty$.

06 · driving

The complex-drive trick

The steady state still has to be found. The fast way is to solve a complex equation and take its real part — legitimate for exactly one reason.

the complex-drive trick

Suppose $z : \mathbb{R}\to\mathbb{C}$ satisfies $Lz = F_0e^{i\omega t}$. Then $x = \operatorname{Re}z$ satisfies $Lx = F_0\cos\omega t$.

proof

$L$ has real coefficients and $\operatorname{Re}$ is $\mathbb{R}$-linear and commutes with $\mathrm{d}/\mathrm{d}t$ (both act componentwise), so $L(\operatorname{Re}z) = \operatorname{Re}(Lz) = \operatorname{Re}\big(F_0e^{i\omega t}\big) = F_0\cos\omega t$. $\square$

the trick has a boundary

It needs $L$ linear with real coefficients. On anything nonlinear it fails at once: $\operatorname{Re}(z^2) \ne (\operatorname{Re}z)^2$. Energies and powers must be computed from the real parts, or by the $\tfrac12\operatorname{Re}(z\bar w)$ rule of the power slide.

Drive and steady response as two arrows in $\mathbb{C}$, turning together at angular speed $\omega$, the response trailing by a fixed angle $\varphi$. What the instrument does is the shadow on the real axis: $\operatorname{Re}$ of each arrow. One instant, with $\varphi = 60^\circ$; the two lengths are in different units, so only their ratio $|\chi|$ means anything.

What to look for in the figure. Both arrows rotate rigidly, so their angle apart and their length ratio never change: that pair of numbers, $\varphi$ and $|\chi|$, is all the steady state contains, and it is what the next slide computes. The physical drive and the physical response are the two dashed shadows, which slide back and forth along the real axis as the arrows turn — two sinusoids, the second lagging the first by $\varphi$.

Why bother. The drive $\cos\omega t$ is an awkward input to $L$: its derivatives cycle between $\cos$ and $\sin$, and matching coefficients takes a $2\times2$ system. The drive $e^{i\omega t}$ is an eigenvector of $\mathrm{d}/\mathrm{d}t$, so $L$ acts on it by multiplication by a single number, $p(i\omega)$, and the equation becomes a division. The real part is taken once, at the end.

Where the trick came from. Steinmetz introduced complex "phasors" for a.c. circuit analysis in 1893 and turned a subject of trigonometric identities into one of algebra; the same manoeuvre is behind every impedance calculation. Its mathematical content is the one line proved above — nothing deeper — but it halves the length of every computation in this section.

06 · driving

The steady state, in one line

the frequency response

Try $z = A e^{i\omega t}$ in $Lz = F_0e^{i\omega t}$. Then $Lz = p(i\omega)Ae^{i\omega t}$, so $A = F_0\chi(\omega)$ with

$$\chi(\omega) = \frac{1}{p(i\omega)} = \frac{1}{\omega_0^2 - \omega^2 + i\gamma\omega}.$$

$p(i\omega) \ne 0$ whenever $\gamma > 0$, so this always works. Writing $\chi = |\chi|e^{-i\varphi}$,

$$\boxed{\ x_p(t) = \frac{F_0\cos(\omega t - \varphi)}{\sqrt{(\omega_0^2-\omega^2)^2 + \gamma^2\omega^2}}, \qquad \varphi = \arg\!\big(\omega_0^2-\omega^2 + i\gamma\omega\big) \in (0,\pi).\ }$$

dimensions

$[\chi] = \mathrm{T^2}$ and $[F_0] = [x]\mathrm{T^{-2}}$, so the amplitude has the dimensions of $x$. ✓

limit $\omega \to 0$

$A \to F_0/\omega_0^2$, $\varphi \to 0$: the static deflection, in phase. For the galvanometer, $\alpha i/\kappa$. ✓

limit $\omega \to \infty$

$|A| \to F_0/\omega^2$, $\varphi \to \pi$: the mass cannot keep up and moves exactly against the drive. ✓

Why $\varphi \in (0,\pi)$ and not $(-\pi,\pi]$. The imaginary part of $p(i\omega)$ is $\gamma\omega$, strictly positive for $\omega > 0$ and $\gamma > 0$; a complex number with positive imaginary part has argument strictly between $0$ and $\pi$. So the response always lags, never leads — by an amount that sweeps the whole interval as $\omega$ runs over $(0,\infty)$. The physical reason the lag is never negative is dissipation: the power slide shows the drive's mean power into the oscillator is $\tfrac12F_0^2\omega|\chi|\sin\varphi$, and it must equal the mean power dissipated, $\gamma\langle\dot x^2\rangle \ge 0$, so $\sin\varphi > 0$. A lead at one frequency would mean the oscillator feeding energy back into the drive.

The computation, spelled out. $\frac{\mathrm{d}^2}{\mathrm{d}t^2}Ae^{i\omega t} = -\omega^2Ae^{i\omega t}$ and $\frac{\mathrm{d}}{\mathrm{d}t}Ae^{i\omega t} = i\omega Ae^{i\omega t}$, so $Lz = (-\omega^2 + i\gamma\omega + \omega_0^2)Ae^{i\omega t}$. Setting that equal to $F_0e^{i\omega t}$ and cancelling the never-zero exponential gives $A$ immediately. Compare the alternative: substitute $a\cos\omega t + b\sin\omega t$, expand, match two coefficients, solve a $2\times2$ system, then convert $(a,b)$ to amplitude and phase. Same answer, four times the algebra. That is what $\mathbb{C}$ is for here.

The modulus and argument, from $\chi = 1/w$: $|1/w| = 1/|w|$ and $\arg(1/w) = -\arg w$. With $w = \omega_0^2-\omega^2+i\gamma\omega$ this gives the boxed formulas directly, and it is why $\varphi$ is defined as the argument of $w$ rather than of $\chi$ — so that it is a lag with a positive sign.

Against unit 02. The steady state is a closed orbit in phase space — an ellipse traversed once per drive period — and every other orbit spirals onto it. In unit 02's vocabulary it is an attracting periodic orbit of the (non-autonomous, hence three-dimensional) system. Linear systems do not have limit cycles, as unit 02 noted; this is not one, because the drive is external.

06 · driving

$\chi$ in the plane: the lag is an angle, the gain a distance

The point $p(i\omega) = (\omega_0^2 - \omega^2) + i\gamma\omega$ as $\omega$ runs from $0$ up, for $\gamma = 5$ ($R = 75\ \Omega$). Its angle from the positive real axis is the lag $\varphi$; its distance from $0$ is $1/|\chi|$. The dashed circle touches the curve at its nearest point.

Since $\chi = 1/p(i\omega)$, everything about the steady state is the position of one point in $\mathbb{C}$. As $\omega$ grows it traces a parabola, $\operatorname{Re} = \omega_0^2 - (\operatorname{Im}/\gamma)^2$:

it stays in the upper half-plane, so $\varphi \in (0,\pi)$: the response always lags;
it crosses the imaginary axis at $\omega = \omega_0$, so $\varphi = \pi/2$ there, whatever $\gamma$;
its nearest point to $0$ is the largest $|\chi|$ — the amplitude peak, at $\omega = 7.18$, not at $\omega_0$;
as $\omega \to \infty$ it runs off to the left: $\varphi \to \pi$, $|\chi| \to 0$.

Every result of this section, in one picture. The phase-lag slide's crossing at $\pi/2$ is the parabola meeting the imaginary axis at height $\gamma\omega_0$. The response-curve slide's peak is the point of closest approach: minimising $|p(i\omega)|^2 = (\omega_0^2-u)^2 + \gamma^2u$ over $u = \omega^2$ is the computation done there, and with $\gamma = 5$ it gives $u = 64 - 12.5 = 51.5$, $\omega = 7.176$, at the point $12.5 + 35.88\,i$, a distance $\gamma\omega_\mathrm{d} = 37.99$ from the origin. And the parabola never touches $0$ when $\gamma > 0$, which is the slide's "$p(i\omega) \ne 0$, so this always works".

Why $\gamma = 5$ and not the galvanometer's $0.4$. At $\gamma = 0.4$ the same parabola is so narrow — it reaches only $3.2\,i$ at $\omega_0$ while starting at $64$ — that it hugs the real axis and the nearest point sits almost on the imaginary axis. That is exactly the high-$Q$ statement that the peak is "at" $\omega_0$, but it makes a poor drawing. The geometry is the same for every $\gamma$.

With $\gamma = 0$ the curve collapses onto the real axis and passes through $0$ at $\omega_0$: the point the ansatz cannot divide by, which is the undamped-resonance slide.

06 · driving · example

Worked: the tuned galvanometer, in numbers

worked example 2 — a null detector at resonance

$R = 995\ \Omega$, so $\gamma = 0.4$ and $Q = 20$; drive $i(t) = i_0\cos\omega t$ with $i_0 = 1\ \mu\mathrm{A}$. The equation is $\ddot\theta + 0.4\dot\theta + 64\,\theta = F_0\cos\omega t$ with $F_0 = \alpha i_0/I = (2\times10^{-3})(10^{-6})/10^{-8} = 0.2\ \mathrm{rad\,s^{-2}}$.

Step 1 — at $\omega = \omega_0 = 8$. $\omega_0^2-\omega^2 = 0$, so $\chi = 1/(i\gamma\omega_0)$, $|\chi| = 1/(0.4\times8) = 0.3125\ \mathrm{s^2}$ and $\varphi = \pi/2$.

$$\theta_{\max} = F_0|\chi| = 0.0625\ \mathrm{rad} = 3.581^\circ .$$

Step 2 — compare with steady current. The same $1\ \mu\mathrm{A}$ held constant gives $\alpha i_0/\kappa = 3.125\times10^{-3}$ rad. Ratio: $0.0625/0.003125 = \mathbf{20} = Q$, exactly.

Step 3 — off resonance. At $\omega = 7.9$: $\omega_0^2-\omega^2 = 1.59$, $\gamma\omega = 3.16$, so $|\chi| = 1/\sqrt{1.59^2+3.16^2} = 0.28269$ and $\varphi = \arctan(3.16/1.59) = 1.1046$ rad $= 63.29^\circ$. The amplitude has already fallen $9.5\%$ for a $1.25\%$ change in frequency.

Step 4 — the check. Substituting $\theta = 0.0625\cos(8t-\pi/2) = 0.0625\sin 8t$: $\ddot\theta = -4\sin 8t$, $0.4\dot\theta = 0.2\cos 8t$, $64\theta = 4\sin 8t$. Sum: $0.2\cos 8t = F_0\cos\omega t$. ✓

What the instrument is for. A resistance bridge fed with alternating current is balanced by adjusting until the detector reads zero. A galvanometer tuned to the supply frequency gives $Q$ times the deflection an untuned one would, so it detects an imbalance $Q$ times smaller: here a microamp becomes a readable $3.6^\circ$ instead of an invisible $0.18^\circ$. Real vibration galvanometers are built with a stiffer suspension so that $\omega_0$ matches a $50$ Hz supply; the principle and the factor $Q$ are identical.

Why the gain is exactly $Q$. $|\chi(\omega_0)|/|\chi(0)| = \big(1/\gamma\omega_0\big)\big/\big(1/\omega_0^2\big) = \omega_0/\gamma = Q$. No approximation anywhere — this is the first of the two exact readings of $Q$ promised when $Q$ was defined. Note that it holds for every $\gamma$, including overdamped ones where "resonance" is a misnomer: at $Q = 0.5$ the "gain" is $0.5$, that is, a loss.

The price of the gain. A high-$Q$ detector is slow: worked example 1 gave $22.8$ s to settle at $\gamma = 0.4$, and the build-up of the steady state takes the same time, since the transient that is dying is the same transient. Gain and speed trade against each other through the one parameter $\gamma$, and that trade is the whole of instrument design. The other half of the bargain is a bonus: the same tuning rejects contamination at the third harmonic by $|\chi(\omega_0)|/|\chi(3\omega_0)| = 0.31250/0.0019528 = 160.0$, so a null detector at resonance is also a filter.

06 · driving

The response curve, and where the peak really is

Gain over the static deflection, $|\chi|\,\omega_0^2$. The peaks of $\gamma = 2$ and $\gamma = 5$ (dots) sit below $\omega_0$ (dashed), the second visibly; $\gamma = 0.4$ runs off the top to $20$; $\gamma = 16$, with $Q < 1/\sqrt2$, has no peak at all.

resonance, located

$|\chi|$ is maximal where $D(\omega) = (\omega_0^2-\omega^2)^2+\gamma^2\omega^2$ is minimal. With $u = \omega^2$, $\;\mathrm{d}D/\mathrm{d}u = -2(\omega_0^2-u)+\gamma^2 = 0$:

$$\omega_{\text{peak}} = \sqrt{\omega_0^2 - \tfrac12\gamma^2},$$ $$|\chi|_{\max} = 1/\gamma\omega_\mathrm{d}.$$

A peak exists only if $\gamma^2 < 2\omega_0^2$, i.e. $Q > 1/\sqrt2$; otherwise $|\chi|$ falls monotonically from $\omega = 0$.

three different frequencies

$\omega_0 > \omega_\mathrm{d} = \sqrt{\omega_0^2-\gamma^2/4} > \omega_{\text{peak}} = \sqrt{\omega_0^2-\gamma^2/2}$. At $Q = 20$: $8$, $7.99750$, $7.99500$. At $Q = 1.6$: $8$, $7.5993$, $7.1764$. They are not interchangeable.

The maximum value, derived. At $u = \omega_0^2-\gamma^2/2$, $D = (\gamma^2/2)^2 + \gamma^2(\omega_0^2-\gamma^2/2) = \gamma^2\omega_0^2 - \gamma^4/4 = \gamma^2(\omega_0^2-\gamma^2/4) = \gamma^2\omega_\mathrm{d}^2$, so $|\chi|_{\max} = 1/\gamma\omega_\mathrm{d}$. At $Q = 20$ that is $0.31260\ \mathrm{s^2}$ against $0.31250$ at $\omega_0$ — the peak is $0.031\%$ higher than the value at $\omega_0$ and sits $0.063\%$ below it in frequency. For a light damping the distinction is invisible; for a car suspension at $Q \approx 1$ it is not.

A convention trap that costs people hours. Many books write the equation as $\ddot x + 2\beta\dot x + \omega_0^2x = f$, so that $\beta = \gamma/2$. Their peak is at $\sqrt{\omega_0^2-2\beta^2}$, their $\omega_\mathrm{d}$ is $\sqrt{\omega_0^2-\beta^2}$, their $Q$ is $\omega_0/2\beta$ and their decay envelope is $e^{-\beta t}$. Others use $\zeta = \gamma/2\omega_0$, the damping ratio, so critical is $\zeta = 1$ and $Q = 1/2\zeta$. Every formula in this deck is in the $\gamma$ convention; when reading elsewhere, find the envelope's exponent first and calibrate everything from it.

Why the peak moves down and not up. Damping removes energy fastest where the velocity is largest, and the velocity amplitude is $\omega|\chi|$; the extra factor of $\omega$ penalises the high-frequency side of the curve more than the low, so the maximum of $|\chi|$ shifts down. The velocity's own maximum, with that factor restored, sits exactly at $\omega_0$ — which is the next slide, and the reason a physicist says "resonance is at $\omega_0$" without being wrong.

06 · driving

Velocity, power, and a width that is exactly $\gamma$

the power curve

The mean power the drive delivers, per unit mass, is

$$\bar P = \tfrac12\gamma\,\omega^2|\chi|^2F_0^2 = \frac{\tfrac12\gamma F_0^2}{g(\omega)^2 + \gamma^2},$$

where $g(\omega) = \dfrac{\omega_0^2}{\omega}-\omega$.

It is maximal exactly at $\omega = \omega_0$, the only positive zero of $g$ — for every $\gamma$, with no smallness assumption.

half-power width

$\bar P = \tfrac12\bar P_{\max}$ where $g^2 = \gamma^2$, i.e. $\omega_\mp = \mp\tfrac\gamma2 + \sqrt{\omega_0^2+\gamma^2/4}$. Hence

$$\Delta\omega = \omega_+ - \omega_- = \gamma \ \text{exactly},\quad \frac{\omega_0}{\Delta\omega} = Q .$$

square the real part, not the complex one

$\bar P = \langle \gamma\dot x\cdot \dot x\rangle$ must be computed from the real $\dot x$. The safe rule: for $z = Ae^{i\omega t}$, $w = Be^{i\omega t}$, $\langle\operatorname{Re}z\operatorname{Re}w\rangle = \tfrac12\operatorname{Re}(A\bar B)$. With $A = B = i\omega F_0\chi$ that gives the formula above.

Full derivation of the averaging rule. Write $A = |A|e^{i a}$, $B = |B|e^{i b}$. Then $\operatorname{Re}z = |A|\cos(\omega t+a)$ and $\operatorname{Re}w = |B|\cos(\omega t+b)$; the product is $\tfrac12|A||B|[\cos(a-b) + \cos(2\omega t + a + b)]$, and the second term averages to zero over a period. So the mean is $\tfrac12|A||B|\cos(a-b) = \tfrac12\operatorname{Re}(A\bar B)$. Taking $z = w$ gives the familiar $\langle(\operatorname{Re}z)^2\rangle = \tfrac12|A|^2$.

Why the power, and not the amplitude, is the physicist's resonance curve. Because it is what an experiment measures: a spectral line's width, an absorption peak, a circuit's bandwidth, the energy a shaken structure takes up. Its centre is exactly $\omega_0$ and its width is exactly $\gamma$, with no small-$\gamma$ caveat, so those two numbers can be read off a measured curve and turned directly into the system's constants. The amplitude curve's peak is at a slightly different place, which is a fact about the amplitude, not a correction to the physics.

Energy balance, as a check. In the steady state the mean power in must equal the mean power dissipated, since the stored energy is periodic. The formula above is the dissipation $\langle\gamma\dot x^2\rangle$, and computing $\langle f\dot x\rangle$ instead gives $\tfrac12 F_0\omega|\chi|F_0\sin\varphi$, which is the same thing because $\sin\varphi = \gamma\omega|\chi|$. Worth doing once.

06 · driving

The half-power width, on the galvanometer

Mean absorbed power, each curve scaled to its own maximum. Both peak exactly at $\omega_0 = 8$; at half height (dashed) each is exactly $\gamma$ wide — $0.4$ for the tuned galvanometer, $2$ for $R = 195\ \Omega$. Five times the damping, five times the width.

the galvanometer, $Q = 20$

$\omega_\pm = \mp0.2 + \sqrt{64.04} = 7.8025$ and $8.2025\ \mathrm{rad\,s^{-1}}$; difference $0.4000 = \gamma$ ✓. In frequency, a width of $0.0637$ Hz about $1.2732$ Hz.

And the phase at those two points: $\tan\varphi = \gamma\omega/(\omega_0^2-\omega^2) = \pm1$, so $\varphi = \pi/4$ and $3\pi/4$. The half-power points are the quarter-turn points of the phase.

What to look for in the figure. Neither curve is symmetric about $\omega_0$ on a linear frequency axis — the broader one visibly leans to the right, since its half-power points $\mp\gamma/2 + \sqrt{\omega_0^2+\gamma^2/4}$ are $7.06$ and $9.06$, not $7$ and $9$ — and yet the width between them is exactly $\gamma$ in both cases. Symmetry holds in the variable $g(\omega) = \omega_0^2/\omega - \omega$, not in $\omega$ itself.

The shape near the peak. For $\gamma \ll \omega_0$ and $\omega$ near $\omega_0$, write $\omega_0^2-\omega^2 = (\omega_0-\omega)(\omega_0+\omega) \approx 2\omega_0(\omega_0-\omega)$; then $\bar P \propto \big[4(\omega-\omega_0)^2 + \gamma^2\big]^{-1}$, a Lorentzian of full width $\gamma$ at half maximum. That is the line shape of a damped radiating atom, and unit 20 will show it is the Fourier transform of the exponentially decaying ring-down — the width in frequency and the decay rate in time being the same number seen twice.

06 · driving

The phase lag, and what it is good for

The lag for four dampings. Every curve crosses $\pi/2$ at $\omega_0$ (dot), whatever the damping; the lighter the damping, the steeper the crossing — its slope there is $2/\gamma$.

$\varphi(\omega_0) = \pi/2$ always

$p(i\omega_0) = \omega_0^2-\omega_0^2+i\gamma\omega_0 = i\gamma\omega_0$, purely imaginary with positive imaginary part. So the crossing is exact and $\gamma$-independent.

Three regions, and each has a mechanical meaning:

$\omega \ll \omega_0$: $\varphi \approx 0$, the drive fights the spring, and $x \approx f/\omega_0^2$;
$\omega = \omega_0$: $\varphi = \pi/2$, drive exactly in phase with velocity, so it does maximal work — this is why the power peaks here;
$\omega \gg \omega_0$: $\varphi \approx \pi$, the drive fights inertia, and $x \approx -f/\omega^2$.

Measuring $\varphi$ locates $\omega_0$ more reliably than measuring the amplitude peak, because the crossing is exact and the curve is steepest there.

The steepness, computed. Near $\omega_0$, $\cot\varphi = (\omega_0^2-\omega^2)/\gamma\omega \approx -2(\omega-\omega_0)/\gamma$, so $\mathrm{d}\varphi/\mathrm{d}\omega|_{\omega_0} = 2/\gamma = 2Q/\omega_0$. At $Q = 20$ the phase sweeps through $\pi/2$ at $5\ \mathrm{rad}$ per $\mathrm{rad\,s^{-1}}$ — a $1\%$ frequency error shows up as a $21.7^\circ$ phase error, which is why phase-sensitive detection is the standard way to lock an oscillator to a resonance. The whole sweep from $\pi/4$ to $3\pi/4$ takes place in a band of width exactly $\gamma$, as the previous slide's coincidence showed.

The sign of the lag is not a convention. With the convention $f = F_0\cos\omega t$ and response $\cos(\omega t - \varphi)$, $\varphi > 0$ means the response happens later. That it is positive for every $\omega$ and every $\gamma > 0$ is forced by $\operatorname{Im}p(i\omega) = \gamma\omega > 0$, which is the energy balance of the power slide: the drive does positive mean work. It is not by itself a statement about causality — a steady sinusoid has no beginning, so a phase at one frequency cannot say whether a response preceded its cause. Causality is a constraint on the whole function $\chi$, and section 4's slide on causality says what kind.

Where this is used in this course. Unit 21's driven pendulum, where the phase lag is the slow variable of the averaged equations; unit 24, where the phase of a radiating dipole relative to the driving field decides whether the medium's index of refraction is above or below one; unit 37, where a relativistic precession is read as a slow phase drift of exactly this kind.

06 · driving

No damping, driven at $\omega_0$: the ansatz fails

Set $\gamma = 0$ and $\omega = \omega_0$. Then $p(i\omega_0) = 0$, so $A = F_0/p(i\omega_0)$ is undefined — and no $Ae^{i\omega_0t}$ can work, because $e^{i\omega_0t} \in \ker L$ and $L$ sends it to $0$, never to $F_0e^{i\omega_0t}$.

the secular term

Use the identity $L(te^{\lambda t}) = p(\lambda)te^{\lambda t} + p'(\lambda)e^{\lambda t}$ at $\lambda = i\omega_0$, where $p = 0$ and $p' = 2i\omega_0$:

$$z = \frac{F_0}{2i\omega_0}\,t\,e^{i\omega_0t}\ \Longrightarrow\ Lz = F_0e^{i\omega_0t},$$ $$x_p = \operatorname{Re}z = \frac{F_0}{2\omega_0}\,t\,\sin\omega_0 t .$$

check, by substitution

$\ddot x_p = \frac{F_0}{2\omega_0}\big(2\omega_0\cos\omega_0t - \omega_0^2 t\sin\omega_0t\big)$, so $\ddot x_p + \omega_0^2x_p = F_0\cos\omega_0 t$. ✓ Dimensions: $[F_0 t/\omega_0] = [x]$. ✓

what "infinite response" really means

The amplitude grows linearly, not instantly, at rate $F_0/2\omega_0$. Undamped resonance is a statement about long times, and long before them the amplitude leaves the region where section 1's Taylor expansion was valid. Real systems do not blow up; they stop being harmonic oscillators.

Why the failure is structural, not accidental. $Lz = F_0e^{i\omega t}$ asks for a preimage of $e^{i\omega t}$ under $L$; the ansatz searches only the one-dimensional space $\mathbb{C}\cdot e^{i\omega t}$, which $L$ maps to itself by multiplication by $p(i\omega)$. That map is onto exactly when $p(i\omega) \ne 0$ and is the zero map when $p(i\omega) = 0$. So the ansatz fails precisely when the drive is already in the kernel — for every linear problem, not just this one. The same phenomenon reappears whenever an inhomogeneous term is in the kernel of the operator: unit 21's secular terms in perturbation series are this, and the Poincaré–Lindstedt method is the standard cure.

Producing the answer as a limit. With $\gamma = 0$ and $\omega \ne \omega_0$ the steady state is $F_0\cos\omega t/(\omega_0^2-\omega^2)$; adding the free solution $-F_0\cos\omega_0t/(\omega_0^2-\omega^2)$ (legitimate, it is in $\ker L$) gives $F_0(\cos\omega t-\cos\omega_0t)/(\omega_0^2-\omega^2)$, which starts from rest, and whose limit as $\omega\to\omega_0$ is $F_0 t\sin\omega_0t/2\omega_0$ by l'Hôpital in the variable $\omega$. At finite detuning that expression is a beat: two frequencies whose difference is slow, so the slow factor $\sin\big(\tfrac{\omega_0-\omega}{2}t\big)$ has period $4\pi/|\omega_0-\omega|$ and the amplitude swells and collapses every $2\pi/|\omega_0-\omega|$. Resonance is that time going to infinity.

The bridge, for the record. The story that soldiers marching in step collapsed the Broughton suspension bridge in 1831 is true, and the order to break step on a bridge dates from it; the Millennium Bridge in 2000 was a different mechanism — lateral lock-in, a feedback in which the pedestrians' response to the motion drives it, which is a nonlinear instability rather than an external resonance, and is unit 02's Hopf bifurcation, not this slide.

06 · driving

Secular growth meets the damped ceiling

The galvanometer driven exactly at $\omega_0$ from rest, with $F_0 = 0.2\ \mathrm{rad\,s^{-2}}$. Undamped, the swing grows between the straight lines $\pm F_0t/2\omega_0$ for ever. With $\gamma = 0.4$ the envelope leaves along the same lines and bends over to the ceiling $F_0/\gamma\omega_0 = 0.0625$ rad.

the two limits agree

With small $\gamma$ the resonant amplitude is $F_0/\gamma\omega_0$, reached after a time $\sim 2/\gamma$. The undamped formula grows to $F_0 t/2\omega_0$, which reaches the same value at $t = 2/\gamma$. The damped steady state and the secular growth are the same physics on either side of the time $2/\gamma$.

The figure is that box drawn: the straight line is the damped envelope's tangent at $t = 0$, and it reaches the ceiling at $t = 2/\gamma = 5$ s. Damping changes nothing at first; it decides where the growth stops.

06 · driving

Variation of parameters: any drive at all

variation of parameters

Let $\{x_1,x_2\}$ be a basis of $\ker L$ with Wronskian $W$, and $f$ continuous. Put

$$u_1(t) = -\int_{t_0}^{t}\frac{x_2(s)f(s)}{W(s)}\,\mathrm{d}s,\qquad u_2(t) = \ \ \int_{t_0}^{t}\frac{x_1(s)f(s)}{W(s)}\,\mathrm{d}s .$$

Then $x_p = u_1x_1 + u_2x_2$ satisfies $Lx_p = f$ with $x_p(t_0) = \dot x_p(t_0) = 0$; equivalently

$$\boxed{\ x_p(t) = \int_{t_0}^{t}\! K(t,s)f(s)\,\mathrm{d}s,\ \ K(t,s) = \frac{x_1(s)x_2(t) - x_2(s)x_1(t)}{W(s)}.\ }$$

proof

By the fundamental theorem of calculus $\dot u_1x_1 + \dot u_2x_2 = \frac{-x_2fx_1 + x_1fx_2}{W} = 0$, so $\dot x_p = u_1\dot x_1 + u_2\dot x_2$ with no $\dot u$ terms. Differentiating again, $\ddot x_p = \dot u_1\dot x_1 + \dot u_2\dot x_2 + u_1\ddot x_1 + u_2\ddot x_2$. Since $Lx_i = 0$, $\;Lx_p = \dot u_1\dot x_1+\dot u_2\dot x_2 = \frac{f}{W}(x_1\dot x_2 - x_2\dot x_1) = f$. At $t = t_0$ both integrals vanish. $\square$

what has been achieved

A closed form for every continuous drive, with no ansatz. Dimensions: $[K] = [x]^2/[W] = \mathrm{T}$, so $\int Kf\,\mathrm{d}s$ has the dimensions of $x$. ✓

And now look at $K$: for constant coefficients it depends only on $t - s$. That is the whole of section 4.

Where the constraint $\dot u_1x_1 + \dot u_2x_2 = 0$ comes from. It is imposed, not derived. Two unknown functions are being used to satisfy one equation, so one condition is free; choosing this one kills the second derivatives of the $u_i$ before they appear and reduces everything to a $2\times2$ linear system for $(\dot u_1,\dot u_2)$ whose matrix is exactly the Wronskian matrix — invertible precisely because $\{x_1,x_2\}$ is a basis. Lagrange introduced the method in 1808, for planetary perturbations: the "parameters" being varied were the orbital elements, held constant in the unperturbed problem. Unit 10 uses it in that original sense.

The general solution, assembled. Adding any element of $\ker L$ to $x_p$ and matching $(x(t_0),\dot x(t_0)) = (x_0,v_0)$ gives the unique solution of the initial value problem. Together with $\dim\ker L = 2$ this is a complete solution theory for the equation: existence, uniqueness and a formula. Very few equations in this course get all three.

Why continuity of $f$ is enough. The integrands are continuous, so the $u_i$ are $C^1$ by the fundamental theorem; then $x_p$ is $C^1$, and the displayed computation shows $\ddot x_p$ exists and is continuous, so $x_p$ is $C^2$ and is a classical solution. A discontinuous $f$ — a switch closing, as in problem P7 — is handled by applying the theorem on each interval of continuity and matching $x$ and $\dot x$ at the join; the result is $C^1$ but not $C^2$ there, and unit 20 will say what object it solves the equation in the sense of.

section 4

One response to everything

Hit it once, record what it does, and you have its answer to every drive there will ever be. The claim is exact, and it is a theorem.

06 · the impulse response

An impulse is a jump in the velocity

Let a drive act only on $[0,\varepsilon]$ with $\int_0^\varepsilon f = 1$, starting from rest. Integrate $Lx = f$ across the pulse:

$$\dot x(\varepsilon) - \dot x(0) + \gamma\big(x(\varepsilon)-x(0)\big) + \omega_0^2\!\!\int_0^\varepsilon\!\! x = 1 .$$

$x$ is continuous and $x(0) = 0$, so the last two terms are $O(\varepsilon)$ if $\dot x$ stays bounded — which it does, by $|\dot x| \le \int|f| $ plus $O(\varepsilon)$. Hence as $\varepsilon \to 0$:

the limit

$x(0^+) = 0$ and $\dot x(0^+) = 1$. A unit impulse leaves the position alone and raises the velocity by one. Everything about the pulse's shape is forgotten.

the galvanometer, kicked

A charge $q$ through the coil in a time $\ll T_0$ delivers angular impulse $\alpha q$, so

$$\dot\theta(0^+) = \frac{\alpha q}{I} = \big(2\times10^5\ \mathrm{rad\,s^{-1}\,C^{-1}}\big)\,q .$$

For $q = 1\ \mu\mathrm{C}$: $\dot\theta(0^+) = 0.2\ \mathrm{rad\,s^{-1}}$, and $\theta(0^+) = 0$ — the needle has not moved yet, but it is on its way.

"$\ll$" needs a number

Short compared with $T_0 = 0.785$ s. A capacitor discharge with time constant $10$ ms qualifies: problem P9 computes the error, and it is $0.32\%$.

The estimate, in full. On $[0,\varepsilon]$ write the equation as $\ddot x = f - \gamma\dot x - \omega_0^2x$. Suppose $|\dot x| \le M$ on the interval; then $|x| \le M\varepsilon$, and integrating gives $|\dot x(t)| \le \int_0^t f + \gamma M\varepsilon + \omega_0^2M\varepsilon^2 \le 1 + \gamma M\varepsilon + \omega_0^2M\varepsilon^2$. For $\varepsilon$ small enough that $\gamma\varepsilon + \omega_0^2\varepsilon^2 \le \tfrac12$, this forces $M \le 2$, so $\dot x$ is bounded independently of $\varepsilon$ (given $f \ge 0$, say). Then $x(\varepsilon) = O(\varepsilon)$ and $\int_0^\varepsilon x = O(\varepsilon^2)$, and the displayed identity gives $\dot x(\varepsilon) = 1 + O(\varepsilon)$. That is the whole content of "an impulse is a velocity jump", and it holds for any pulse shape with the same integral.

The physical statement. $\int F\,\mathrm{d}t$ is the impulse, and unit 00's second law integrates to $m\Delta v = \int F\,\mathrm{d}t$ — momentum change equals impulse, regardless of how the force was distributed in time. This slide is that statement with the spring and the damper shown not to interfere in the limit. It is also why a ballistic galvanometer measures charge: the swing depends on $\int i\,\mathrm{d}t$ and on nothing else about the current.

What unit 20 will do instead. It will define the Dirac delta as a distribution, write $LG = \delta$, and recover the jump condition as a theorem rather than a definition. Everything on this slide is then a special case of a much more general machine. Nothing here is wrong or provisional — it is the same object, defined with tools that have arrived.

06 · the impulse response

The Green's function, defined and computed

the impulse response $G$

$G : \mathbb{R}\to\mathbb{R}$ is the impulse response of $L$ if

$G(t) = 0$ for $t \le 0$ (causality: nothing before the kick);
$G$ is continuous on $\mathbb{R}$ and $C^2$ on $(0,\infty)$, with $LG = 0$ there;
$G(0^+) = 0$ and $\dot G(0^+) = 1$ (a unit jump in the derivative).

existence, uniqueness, formula

Exactly one such $G$ exists, namely $G(t) = \big(e^{\lambda_+t}-e^{\lambda_-t}\big)/(\lambda_+-\lambda_-)$ for $t > 0$, which in the three regimes reads

$$G(t) = \frac{e^{-\gamma t/2}\sin\omega_\mathrm{d}t}{\omega_\mathrm{d}},\qquad t\,e^{-\omega_0 t},\qquad \frac{e^{\lambda_+t}-e^{\lambda_-t}}{\lambda_+-\lambda_-} .$$

proof

On $(0,\infty)$, $G \in \ker L$, which is two-dimensional; the two conditions at $0^+$ fix $\mathrm{ev}(G) = (0,1)$, and $\mathrm{ev}$ is a bijection. So $G$ is unique, and the displayed function has $G(0)=0$, $\dot G(0) = (\lambda_+-\lambda_-)/(\lambda_+-\lambda_-) = 1$. The underdamped form is the same expression with $\lambda_\pm = -\gamma/2\pm i\omega_\mathrm{d}$ and $\sin\vartheta = (e^{i\vartheta}-e^{-i\vartheta})/2i$; the critical form is the limit, and is checked directly: $\frac{\mathrm{d}}{\mathrm{d}t}te^{-\omega_0t} = (1-\omega_0t)e^{-\omega_0t} \to 1$. $\square$

checks

Dimensions: $[G] = \mathrm{T}$, since $\dot G(0^+) = 1$ is dimensionless per unit of $x$. Limit $\gamma\to0$: $G = \sin\omega_0t/\omega_0$ — a unit velocity kick gives amplitude $1/\omega_0$, as energy demands. ✓

Why a jump in $\dot G$ and not in $G$. Because $L$ is second order, and the highest derivative is what absorbs the concentrated force. Integrating $LG = 0$ twice across the origin shows that $G$ itself must be continuous — a jump in $G$ would make $\dot G$ an infinite spike and $\ddot G$ worse, with nothing on the right-hand side to balance it. The rule generalises: for an $n$-th order operator, $G$ and its first $n-2$ derivatives are continuous and the $(n-1)$-th jumps by $1/a_n$, with $a_n$ the leading coefficient.

The name. George Green introduced these functions in his 1828 Essay on the Application of Mathematical Analysis to the Theories of Electricity and Magnetism, self-published in Nottingham while he was running a mill; the essay sold fifty-one copies and was rediscovered by Kelvin in 1845. Unit 23 uses Green's own case, the Laplacian, and unit 24 the wave operator; both are this construction with more dimensions.

The critical case as a limit, safely. $\frac{e^{\lambda_+t}-e^{\lambda_-t}}{\lambda_+-\lambda_-}$ is a difference quotient of $\lambda \mapsto e^{\lambda t}$, so as $\lambda_\pm \to \lambda$ it tends to $\frac{\partial}{\partial\lambda}e^{\lambda t} = te^{\lambda t}$. That is a clean way to remember the formula; the proof above does not rely on it, it verifies the critical case directly.

$G$ is the whole system. Two numbers, $\gamma$ and $\omega_0$, are in $G$; and by the Duhamel slide, $G$ determines the response to everything. Measuring a ring-down after one tap is therefore a complete characterisation of a linear system — which is exactly how a loudspeaker, a room, or a seismometer is calibrated in practice.

06 · the impulse response

What $G$ looks like, in all three regimes

All three leave $G = 0$ with slope $1$ (dashed) — the unit jump in $\dot G$ — and differ in everything after. Dots mark the peaks, which are the numbers in the box.

read off the picture

Light damping: rings for $\sim Q$ radians, peak $\approx 1/\omega_0 = 0.125$. Critical: one hump, peak $1/(e\omega_0) = 0.04598$ at $t = 1/\omega_0 = 0.125$ s. Heavy: peak $0.01204$ — the shorted instrument barely notices the kick at all.

more damping, smaller response

Damping is not a cosmetic change to the shape: it costs amplitude. From $\gamma = 0.4$ to $\gamma = 80$, the peak response to the same kick falls by a factor $10$. A ballistic galvanometer is therefore used open-circuit, and a dead-beat one is not used ballistically.

The critical peak, exactly. $G = te^{-\omega_0t}$ has $\dot G = (1-\omega_0t)e^{-\omega_0t}$, zero at $t = 1/\omega_0 = 0.125$ s, where $G = 1/(e\omega_0) = 0.0459849$. The overdamped peak is at $t = \ln(\lambda_-/\lambda_+)/(\lambda_+-\lambda_-)$; with $\lambda_\pm = -0.80816, -79.192$ that is $t = 0.05849$ s and $G = 0.012045$. And the underdamped peak is at $t_\ast = \arctan(2\omega_\mathrm{d}/\gamma)/\omega_\mathrm{d} = 0.19328$ s with $G = 0.120260$ — computed in worked example 4.

The scaling behind "a factor of ten". For heavy damping $\lambda_+ \approx -\omega_0^2/\gamma$ and $\lambda_- \approx -\gamma$, so $G(t) \approx (e^{-\omega_0^2t/\gamma}-e^{-\gamma t})/\gamma$, whose peak is $O(1/\gamma)$: the response to an impulse falls off like $1/\gamma$ for large damping, while for light damping it saturates at $1/\omega_0$. The crossover is at $\gamma \sim \omega_0$, as dimensional analysis would have said in advance.

Reading $G$ as an experiment. Every curve here is directly observable: tap the system once, record the response, and you have plotted $G$. The three curves are the three things an instrument can do, and choosing between them is the entire design decision, made by turning one resistor.

06 · the impulse response

Duhamel: convolution solves everything

Duhamel's principle

Let $f$ be continuous, with $f(s) = 0$ for $s < t_0$. Then

$$x(t) = \int_{t_0}^{t} G(t-s)\,f(s)\,\mathrm{d}s \;=\; \int_{-\infty}^{\infty} G(t-s)\,f(s)\,\mathrm{d}s \;=\; (G * f)(t)$$

is the unique solution of $Lx = f$ with $x(t_0) = \dot x(t_0) = 0$: a convolution, upper limit $t$ as $G$ vanishes on the negatives.

proof

Take $x_i = e^{\lambda_\pm t}$ in $K$, so $W(s) = (\lambda_--\lambda_+)e^{(\lambda_++\lambda_-)s}$:

$$K(t,s) = \frac{e^{\lambda_+s}e^{\lambda_-t}-e^{\lambda_-s}e^{\lambda_+t}}{(\lambda_--\lambda_+)e^{(\lambda_++\lambda_-)s}}$$ $$= \frac{e^{\lambda_+(t-s)}-e^{\lambda_-(t-s)}}{\lambda_+-\lambda_-} = G(t-s). \ \square$$

The kernel of variation of parameters, built from the exponential basis, collapses to a function of $t - s$ alone, and that function is $G$. Nothing else is needed: the next slide draws what the integral is doing.

Why $K$ depends only on $t-s$. Because $L$ has constant coefficients, it commutes with time translation; the system is linear time-invariant, and a linear time-invariant map is exactly a convolution. The proof above is the computational face of that; the structural face is: the response to a kick at time $s$ must be the response to a kick at time $0$, shifted. Unit 20 proves the converse — every reasonable linear time-invariant operator is convolution with something — and turns it into the statement that the Fourier transform diagonalises them.

The repeated-root case. The computation above assumed $\lambda_+ \ne \lambda_-$. For $\gamma = 2\omega_0$ take $x_1 = e^{\lambda t}$, $x_2 = te^{\lambda t}$, whose Wronskian is $W(s) = e^{2\lambda s}$; then $K(t,s) = \big(e^{\lambda s}te^{\lambda t} - se^{\lambda s}e^{\lambda t}\big)/e^{2\lambda s} = (t-s)e^{\lambda(t-s)} = G(t-s)$, as required. Both cases, and the overdamped one, are covered.

Non-zero initial conditions. Add the element of $\ker L$ with the right $\mathrm{ev}$: $x(t) = x_h(t) + (G*f)(t)$ where $x_h \in \ker L$ has $\mathrm{ev}(x_h) = (x_0,v_0)$. Since $G*f$ and its derivative vanish at $t_0$, the initial conditions are met, and by uniqueness this is the whole answer. Explicitly $x_h(t) = x_0\big(\dot G(t-t_0) + \gamma G(t-t_0)\big) + v_0G(t-t_0)$, which is a good exercise and is problem P10's second part.

Duhamel's name is attached to it from his 1833 work on heat conduction with time-dependent boundary conditions; the trick of building a general solution by superposing responses to instantaneous sources is his. Unit 24 uses the identical argument for the wave equation, where the "kick" is a flash of light and $G$ is supported on the light cone.

06 · the impulse response

Duhamel, drawn: every sliver is a kick

A half-second pulse of drive, cut into five slivers. Each sliver, of area $f(s_k)\Delta s$, is an impulse at its own time $s_k$ and launches its own scaled copy of $G$ (thin). Their sum (dashed) already lies close to $G * f$ (solid); as $\Delta s \to 0$ it is exactly the integral. $\gamma = 2$.

the statement in words

Each sliver $f(s)\,\mathrm{d}s$ of drive is an impulse, contributing $G(t-s)f(s)\,\mathrm{d}s$ later; superposition adds them. The integral is that sentence.

checks

Dimensions: $\mathrm{T}\times[x]\mathrm{T^{-2}}\times\mathrm{T} = [x]$ ✓. Commutativity: $s \mapsto t-s$ turns $G*f$ into $f*G$ ✓. Limit: $f$ concentrated at $t_0$ gives $x \to G(t-t_0)$ ✓.

06 · the impulse response · example

Worked: the steady state, a second time

worked example 3 — the same answer by unrelated machinery

Drive with $f(s) = F_0\cos\omega s$ for all $s \in \mathbb{R}$ — switched on in the infinite past, so the transient is long gone. Duhamel gives $x(t) = \int_{-\infty}^{t}G(t-s)F_0\cos\omega s\,\mathrm{d}s$; substituting $\tau = t-s$,

$$x(t) = F_0\!\int_0^\infty\! G(\tau)\cos\big(\omega(t-\tau)\big)\mathrm{d}\tau = F_0\operatorname{Re}\Big[e^{i\omega t}\underbrace{\int_0^\infty\! G(\tau)e^{-i\omega\tau}\mathrm{d}\tau}_{\textstyle =\,\hat G(\omega)}\Big].$$

Step 1 — the integral converges: $|G(\tau)| \le e^{-\gamma\tau/2}/\omega_\mathrm{d}$, and $\gamma > 0$. Step 2 — split the sine: $G(\tau) = \big(e^{\lambda_+\tau}-e^{\lambda_-\tau}\big)/(\lambda_+-\lambda_-)$ with $\lambda_\pm = -\gamma/2\pm i\omega_\mathrm{d}$, so by section 1's lemma,

$$\hat G = \frac{1}{\lambda_+-\lambda_-}\left[\frac{-1}{\lambda_+-i\omega} + \frac{1}{\lambda_--i\omega}\right]$$ $$= \frac{1}{(\lambda_+-i\omega)(\lambda_--i\omega)} .$$

Step 3 — expand: $(i\omega-\lambda_+)(i\omega-\lambda_-) = p(i\omega)$, and the sign flips cancel, so

$$\hat G(\omega) = \frac{1}{p(i\omega)} = \chi(\omega).$$

Step 4 — conclude: $x(t) = F_0\operatorname{Re}\big[\chi(\omega)e^{i\omega t}\big]$, which is section 3's boxed answer, character for character.

The check, numerically. At $\gamma = 0.4$, $\omega_0 = 8$, $\omega = 5$: integrating $\int_0^\infty G(\tau)e^{-5i\tau}\mathrm{d}\tau$ gives $0.0255738 - 0.0013115\,i$, and $\chi(5) = 1/(39 + 2i) = 0.0255738 - 0.0013115\,i$. ✓

Why this is worth a slide. Section 3 guessed an ansatz and verified it. Section 4 built $G$ from a jump condition, proved a convolution theorem, and evaluated an integral. The two routes share no step, and they produce the same function. In a subject where a sign or a factor of two is the usual failure mode, an independent second derivation is the strongest check available, and getting into the habit of finding one is the point of this example.

Step 2, in full. $\int_0^\infty e^{(\lambda-i\omega)\tau}\mathrm{d}\tau = \big[e^{(\lambda-i\omega)\tau}/(\lambda-i\omega)\big]_0^\infty = -1/(\lambda-i\omega)$, the upper limit vanishing because $\operatorname{Re}(\lambda-i\omega) = -\gamma/2 < 0$. So the bracket is $\frac{-1}{\lambda_+-i\omega}+\frac{1}{\lambda_--i\omega} = \frac{-(\lambda_--i\omega)+(\lambda_+-i\omega)}{(\lambda_+-i\omega)(\lambda_--i\omega)} = \frac{\lambda_+-\lambda_-}{(\lambda_+-i\omega)(\lambda_--i\omega)}$, and the prefactor $1/(\lambda_+-\lambda_-)$ cancels the numerator. Finally $p(\mu) = (\mu-\lambda_+)(\mu-\lambda_-)$ gives $(\lambda_+-i\omega)(\lambda_--i\omega) = (i\omega-\lambda_+)(i\omega-\lambda_-) = p(i\omega)$.

$\hat G$ has a name. It is the Fourier transform of $G$, and this slide has computed one without having the theory: the integral is elementary because $G$ is a sum of exponentials. Unit 20 builds the transform properly and shows that what has been proved here — $\widehat{G*f} = \hat G\hat f$ in the case $f = e^{i\omega t}$ — is a general theorem, the convolution theorem, and that it is the reason Fourier methods work on linear equations.

06 · the impulse response

Causality, and the promissory note to unit 20

causality

$G(t) = 0$ for $t < 0$: the response at time $t$ depends on the drive at times $s \le t$ only. In the convolution, the upper limit is $t$ and not $+\infty$.

what causality does to $\chi$

Because the integral $\hat G(\omega) = \int_0^\infty G(\tau)e^{-i\omega\tau}\mathrm{d}\tau$ runs over $\tau \ge 0$ only, it still converges when $\omega$ is given a negative imaginary part — indeed, underdamped, for every complex $\omega$ with $\operatorname{Im}\omega < \gamma/2$. So $\chi$ is defined and has no singularity anywhere in the closed lower half-plane; its two poles,

$$\omega = \pm\omega_\mathrm{d} + i\gamma/2,$$

sit in the upper half-plane, at a height $\gamma/2$ above the real axis.

stated, not proved

That a function given by such an integral is analytic there, and that analyticity in a half-plane forces a relation between $\operatorname{Re}\chi$ and $\operatorname{Im}\chi$ — the Kramers–Kronig relations — needs complex analysis, which is unit 20. Nothing on this slide beyond the convergence claim is proved here.

Sign conventions differ: with $e^{-i\omega t}$ for the drive, the poles land in the lower half-plane. The physics — poles on the far side from where the contour must be closed — is convention-independent.

The convergence claim, checked. With $\omega = \omega_r + i\omega_i$, $|e^{-i\omega\tau}| = e^{\omega_i\tau}$, and $|G(\tau)| \le e^{-\gamma\tau/2}/\omega_\mathrm{d}$, so the integrand is bounded by $e^{(\omega_i-\gamma/2)\tau}/\omega_\mathrm{d}$, integrable on $[0,\infty)$ exactly when $\omega_i < \gamma/2$. (Overdamped, $0 \le G(\tau) \le e^{\lambda_+\tau}/(\lambda_+-\lambda_-)$ gives the smaller half-plane $\omega_i < |\lambda_+|$, whose edge is the nearest pole, at $\omega = i|\lambda_+|$; critically damped, $\omega_i < \omega_0 = \gamma/2$. The lower half-plane is inside in every case.) Had $G$ been non-zero for $\tau < 0$ as well, the integral would have run over the whole line and no half-plane would have been distinguished. That asymmetry — one half-plane, not both — is the mathematical shadow of "effects follow causes", and it is worth noticing that it arrived here with no physics beyond the definition of $G$.

What unit 20 will cash. Closing a contour in the half-plane where $\chi$ has no poles gives $\int\chi(\omega)e^{i\omega t}\mathrm{d}\omega = 0$ for $t < 0$ — that is, causality of $G$ recovered from analyticity of $\chi$, the converse of what this slide observed. Pushing the same contour argument gives the Kramers–Kronig relations, and the $i\varepsilon$ prescription that unit 24 needs for the retarded Green's function of the wave operator is the same choice of half-plane made in four dimensions.

06 · the impulse response

Where the poles of $\chi$ sit

The complex $\omega$-plane for $\gamma = 5$. The real axis is every frequency you can drive at. Shaded: where the integral for $\hat G$ converges, everything below height $\gamma/2$. The two poles of $\chi$ sit on its edge, above the axis; for the galvanometer at $Q = 20$ they are only $0.2$ above it (circles).

The poles are the zeros of $p(i\omega)$: $\omega^2 - i\gamma\omega - \omega_0^2 = 0$, so

$$\omega = \pm\omega_\mathrm{d} + \tfrac{i\gamma}{2} .$$

They are the free motion's own exponents, $\lambda_\pm = i\omega$: the real part is the ringing frequency, the height above the axis the decay rate.

The slogan: dissipation and dispersion are the same function, and causality is why. It is how the refractive index of a medium at one colour constrains its absorption at all the others.

What to look for in the figure. Causality is the shading: it reaches all the way down, so the whole lower half-plane is free of singularities, and the poles are pushed above the real axis. Light damping brings them down towards the axis — that is why a high-$Q$ response is a sharp spike on the real line, which passes very close to a pole — and $\gamma = 0$ puts them on it, where the undamped-resonance slide found the ansatz breaking. Nothing about analyticity is claimed here; the figure shows only where the defining integral converges and where $p(i\omega)$ vanishes.

Where the poles are, and what they mean. $p(i\omega) = 0$ gives $\omega^2 - i\gamma\omega - \omega_0^2 = 0$, hence $\omega = \tfrac{i\gamma}{2}\pm\omega_\mathrm{d}$: the real part is the ringing frequency and the imaginary part is the decay rate. Poles of the frequency response are therefore the system's own modes, read off the complex plane. The quasinormal modes of a perturbed black hole are this statement for a very different operator, and they are quoted in exactly this form: a complex frequency whose imaginary part is a ring-down time.

06 · the running example, closed

Worked: the ballistic galvanometer measures a charge

worked example 4 — one kick, one number

Setting. Discharge $q = 1\ \mu\mathrm{C}$ through the coil in a few milliseconds, with $R = 995\ \Omega$ so $\gamma = 0.4$, $\omega_\mathrm{d} = 7.99750$.

Step 1 — the drive is an impulse: $\int(\alpha/I)i\,\mathrm{d}t = \alpha q/I = 0.2\ \mathrm{rad\,s^{-1}}$, so by Duhamel with $f$ concentrated at $0$,

$$\theta(t) = \frac{\alpha q}{I}\,G(t) = 0.025008\,e^{-0.2t}\sin 7.9975t$$

Step 2 — the first swing. $\dot\theta = 0$ when $\tan\omega_\mathrm{d}t = 2\omega_\mathrm{d}/\gamma$, giving $t_\ast = \arctan(39.9875)/7.9975 = 0.19328$ s.

Step 3 — the number.

$$\theta_{\max} = 0.2\times0.120260 = 0.024052\ \mathrm{rad},$$

that is $1.3781^\circ$.

Step 4 — why this is a measurement. $\theta_{\max} \propto q$ with a constant that contains only $\alpha$, $I$, $\gamma$ and $\omega_0$ — instrument constants. Read the first swing, divide, and you have the charge. Nothing about the discharge's shape or duration enters.

checks

Undamped ideal: $\alpha q/(I\omega_0) = 0.025$ rad; the damping costs exactly $e^{-\gamma t_\ast/2} = 0.96208$ of it, and $0.025\times0.96208 = 0.024052$ ✓. Dimensions: $[\alpha q/I] = \mathrm{s^{-1}}$, times $[G] = \mathrm{s}$: an angle ✓.

Step 2, in full. $\theta \propto e^{-\gamma t/2}\sin\omega_\mathrm{d}t$, so $\dot\theta \propto e^{-\gamma t/2}\big(\omega_\mathrm{d}\cos\omega_\mathrm{d}t - \tfrac\gamma2\sin\omega_\mathrm{d}t\big)$, which vanishes when $\tan\omega_\mathrm{d}t = 2\omega_\mathrm{d}/\gamma$. The first positive root is $\arctan(2\omega_\mathrm{d}/\gamma)/\omega_\mathrm{d}$, slightly less than the quarter period $T_\mathrm{d}/4 = 0.19641$ s — the needle turns round just before it would have done undamped, which is the correct sign.

The exact throw formula. Combining, $\theta_{\max} = \dfrac{\alpha q}{I\omega_0}\exp\!\Big(\!-\dfrac{\gamma}{2\omega_\mathrm{d}}\arctan\dfrac{2\omega_\mathrm{d}}{\gamma}\Big)$, since $\sin(\arctan u) = u/\sqrt{1+u^2}$ collapses the algebra to a factor $\omega_\mathrm{d}/\omega_0$ that cancels one already present. As $\gamma\to0$ the exponent $\to0$ and the throw $\to \alpha q/I\omega_0$. The correction factor at $Q = 20$ is $0.962$: a $3.8\%$ effect, far too large to neglect, and in practice removed either by measuring $\delta$ and correcting, or by working open-circuit.

How the real instrument was used. Ballistic galvanometers measured capacitance, magnetic flux (by yanking a search coil out of a field, where $\int i\,\mathrm{d}t = \Delta\Phi/(R+r)$), and the charge stored by a dielectric. The whole method rests on the impulse theorem — that the swing depends on the time integral of the current and on nothing else — and it is a textbook case of a mathematical limit being turned into an instrument.

Where the running example ends up. One equation, one knob. Turning it from $R=0$ to $R=\infty$ moved the instrument from sluggish, through dead-beat at exactly $20\ \Omega$, to a $Q = 20$ resonator, and each setting is the right one for a different measurement: a pointer instrument at critical, a tuned bridge detector at high $Q$, a shorted coil for transport. The mathematics of section 2 is a specification sheet.

06 · summary

What this unit bought

you may now say

the solution space of $Lx=0$ is a plane, by unit 01 — and a verified guess of two independent solutions is therefore the general solution;
a steady state exists, is unique, and equals $F_0\operatorname{Re}[\chi e^{i\omega t}]$, with $|\chi|$ and $\arg\chi$ both exactly known;
the power resonance is centred exactly at $\omega_0$ with width exactly $\gamma$, so a measured curve gives both constants;
the response to any continuous drive is $G*f$, with $G$ fixed by a jump in the derivative — the whole system in one function.

and you may not say

"second order, so two constants" — that is false without linearity (unit 01's dome);
"resonance is at $\omega_0$" without saying which resonance: amplitude, velocity and free ringing sit at three different frequencies;
"$Q$ is $2\pi$ times the energy over the loss per cycle" as though it were exact — at $Q = 20$ that reading is $16.5\%$ out;
anything at all about poles, analyticity or Kramers–Kronig: that is unit 20, and this unit only points at it.

Next: unit 07 replaces $\omega_0^2$ by a matrix, and the spectral theorem tells you which basis to use; unit 08 finds this equation again, unexpectedly, inside the Kepler problem.

The unit in one sentence: a linear constant-coefficient operator is diagonalised by exponentials, and everything else — regimes, resonance, phase, the Green's function — is bookkeeping about where the roots of one quadratic sit in the complex plane.

Two habits to carry. First: when an ansatz fails, ask whether the thing you are trying to hit is in the kernel; if it is, multiply by $t$. That single rule covers critical damping, undamped resonance, and every secular term in unit 21. Second: when you have an answer, look for a second route to it. Worked example 3 did the same calculation twice by unrelated machinery, and that is the cheapest insurance available in physics.

A word on what has been left for later. The drive was always given; unit 12 will derive it from a Lagrangian. The damping was postulated; a real derivation needs statistical mechanics and is outside this course. The delta function was avoided; unit 20 supplies it. And the assumption that the oscillator is one degree of freedom is dropped in the very next unit.

06 · checkpoint

Done when

checkpoint

You construct $G(t-t')$ for $\ddot x + \gamma\dot x + \omega_0^2x = f$ from its defining property — zero before the kick, continuous, a unit jump in the derivative — solve an arbitrary drive by convolution, verify the answer against a directly computed piecewise solution including at the corners, and say why the solution space is two-dimensional by citing unit 01's theorem, not by counting constants.

How to fail it, unambiguously

Any of: $G$ is written down from memory rather than derived from the two conditions at $0^+$; the convolution's limits are $\pm\infty$ with no remark on why the upper one is $t$; the two solutions of the square-pulse problem are not checked to agree at $t = T$ in both value and slope; or "two arbitrary constants" appears as the reason for $\dim\ker L = 2$.

Then

./mr problems 06 — P7 is the square pulse, P10 the Green's function identity, P12 the causality argument.
Hints, then the worked solution, fold under each problem — only after an attempt.
./mr done 06 when the checkpoint is met.

Reading, all optional: French, Vibrations and Waves, ch. 3–4; Arnold, Ordinary Differential Equations, ch. 3 §§15–17; Coddington & Levinson ch. 3; Pippard, The Physics of Vibration, for the instruments.

The falsifiable part is "by citing unit 01's theorem". It is easy to produce two solutions, add two constants and move on; that is not the checkpoint. The checkpoint is the sentence: the evaluation map $x \mapsto (x(t_0),\dot x(t_0))$ is injective by uniqueness and surjective by global existence, both for the linear field $\dot y = Ay$, hence an isomorphism onto $\mathbb{R}^2$. If that sentence is not available, unit 07's $2n$-dimensional version will not be either.

Next units: 07 couples oscillators and needs every structural fact here; 08 turns the orbit equation into $u'' + u = \text{const}$, which is this equation driven by a constant, and unit 37 adds one small term to it to produce the perihelion precession; 20 retells section 4 with distributions and contour integrals and cashes the causality slide; 21 perturbs this solution and meets the secular term again, this time as a defect to be cured.