---
title: "Dispersion and Wave Packets"
book: "University Physics — Year 2"
subject: physics
language: en
chapter: 8
exercises: 12
source: https://one-course.com/books/physics/4/en/chapter/8-dispersion-and-wave-packets
---

# Chapter 8 — Dispersion and Wave Packets

Drop a stone in a pond and watch the ring of ripples: new crests are born at its inner edge, run outward through the ring, and die at its outer edge — the ring as a whole moves slower than the crests in it. Sit on a beach after a distant storm and the long, slow swell arrives a day before the short chop: the sea sorts waves by wavelength. The first transatlantic telegraph cable, in 1858, turned crisp dots and dashes into a smear that took minutes to read. In all three cases the speed of a wave depends on its frequency: the medium is *dispersive*. This chapter introduces the [dispersion relation](#def-b2-dispersion-wave-packets-relation), the [complex wavenumber](#def-b2-dispersion-wave-packets-complex) that carries both propagation and absorption, the [group velocity](#prop-b2-dispersion-wave-packets-group) at which a packet — and its energy, and its information — actually travels, and the line along which telegraphy, television and the internet have travelled: the [coaxial cable](#prop-b2-dispersion-wave-packets-coax).

## 8.1 The dispersion relation

**Definition 8.1 (Dispersion relation; phase velocity).**

A linear, homogeneous, time-invariant medium admits the plane monochromatic waves $\underline s = A\,\eu^{\iu(\omega t - kx)}$ (complex notation; the physical signal is the real part) provided $\omega$ and $k$ satisfy the *dispersion relation* of the medium, $\mathcal D(\omega, k)
= 0$, solved as $k(\omega)$ or $\omega(k)$. For real $k$ the wave travels without deforming at the *phase velocity*

$$
v_\varphi = \frac\omega k .
$$

The medium is *non-dispersive* when $v_\varphi$ does not depend on $\omega$ (then $\omega = ck$ and every signal propagates undeformed — the [d’Alembert equation](https://one-course.com/books/physics/4/en/chapter/6-waves-on-strings-and-rods-the-dalembert-equation#thm-b2-waves-on-strings-equation)) and *dispersive* otherwise.

**Proof.** Inserting $\eu^{\iu(\omega t - kx)}$ into a linear equation with constant coefficients turns each $\partial_t$ into $\iu\omega$ and each $\partial_x$ into $-\iu k$, leaving an algebraic relation. The string, the sound wave and the rod gave $\omega^2 = c^2k^2$. ∎

**Example 8.2 (Dispersive and non-dispersive).**

(i) A string on an elastic bed (restoring force $-Ky$ per unit length, or a chain of pendulums), $\mu\partial_t^2y = T\partial_x^2y - Ky$: $\omega^2 = \omega_c^2
+ c^2k^2$ with $\omega_c = \sqrt{K/\mu}$ — the *Klein–Gordon* relation, the same as a plasma’s and a waveguide’s (Chapters [14](https://one-course.com/books/physics/4/en/chapter/14-electromagnetic-waves-in-plasmas-conductors-and-dielectrics#ch-b2-waves-in-media) and [16](https://one-course.com/books/physics/4/en/chapter/16-guided-waves-and-cavities#ch-b2-guided-waves)): $v_\varphi = c/\sqrt{1 -
\omega_c^2/\omega^2} > c$, and no real $k$ below the *cut-off* $\omega_c$. (ii) Deep-water gravity waves: $\omega^2 = gk$ (admitted), $v_\varphi =
\sqrt{g/k} = \sqrt{g\lambda/2\pi}$: long waves are faster — $12.5\,\mathrm{m}/\mathrm{s}$ for $100\,\mathrm{m}$, $40\,\mathrm{m}/\mathrm{s}$ for $1\,\mathrm{km}$. (iii) The [chain of atoms](https://one-course.com/books/physics/4/en/chapter/6-waves-on-strings-and-rods-the-dalembert-equation#prop-b2-waves-on-strings-chain), $\omega =
2\omega_0|\sin(ka/2)|$ ([Chapter 6](https://one-course.com/books/physics/4/en/chapter/6-waves-on-strings-and-rods-the-dalembert-equation#ch-b2-waves-on-strings)). (iv) Light in glass, $k = n(\omega)\omega/c$, which is why a prism spreads a spectrum.

![Left: three dispersion relations — a straight line for a non-dispersive medium, the Klein–Gordon hyperbola with its cut-off, the parabola of deep-water waves. Right: for the Klein–Gordon relation the phase velocity exceeds c and the group velocity stays below it, with v_ v_g = c2.](https://one-course.com/images/onecourse/chapters/physics-4/b2-dispersion-wave-packets/fig-0ac45db78980.svg)

![Left: three dispersion relations — a straight line for a non-dispersive medium, the Klein–Gordon hyperbola with its cut-off, the parabola of deep-water waves. Right: for the Klein–Gordon relation the phase velocity exceeds c and the group velocity stays below it, with v_ v_g = c2.](https://one-course.com/images/onecourse/chapters/physics-4/b2-dispersion-wave-packets/fig-3e9b1376f9c9.svg)

*Left: three [dispersion relations](#def-b2-dispersion-wave-packets-relation) — a straight line for a non-dispersive medium, the Klein–Gordon hyperbola with its cut-off, the parabola of [deep-water waves](#ex-b2-dispersion-wave-packets-examples). Right: for the Klein–Gordon relation the [phase velocity](#def-b2-dispersion-wave-packets-relation) exceeds $c$ and the [group velocity](#prop-b2-dispersion-wave-packets-group) stays below it, with $v_\varphi v_g = c^2$.*

**Definition 8.3 (Complex wavenumber: propagation and attenuation).**

When the [dispersion relation](#def-b2-dispersion-wave-packets-relation) gives, for real $\omega$, a complex $\underline k = k' - \iu k''$, the wave is

$$
\underline s = A\,\eu^{-k''x}\,\eu^{\iu(\omega t - k'x)} :
$$

it propagates at $v_\varphi = \omega/k'$ and its amplitude decays as $\eu^{-x/\delta}$ with the *attenuation length* $\delta = 1/k''$ (it must decay in its direction of propagation: $k'$ and $k''$ of the same sign). If $\underline k$ is purely imaginary, $\underline k = -\iu\kappa$, the wave is *evanescent*: $A\,\eu^{-\kappa x}\eu^{\iu\omega t}$, a standing oscillation whose amplitude dies over the distance $1/\kappa$ without propagating anything — the case of the Klein–Gordon medium below its cut-off, $\kappa = \sqrt{\omega_c^2 - \omega^2}/c$, and of a metal at optical frequencies.

**Example 8.4 (A damped string).**

A string in a viscous fluid, $\mu\partial_t^2y + \alpha\partial_ty = T\partial_x^2y$: $\underline k^2 = (\omega^2 - \iu\alpha\omega/\mu)/c^2$. For weak damping ($\alpha \ll \mu
\omega$), $\underline k \approx \dfrac\omega c\Bigl(1 - \iu\dfrac{\alpha}{2\mu\omega}\Bigr)$: the wave keeps its speed and loses amplitude over $\delta = 2\mu c/\alpha = 2Z/\alpha$, independent of frequency — a *dissipative*, non-dispersive loss, $8.7\,\mathrm{dB}$ per length $\delta$.

## 8.2 Wave packets and group velocity

**Proposition 8.5 (Beats; group velocity).**

Two waves of equal amplitude and neighbouring frequencies $\omega \pm
\delta\omega$, wavenumbers $k \pm \delta k$, superpose into

$$
s = 2A\cos(\delta\omega\,t - \delta k\,x)\cos(\omega t - kx) :
$$

a carrier at $(\omega, k)$ moving at $v_\varphi = \omega/k$, modulated by a slow envelope moving at $\delta\omega/\delta k$. A *[wave packet](#prop-b2-dispersion-wave-packets-group)* — a superposition of monochromatic waves with wavenumbers in a narrow band around $k_0$ — travels, as a whole, at the *[group velocity](#prop-b2-dispersion-wave-packets-group)*

$$
v_g = \frac{\dd\omega}{\dd k}\Bigr|_{k_0} ,
$$

which is the velocity of its energy and of the information it carries. In a non-dispersive medium $v_g = v_\varphi = c$; otherwise the packet deforms as it goes (it *spreads*), the faster the larger $\dd^2\omega/\dd k^2$ and the narrower its spectrum’s complement, its spatial width.

**Proof.** The sum-to-product formula gives the beats. For a packet, write $s =
\int a(k)\eu^{\iu(\omega(k)t - kx)}\dd k$ (a Fourier superposition — the Fourier integral is studied in the Year 3 mathematics volume; here only its interpretation as a continuous sum of sinusoids is used), with $a(k)$ peaked at $k_0$. Expand $\omega(k) \approx \omega_0 + v_g(k - k_0)$: then $s \approx \eu^{\iu(\omega_0t - k_0x)}\int a(k)\eu^{\iu(k - k_0)(v_gt - x)}\dd k$, a carrier times an envelope that is a function of $x - v_gt$ only — the envelope moves at $v_g$. The next term, $\tfrac12\omega''(k_0)(k - k_0)^2t$, dephases the components in time and spreads the envelope. The energy of a packet is located where its envelope is; so is any signal. ∎

![Beats: two neighbouring frequencies produce a fast carrier (moving at the phase velocity) under a slow envelope (moving at the group velocity); in a dispersive medium the two speeds differ and the crests slide through the envelope.](https://one-course.com/images/onecourse/chapters/physics-4/b2-dispersion-wave-packets/fig-82cfe920136c.svg)

*Beats: two neighbouring frequencies produce a fast carrier (moving at the [phase velocity](#def-b2-dispersion-wave-packets-relation)) under a slow envelope (moving at the [group velocity](#prop-b2-dispersion-wave-packets-group)); in a [dispersive medium](#def-b2-dispersion-wave-packets-relation) the two speeds differ and the crests slide through the envelope.*

**Example 8.6 (Deep water: the crests outrun the group).**

$\omega = \sqrt{gk}$: $v_g = \tfrac12\sqrt{g/k} = \tfrac12v_\varphi$. A group of swell moves at half the speed of its crests: watching a wave train, you see crests appear at the back of the group, travel through it and vanish at the front — exactly the pond’s ring. A storm $3000\,\mathrm{km}$ away sends its $15\,\mathrm{s}$ swell ($\lambda = gT^2/2\pi = 350\,\mathrm{m}$, $v_g =
gT/4\pi = 11.7\,\mathrm{m}/\mathrm{s}$) in three days, its $8\,\mathrm{s}$ waves ($v_g =
6.2\,\mathrm{m}/\mathrm{s}$) in five and a half: from the arrival times of the different periods, oceanographers locate the storm.

![The wake of a motorboat on a calm lake: the feathered pattern stays inside a wedge of fixed angle because the energy of each water wave travels at half the speed of its crests — dispersion made visible.](https://one-course.com/images/onecourse/chapters/physics-4/b2-dispersion-wave-packets/img-448c85f67c55.jpg)

*The wake of a motorboat on a calm lake: the feathered pattern stays inside a wedge of fixed angle because the energy of each water wave travels at half the speed of its crests — dispersion made visible.*

**Example 8.7 (Klein–Gordon: faster than light, and not).**

For $\omega^2 = \omega_c^2 + c^2k^2$: $v_\varphi = \omega/k$ and $v_g = c^2k/\omega$, so $v_\varphi v_g = c^2$: the [phase velocity](#def-b2-dispersion-wave-packets-relation) exceeds $c$ (in the ionosphere, a $10\,\mathrm{MHz}$ wave’s crests move faster than light) but the [group velocity](#prop-b2-dispersion-wave-packets-group), which carries the signal, stays below it. A [phase velocity](#def-b2-dispersion-wave-packets-relation) transports no energy and no information — the crests are like the spot of a lighthouse beam sweeping a distant cloud.

**Remark 8.8 (Spreading).**

A packet of spatial width $\Delta x$ contains wavenumbers over $\Delta k \sim
1/\Delta x$ (the Fourier reciprocity, admitted); its components’ group velocities differ by $\Delta v_g \approx |\omega''|\Delta k$, so after a time $t$ it has spread by $|\omega''|\Delta k\,t$: it doubles its width after $t_{\text{sp}}
\sim \Delta x^2/|\omega''|$. A packet of ten [deep-water waves](#ex-b2-dispersion-wave-packets-examples) of $100\,\mathrm{m}$ ($\Delta x \approx 1\,\mathrm{km}$, $\omega'' = -\tfrac14\sqrt{g/k^3} = -50\,\mathrm{m}^{2}/\mathrm{s}$) doubles in about six hours; a light pulse of $1\,\mathrm{ns}$ in an optical fibre spreads by tens of picoseconds per kilometre — the limit of the bit rate of long links. The quantum [wave packet](#prop-b2-dispersion-wave-packets-group) of [Chapter 30](https://one-course.com/books/physics/4/en/chapter/30-the-schrodinger-equation-and-wave-functions#ch-b2-schrodinger-wave-functions) spreads for the same reason.

## 8.3 The coaxial cable

**Proposition 8.9 (Telegrapher’s equations; lossless line).**

A [coaxial cable](#prop-b2-dispersion-wave-packets-coax) (or any two-conductor line) has, per unit length, an inductance $\Lambda$ and a capacitance $\Gamma$. The voltage $v(x, t)$ between the conductors and the current $i(x, t)$ in the inner one obey the *[telegrapher’s equations](#prop-b2-dispersion-wave-packets-coax)*

$$
\frac{\partial v}{\partial x} = -\Lambda\frac{\partial i}{\partial t} , \qquad
\frac{\partial i}{\partial x} = -\Gamma\frac{\partial v}{\partial t} ,
$$

hence the [d’Alembert equation](https://one-course.com/books/physics/4/en/chapter/6-waves-on-strings-and-rods-the-dalembert-equation#thm-b2-waves-on-strings-equation) for $v$ and $i$ with

$$
c = \frac1{\sqrt{\Lambda\Gamma}} , \qquad
v = Z_c\,i \ \text{for a wave toward } +x, \quad Z_c = \sqrt{\frac\Lambda\Gamma} ,
$$

$Z_c$ being the *[characteristic impedance](#prop-b2-dispersion-wave-packets-coax)* of the line. For a [coaxial cable](#prop-b2-dispersion-wave-packets-coax) of radii $a < b$ filled with a dielectric of relative permittivity $\varepsilon_r$, $\Lambda = \dfrac{\mu_0}{2\pi}\ln\dfrac ba$, $\Gamma =
\dfrac{2\pi\varepsilon_0\varepsilon_r}{\ln(b/a)}$, so $c = c_0/\sqrt{\varepsilon_r}$ (about $2 \times 10^{8}\,\mathrm{m}/\mathrm{s}$, two thirds of the speed of light) and $Z_c = \dfrac{60\,\Omega}
{\sqrt{\varepsilon_r}}\ln\dfrac ba$: $50\,\Omega$ or $75\,\Omega$ for the usual cables.

**Proof.** Model the slice $[x, x + \dd x]$ as a series inductance $\Lambda\dd x$ and a shunt capacitance $\Gamma\dd x$: the voltage drop along the slice is $\Lambda\dd x\,\partial_ti$, and the current lost into the capacitance is $\Gamma\dd x
\,\partial_tv$. Cross-differentiate. For $v = f(x - ct)$, the first equation gives $i = f(x - ct)/\Lambda c = v/\sqrt{\Lambda/\Gamma}$. The values of $\Lambda$ and $\Gamma$ are those of the cylindrical capacitor and of the coaxial inductor computed in the Year 1 volume. ∎

![Left: the equivalent circuit of a slice x of a lossless line — series inductance, shunt capacitance — from which the telegrapher’s equations follow. Right: a coaxial cable in section; its and depend only on the ratio of radii and on the dielectric.](https://one-course.com/images/onecourse/chapters/physics-4/b2-dispersion-wave-packets/fig-24ffa98dba3e.svg)

*Left: the equivalent circuit of a slice $\dd x$ of a lossless line — series inductance, shunt capacitance — from which the [telegrapher’s equations](#prop-b2-dispersion-wave-packets-coax) follow. Right: a [coaxial cable](#prop-b2-dispersion-wave-packets-coax) in section; its $\Lambda$ and $\Gamma$ depend only on the ratio of radii and on the dielectric.*

![A coaxial cable: inner core, dielectric, braided outer conductor — a transmission line whose inductance and capacitance per metre set its speed and its characteristic impedance.](https://one-course.com/images/onecourse/chapters/physics-4/b2-dispersion-wave-packets/img-865be02f9b05.jpg)

*A [coaxial cable](#prop-b2-dispersion-wave-packets-coax): inner core, dielectric, braided outer conductor — a [transmission line](#prop-b2-dispersion-wave-packets-coax) whose inductance and capacitance per metre set its speed and its [characteristic impedance](#prop-b2-dispersion-wave-packets-coax).*

**Proposition 8.10 (Lossy line; the distortionless condition).**

With a series resistance $R$ and a shunt conductance $G$ per unit length, the equations become $\partial_xv = -\Lambda\partial_ti - Ri$, $\partial_xi =
-\Gamma\partial_tv - Gv$, and the [dispersion relation](#def-b2-dispersion-wave-packets-relation) $\underline k^2 = (\Lambda\omega -
\iu R)(\Gamma\omega - \iu G)$: the line is dispersive and attenuating. Two limits: (i) $R/\Lambda = G/\Gamma$ (Heaviside’s condition, reached by adding inductance): $\underline k = \omega\sqrt{\Lambda\Gamma} - \iu\sqrt{RG}$, every frequency travels at the same speed with the same [attenuation](#def-b2-dispersion-wave-packets-complex) — the line *attenuates* but does not *distort*. (ii) $\Lambda \approx 0$, $G \approx 0$ (the first submarine cables): $\partial_x^2v = R\Gamma\,\partial_tv$, a *diffusion* equation — a pulse does not propagate but smears, reaching a distance $L$ after a time $\sim R\Gamma L^2$ (Kelvin’s law of squares), a hundred times longer for a cable ten times longer.

**Proof.** Insert $\eu^{\iu(\omega t - kx)}$: $-\iu kv = -(\iu\omega\Lambda + R)i$, $-\iu ki = -(\iu
\omega\Gamma + G)v$; multiply. Under Heaviside’s condition the product is a perfect square. The diffusion limit drops the $\iu\omega\Lambda$ and $G$ terms. ∎

**Example 8.11 (Three cables).**

The 1858 transatlantic cable: a single copper wire in gutta-percha, $R \approx 3\,\Omega/\mathrm{km}$, $\Gamma \approx 0.3\,\text{µ}\mathrm{F}/\mathrm{km}$, $L = 3000\,\mathrm{km}$: $R\Gamma = 9 \times 10^{-10}\,\mathrm{s}/\mathrm{m}^{2}$ and $R\Gamma L^2 \approx 8\,\mathrm{s}$ per dot: a word a minute, and the operators’ attempts to force the signal with $2\,\mathrm{kV}$ destroyed the insulation within weeks. A telephone line loaded with coils every $2\,\mathrm{km}$ to satisfy Heaviside’s condition (1900): clear speech over hundreds of kilometres. A modern $50\,\Omega$ [coaxial cable](#prop-b2-dispersion-wave-packets-coax): $c = 2 \times 10^{8}\,\mathrm{m}/\mathrm{s}$, $\Lambda = 0.25\,\text{µ}\mathrm{H}/\mathrm{m}$, $\Gamma =
100\,\mathrm{pF}/\mathrm{m}$, and an [attenuation](#def-b2-dispersion-wave-packets-complex), due to the skin effect in the conductors ([Chapter 14](https://one-course.com/books/physics/4/en/chapter/14-electromagnetic-waves-in-plasmas-conductors-and-dielectrics#ch-b2-waves-in-media)), growing as $\sqrt f$: a few [decibels](https://one-course.com/books/physics/4/en/chapter/7-sound-waves-in-fluids#def-b2-sound-waves-decibel) per hundred metres at $100\,\mathrm{MHz}$.

**Method 8.12 (Reading a dispersion relation).**

(1) Write the [wave equation](https://one-course.com/books/physics/4/en/chapter/6-waves-on-strings-and-rods-the-dalembert-equation#thm-b2-waves-on-strings-equation) of the medium in complex notation and solve for $\underline k(\omega)$. (2) Real $k$: compute $v_\varphi = \omega/k$ and $v_g = \dd\omega/\dd k$ (or $1/(\dd k/\dd\omega)$); compare: dispersive or not, normal ($v_g < v_\varphi$) or anomalous. (3) Complex $k$: split into propagation ($k'$) and [attenuation](#def-b2-dispersion-wave-packets-complex) ($k''$, length $1/k''$); purely imaginary means evanescent, no transport. (4) A cut-off frequency separates a propagating band from an evanescent one. (5) For a signal, reason on the envelope at $v_g$, never on the crests.

## 8.4 Exercises

**Exercise 8.1 ★.**

[Deep-water waves](#ex-b2-dispersion-wave-packets-examples), $\omega^2 = gk$. (a) [Phase velocity](#def-b2-dispersion-wave-packets-relation), [group velocity](#prop-b2-dispersion-wave-packets-group) and period of a $100\,\mathrm{m}$ swell. (b) A swell of period $14\,\mathrm{s}$: wavelength, $v_\varphi$, $v_g$; time for its energy to cross $5000\,\mathrm{km}$. (c) Why does the chop of a $2\,\mathrm{s}$ period not get there at all?

**Solution of Exercise 8.1.**

(a) $k = 0.063\,\mathrm{rad}/\mathrm{m}$, $\omega = \sqrt{gk} = 0.79\,\mathrm{rad}/\mathrm{s}$, $T = 8.0\,\mathrm{s}$; $v_\varphi = 12.5\,\mathrm{m}/\mathrm{s}$, $v_g = 6.3\,\mathrm{m}/\mathrm{s}$. (b) $\omega = 0.45\,\mathrm{rad}/\mathrm{s}$, $k = \omega^2
/g = 0.021\,\mathrm{rad}/\mathrm{m}$, $\lambda = 310\,\mathrm{m}$; $v_\varphi = gT/2\pi = 22\,\mathrm{m}/\mathrm{s}$, $v_g =
11\,\mathrm{m}/\mathrm{s}$; $5 \times 10^6/11 = 4.6 \times 10^{5}\,\mathrm{s} = 5.3$ days. (c) $\lambda = 6\,\mathrm{m}$, $v_g
= 1.6\,\mathrm{m}/\mathrm{s}$: it would need five weeks, and short waves are damped by [viscosity](https://one-course.com/books/physics/4/en/chapter/4-viscous-flows#def-b2-viscous-flows-viscosity) and broken by the wind long before.

**Exercise 8.2 ★.**

A medium obeys $\omega^2 = \omega_c^2 + c^2k^2$ with $\omega_c = 2\pi \times 1 \times 10^{7}\,\mathrm{rad}/\mathrm{s}$ and $c = 3 \times 10^{8}\,\mathrm{m}/\mathrm{s}$. For $f = 20\,\mathrm{MHz}$: $k$, $v_\varphi$, $v_g$, and their product. For $f = 5\,\mathrm{MHz}$: $\kappa$ and the penetration depth.

**Solution of Exercise 8.2.**

$\omega = 1.26 \times 10^{8}\,\mathrm{rad}/\mathrm{s}$, $\omega_c = 6.3 \times 10^{7}\,\mathrm{rad}/\mathrm{s}$: $k = \sqrt{\omega^2 - \omega_c^2}/c =
0.36\,\mathrm{rad}/\mathrm{m}$, $v_\varphi = 3.5 \times 10^{8}\,\mathrm{m}/\mathrm{s}$, $v_g = c^2k/\omega = 2.6 \times 10^{8}\,\mathrm{m}/\mathrm{s}$, product $c^2$. At $5\,\mathrm{MHz}$, $\omega < \omega_c$: $\kappa = \sqrt{\omega_c^2 - \omega^2}/c =
0.18\,\mathrm{m}^{-1}$, depth $5.5\,\mathrm{m}$.

**Exercise 8.3 ★.**

Two tuning forks at $440\,\mathrm{Hz}$ and $444\,\mathrm{Hz}$: beat frequency, and the period of the loudness variations. Two water waves of wavelengths $100\,\mathrm{m}$ and $110\,\mathrm{m}$: frequencies, wavelength and speed of the envelope; compare with $v_g$ at $105\,\mathrm{m}$.

**Solution of Exercise 8.3.**

Beat at $4\,\mathrm{Hz}$, loudness period $0.25\,\mathrm{s}$. Water: $f = 0.125$ and $0.119\,\mathrm{Hz}$; envelope $\cos(\delta\omega t - \delta kx)$ with $\delta k = 2.9 \times 10^{-3}\,\mathrm{rad}/\mathrm{m}$ (beats $1.1\,\mathrm{km}$ apart) and speed $\delta\omega/\delta k = 6.4\,\mathrm{m}/\mathrm{s}$ — the [group velocity](#prop-b2-dispersion-wave-packets-group) at $105\,\mathrm{m}$, $\tfrac12\sqrt{g\lambda/2\pi} = 6.4\,\mathrm{m}/\mathrm{s}$.

**Exercise 8.4 ★.**

A [coaxial cable](#prop-b2-dispersion-wave-packets-coax): inner conductor $0.9\,\mathrm{mm}$ in diameter, outer $3.0\,\mathrm{mm}$, polyethylene ($\varepsilon_r = 2.3$). $\Lambda$, $\Gamma$, wave speed, [characteristic impedance](#prop-b2-dispersion-wave-packets-coax), delay per metre; length of cable equivalent to a $10\,\mathrm{ns}$ delay. Same cable with air instead of polyethylene.

**Solution of Exercise 8.4.**

$\ln(b/a) = 1.20$: $\Lambda = 0.24\,\text{µ}\mathrm{H}/\mathrm{m}$, $\Gamma = 106\,\mathrm{pF}/\mathrm{m}$, $c = 1/
\sqrt{\Lambda\Gamma} = 2.0 \times 10^{8}\,\mathrm{m}/\mathrm{s}$, $Z_c = 48\,\Omega$, $5.0\,\mathrm{ns}/\mathrm{m}$; $10\,\mathrm{ns}$ $\leftrightarrow$ $2.0\,\mathrm{m}$. Air: $\Gamma = 46\,\mathrm{pF}/\mathrm{m}$, $c = c_0$, $Z_c = 72\,\Omega$.

**Exercise 8.5 ★★.**

*The stiff string.* A real piano string obeys $\mu\partial_t^2y = T\partial_x^2y
- EI\,\partial_x^4y$ ($EI$ the bending stiffness). (a) [Dispersion relation](#def-b2-dispersion-wave-packets-relation), phase and group velocities. (b) For the A$_4$ string of [Chapter 6](https://one-course.com/books/physics/4/en/chapter/6-waves-on-strings-and-rods-the-dalembert-equation#ch-b2-waves-on-strings) ($T = 685\,\mathrm{N}$, $\mu = 6.1\,\mathrm{g}/\mathrm{m}$, $EI =
9.8 \times 10^{-3}\,\mathrm{N}\,\mathrm{m}^{2}$), relative increase of $v_\varphi$ for the eighth harmonic ($k = 8\pi/L$, $L = 0.38\,\mathrm{m}$); check against the inharmonicity found there. (c) Is the dispersion normal or anomalous? What does a sharp pluck look like after a few round trips?

**Solution of Exercise 8.5.**

(a) $\omega^2 = c^2k^2 + (EI/\mu)k^4$; $v_\varphi = \sqrt{c^2 + (EI/\mu)k^2}$; $v_g = (c^2k +
2(EI/\mu)k^3)/\omega$. (b) $c^2 = 1.12 \times 10^{5}\,\mathrm{m}^{2}/\mathrm{s}^{2}$, $k = 66\,\mathrm{rad}/\mathrm{m}$, $(EI/\mu)k^2
= 7.0 \times 10^{3}\,\mathrm{m}^{2}/\mathrm{s}^{2}$: $v_\varphi/c = \sqrt{1.063} = 1.031$, $+3.1\%$ — the $53$ cents of [Chapter 6](https://one-course.com/books/physics/4/en/chapter/6-waves-on-strings-and-rods-the-dalembert-equation#ch-b2-waves-on-strings). (c) $v_g > v_\varphi$: anomalous; the high harmonics run ahead and the attack of a pluck turns into a brief descending chirp.

**Exercise 8.6 ★★.**

*Damped string.* $\mu\partial_t^2y + \alpha\partial_ty = T\partial_x^2y$. (a) Derive the [dispersion relation](#def-b2-dispersion-wave-packets-relation). (b) For $\alpha \ll \mu\omega$, show $k' \approx \omega/c$ and $k'' \approx \alpha/2Z$; [attenuation](#def-b2-dispersion-wave-packets-complex) in [decibels](https://one-course.com/books/physics/4/en/chapter/7-sound-waves-in-fluids#def-b2-sound-waves-decibel) per metre. (c) Numbers: $\mu =
5\,\mathrm{g}/\mathrm{m}$, $T = 50\,\mathrm{N}$, $\alpha = 0.02\,\mathrm{kg}\,\mathrm{m}^{-1}\,\mathrm{s}^{-1}$, $f =
100\,\mathrm{Hz}$: check the approximation and give $\delta$. (d) Is the damped string dispersive? In which regime of $\alpha$ would it become so?

**Solution of Exercise 8.6.**

(a) $-\mu\omega^2 + \iu\alpha\omega = -Tk^2$: $\underline k^2 = (\omega^2/c^2)(1 - \iu\alpha/\mu\omega)$. (b) $\underline k \approx (\omega/c)(1 - \iu\alpha/2\mu\omega)$: $k' = \omega/c$, $k'' = \alpha/2\mu c =
\alpha/2Z$; $8.7k''$ dB/m. (c) $c = 100\,\mathrm{m}/\mathrm{s}$, $Z = 0.5\,\mathrm{kg}/\mathrm{s}$, $\mu\omega = 3.1 \gg
0.02$: $k'' = 0.02\,\mathrm{m}^{-1}$, $\delta = 50\,\mathrm{m}$, $0.17\,\mathrm{dB}/\mathrm{m}$. (d) Not in this limit ($k' = \omega/c$); when $\alpha \gtrsim \mu\omega$ the square root no longer linearizes and $k'$ depends non-linearly on $\omega$: dispersive (the diffusive limit of the cable).

**Exercise 8.7 ★★.**

*Ripples.* Short water waves are ruled by [surface tension](https://one-course.com/books/physics/4/en/chapter/4-viscous-flows#rem-b2-viscous-flows-capillarity): $\omega^2 =
\gamma k^3/\rho$ ($\gamma = 0.072\,\mathrm{N}/\mathrm{m}$); the full relation is $\omega^2 = gk +
\gamma k^3/\rho$. (a) $v_\varphi$ and $v_g$ for pure capillary waves; which is larger? (b) Show that the [phase velocity](#def-b2-dispersion-wave-packets-relation) of the full relation is minimal at $\lambda_m = 2\pi\sqrt{\gamma/\rho g}$ and compute $\lambda_m$ and $v_{\min}$. (c) A boat slower than $v_{\min}$ makes no wake: why? (d) Raindrops on a pond make rings whose crests lag behind the ring: in which regime?

**Solution of Exercise 8.7.**

(a) $v_\varphi = \sqrt{\gamma k/\rho}$, $v_g = \tfrac32v_\varphi$: the group outruns the crests. (b) $v_\varphi^2 = g/k + \gamma k/\rho$ is minimal at $k_m = \sqrt{\rho g/\gamma}$: $\lambda_m = 2\pi\sqrt{\gamma/\rho g} = 1.7\,\mathrm{cm}$, $v_{\min} = (4g\gamma/\rho)^{1/4} =
23\,\mathrm{cm}/\mathrm{s}$. (c) A wake is a pattern stationary with respect to the boat, made of waves whose [phase velocity](#def-b2-dispersion-wave-packets-relation) matches its speed; none is slower than $23\,\mathrm{cm}/\mathrm{s}$. (d) Capillary regime, $v_g > v_\varphi$: crests are born at the front of the ring and die at its back.

**Exercise 8.8 ★★.**

*Spreading.* A packet of initial width $\Delta x_0$ doubles its width after $t_{\text{sp}} \approx \Delta x_0^2/|\omega''(k_0)|$. (a) A group of [deep-water waves](#ex-b2-dispersion-wave-packets-examples), $\lambda = 50\,\mathrm{m}$, $\Delta x_0 = 500\,\mathrm{m}$. (b) A pulse of light in a glass fibre: $\omega'' \approx -0.16\,\mathrm{m}^{2}/\mathrm{s}$; a $10\,\mathrm{ps}$ pulse ($\Delta x_0 =
2\,\mathrm{mm}$): spreading time and the distance travelled meanwhile; the same for a $1\,\mathrm{ns}$ pulse. Why do long optical links use longer pulses or dispersion-compensating fibre?

**Solution of Exercise 8.8.**

(a) $k = 0.126\,\mathrm{rad}/\mathrm{m}$, $\omega'' = -\tfrac14\sqrt{g/k^3} = -18\,\mathrm{m}^{2}/\mathrm{s}$: $t_{\text{sp}}
= 2.5 \times 10^5/18 = 1.4 \times 10^{4}\,\mathrm{s}$, four hours. (b) $10\,\mathrm{ps}$: $t_{\text{sp}} =
4 \times 10^{-6}/0.16 = 25\,\text{µ}\mathrm{s}$, $5\,\mathrm{km}$ at $2 \times 10^{8}\,\mathrm{m}/\mathrm{s}$; $1\,\mathrm{ns}$: $0.04/0.16 =
0.25\,\mathrm{s}$, $50\,000\,\mathrm{km}$. Short pulses (high bit rates) spread within kilometres: long links use compensating fibre or moderate rates.

**Exercise 8.9 ★★.**

*The submarine cable.* $R = 3\,\Omega/\mathrm{km}$, $\Gamma = 0.3\,\text{µ}\mathrm{F}/\mathrm{km}$, $\Lambda$ and $G$ negligible. (a) Show that $v$ obeys a diffusion equation and give its diffusivity. (b) [Dispersion relation](#def-b2-dispersion-wave-packets-relation) $\underline k(\omega)$; [phase velocity](#def-b2-dispersion-wave-packets-relation) and [attenuation](#def-b2-dispersion-wave-packets-complex) length at $1\,\mathrm{Hz}$ and at $10\,\mathrm{Hz}$. (c) Time for a signal to "arrive" over $3000\,\mathrm{km}$ ($\sim R\Gamma L^2$); words per minute. (d) Heaviside’s cure: the inductance per kilometre needed for $R/\Lambda = G/\Gamma$ if $G = 1 \times 10^{-7}\,\mathrm{S}/\mathrm{km}$; why was it hard to add?

**Solution of Exercise 8.9.**

(a) $\partial_x^2v = R\Gamma\,\partial_tv$, $D = 1/R\Gamma = 1.1 \times 10^{12}\,\mathrm{m}^{2}/\mathrm{s}$. (b) $\underline k^2
= -\iu R\Gamma\omega$: $k' = k'' = \sqrt{R\Gamma\omega/2}$; $1\,\mathrm{Hz}$: $k = 1.7 \times 10^{-6}\,\mathrm{m}^{-1}$, $v_\varphi = 3.7 \times 10^{6}\,\mathrm{m}/\mathrm{s}$, $\delta = 600\,\mathrm{km}$; $10\,\mathrm{Hz}$: $v_\varphi = 1.2 \times 10^{7}\,\mathrm{m}/\mathrm{s}$, $\delta = 190\,\mathrm{km}$. (c) $R\Gamma L^2 = 9 \times 10^{-13} \times 9 \times 10^{12} = 8\,\mathrm{s}$ per dot: about a word a minute. (d) $\Lambda = R\Gamma/G = 9\,\mathrm{H}/\mathrm{km}$ — an absurd inductance for a submarine cable; loading coils worked on land telephone lines, whose $G$ and $R\Gamma$ were different.

**Exercise 8.10 ★★★.**

*Energy travels at the [group velocity](#prop-b2-dispersion-wave-packets-group).* String on an elastic bed: $\mu\partial_t^2y = T\partial_x^2y - Ky$, wave $y = A\cos(\omega t - kx)$ with $\omega^2 =
\omega_c^2 + c^2k^2$. (a) Energy density $e = \tfrac12\mu(\partial_ty)^2 + \tfrac12T
(\partial_xy)^2 + \tfrac12Ky^2$ and flux $\mathcal P = -T\partial_xy\,\partial_ty$: check $\partial_te + \partial_x\mathcal P = 0$. (b) Time averages $\langle e\rangle$ and $\langle\mathcal P\rangle$ for the wave. (c) Show that $\langle\mathcal P\rangle/
\langle e\rangle = c^2k/\omega = v_g$, not $v_\varphi$. (d) What happens to the energy flux below the cut-off?

**Solution of Exercise 8.10.**

(a) $\partial_te = \dot y(\mu\ddot y + Ky) + Ty'\dot y' = \dot yTy'' + Ty'\dot y' = \partial_x
(Ty'\dot y) = -\partial_x\mathcal P$. (b) $\langle e\rangle = \tfrac14A^2(\mu\omega^2 + Tk^2 + K) =
\tfrac12\mu\omega^2A^2$ (using $\mu\omega^2 = Tk^2 + K$); $\langle\mathcal P\rangle = \tfrac12Tk\omega
A^2$. (c) Ratio $Tk/\mu\omega = c^2k/\omega = v_g$. (d) $y = A\eu^{-\kappa x}\cos\omega t$: $\mathcal P \propto \sin2\omega t$, zero mean — no energy crosses the medium.

**Exercise 8.11 ★★★.**

*Swell and tsunami.* Water waves of wavelength $\lambda$ on a depth $h$ obey $\omega^2 = gk\tanh kh$ (admitted). (a) Recover the deep-water and the shallow-water ($kh \ll 1$) limits; speed of shallow-water waves; are they dispersive? (b) A tsunami of $\lambda = 200\,\mathrm{km}$ on $4\,\mathrm{km}$ of ocean: which limit, speed, period; time to cross $8000\,\mathrm{km}$. (c) Its amplitude offshore is $0.5\,\mathrm{m}$: energy per unit length of crest (admit $E = \tfrac12\rho gA^2$ per unit area) for a $1000\,\mathrm{km}$ front; what happens when the depth falls to $10\,\mathrm{m}$, if the energy flux $Ev_g$ is conserved (Green’s law, $A \propto h^{-1/4}$)? (d) A $100\,\mathrm{m}$ swell arriving on a beach: at what depth does it start to feel the bottom, and why do its crests turn parallel to the shore?

**Solution of Exercise 8.11.**

(a) $\tanh kh \to 1$: $\omega^2 = gk$; $kh \ll 1$: $\omega = \sqrt{gh}\,k$, speed $\sqrt{gh}$, non-dispersive. (b) $kh = 0.13$: shallow; $c = \sqrt{9.81 \times 4000} = 198\,\mathrm{m}/\mathrm{s}$, $T = 17\,\mathrm{min}$; $8000\,\mathrm{km}$ in $11\,\mathrm{h}$. (c) $\tfrac12\rho gA^2 = 1.2\,\mathrm{kJ}/\mathrm{m}^{2}$ over $200 \times 1000$ km$^2$: $2.5 \times 10^{14}\,\mathrm{J}$. Green: $A \propto h^{-1/4}$, $(4000/10)^{1/4}
= 4.5$: $2.2\,\mathrm{m}$, before steepening and run-up. (d) At $h \approx \lambda/2 =
50\,\mathrm{m}$; the part of a crest in shallower water slows down, so the crest swings round parallel to the depth contours — refraction.

**Exercise 8.12 ★★★.**

*A pulse on a line.* A $50\,\Omega$ cable of length $\ell = 100\,\mathrm{m}$ ($c = 2 \times 10^{8}\,\mathrm{m}/\mathrm{s}$) is fed at $x = 0$ by a generator of internal resistance $50\,\Omega$ and ended at $x = \ell$ by a load $R_L$. (a) Show that a wave arriving at the end is reflected with the voltage coefficient $r = (R_L - Z_c)/(R_L + Z_c)$ (write $v$ and $i$ as incident plus reflected waves and impose Ohm’s law in the load). (b) Values for $R_L = 50$, $0$, $\infty$, $150\,\Omega$. (c) A $1\,\mathrm{ns}$ pulse of $1\,\mathrm{V}$ is sent: what the oscilloscope at the generator shows, and when, for each load; why the generator’s $50\,\Omega$ matter. (d) A fault in a buried cable returns an echo after $1.8\,\text{µ}\mathrm{s}$: where is it? (Time-domain reflectometry.)

**Solution of Exercise 8.12.**

(a) At $x = \ell$: $v = v_i + v_r$, $i = (v_i - v_r)/Z_c$, $v = R_Li$: $r = (R_L -
Z_c)/(R_L + Z_c)$. (b) $0$, $-1$, $+1$, $0.5$. (c) The line’s input looks like $Z_c$, so the scope sees $0.5\,\mathrm{V}$ at $t = 0$, then the echo $0.5r$ V at $2\ell/c = 1\,\text{µ}\mathrm{s}$: nothing, $-0.5\,\mathrm{V}$, $+0.5\,\mathrm{V}$, $+0.25\,\mathrm{V}$; the matched generator absorbs the echo, so there is no second one. (d) $\tfrac12 \times 2 \times 10^8 \times 1.8 \times 10^{-6} = 180\,\mathrm{m}$.

## 8.5 Problem: The cable and the sea

**Problem 8.1.**

Weekend problem — dispersion at work: the telegraph cable that smeared the dots, the swell that announces a storm, the tsunami that crosses an ocean, and the packet that spreads

**Part I — The telegraph cable.** The 1866 transatlantic cable: $R = 2.0\,\Omega/\mathrm{km}$, $\Gamma = 0.25\,\text{µ}\mathrm{F}/\mathrm{km}$, $\Lambda = 0.5\,\mathrm{mH}/\mathrm{km}$, $G$ negligible, length $L = 3500\,\mathrm{km}$.

1. Write the [telegrapher’s equations](#prop-b2-dispersion-wave-packets-coax) with $R$ and derive the [dispersion relation](#def-b2-dispersion-wave-packets-relation) $\underline k^2 = \Gamma\omega(\Lambda\omega - \iu R)$ .
2. At what frequency does $\Lambda\omega$ equal $R$ ? Conclude that for telegraph frequencies (a few hertz) the line is in the diffusive regime.
3. In that regime, $\underline k = \sqrt{-\iu R\Gamma\omega}$ : give $k'$ and $k''$ (recall $\sqrt{-\iu} = (1 - \iu)/\sqrt2$ ); [phase velocity](#def-b2-dispersion-wave-packets-relation) and [attenuation](#def-b2-dispersion-wave-packets-complex) length at $1\,\mathrm{Hz}$ ; at $4\,\mathrm{Hz}$ .
4. [Attenuation](#def-b2-dispersion-wave-packets-complex) in [decibels](https://one-course.com/books/physics/4/en/chapter/7-sound-waves-in-fluids#def-b2-sound-waves-decibel) over the whole cable at $1\,\mathrm{Hz}$ and at $4\,\mathrm{Hz}$ . Which frequency survives? What does that do to a sharp dot?
5. Kelvin’s estimate of the transit time, $R\Gamma L^2$ ; the practical rate of the cable in words per minute if a word is about $10$ dots and dashes and the dots must not overlap.
6. Heaviside proposed to raise $\Lambda$ : what value makes $\Lambda\omega = R$ at $4\,\mathrm{Hz}$ , and what would the speed and [attenuation](#def-b2-dispersion-wave-packets-complex) then be (take $G = 0$ , so the condition $R/\Lambda = G/\Gamma$ cannot be met exactly — use the general relation at that frequency)?
7. A modern optical fibre carries $1 \times 10^{10}\,\mathrm{bit}/\mathrm{s}$ : compare with question 5, in powers of ten.

**Part II — The swell.** Deep water: $\omega^2 = gk$.

8. Phase and [group velocity](#prop-b2-dispersion-wave-packets-group) as functions of the period $T$ .
9. A storm $4000\,\mathrm{km}$ from a coast generates waves of periods from $6\,\mathrm{s}$ to $18\,\mathrm{s}$ : arrival time of each extreme; duration of the swell event.
10. Show that at a coast at distance $D$ the arriving period obeys $1/T = gt/4\pi D$ , $t$ counted from the storm: $1/T$ grows linearly in time. A buoy records $T = 16\,\mathrm{s}$ one day at noon and $T = 12\,\mathrm{s}$ exactly $24\,\mathrm{h}$ later: distance of the storm.
11. Energy per unit area of a swell of amplitude $A$ is $\tfrac12\rho  gA^2$ ; power crossing a line of $1\,\mathrm{m}$ of crest, for $A =  1.5\,\mathrm{m}$ , $T = 12\,\mathrm{s}$ . (Use the [group velocity](#prop-b2-dispersion-wave-packets-group) — why?)
12. A wave-energy farm of $1\,\mathrm{km}$ of front with $30\%$ efficiency: electrical power; compare with a wind turbine ( $3\,\mathrm{MW}$ ).
13. A boat at $5\,\mathrm{m}/\mathrm{s}$ on deep water leaves a stationary wake behind it: which wavelength has a [phase velocity](#def-b2-dispersion-wave-packets-relation) equal to the boat’s speed and so follows it? Why does the pattern trail in a V, narrower than the crests’ own spreading would suggest (think of $v_g = v_\varphi/2$ )?
14. Watching one group of swell from a cliff, describe what an individual crest does during the time the group passes a fixed buoy; how many crests does a group of duration $2\,\mathrm{min}$ contain at $T = 12\,\mathrm{s}$ , and how many crest "lifetimes" does that represent?

**Part III — The tsunami.** $\omega^2 = gk\tanh kh$; Pacific depth $h = 4.0\,\mathrm{km}$.

15. An earthquake lifts a sea-floor patch of $100\,\mathrm{km}$ by $2\,\mathrm{m}$ ; the wave has $\lambda \approx 200\,\mathrm{km}$ : check that it is a shallow-water wave and give its speed, period and the time to reach a coast $6000\,\mathrm{km}$ away.
16. Is it dispersive? Compare with the swell of Part II: does it arrive before or after the swell would? Which arrives as a single pulse?
17. Its height offshore is $0.6\,\mathrm{m}$ : energy per square metre; total energy of a $500\,\mathrm{km}$ -long front spread over one wavelength; compare with a $1\,\mathrm{Mt}$ explosion ( $4.2 \times 10^{15}\,\mathrm{J}$ ).
18. Green’s law: height when the depth falls to $20\,\mathrm{m}$ near the shore (energy flux $\tfrac12\rho gA^2v_g$ conserved, $v_g = \sqrt{gh}$ ); speed and wavelength there; why the wave then steepens and breaks like a wall.
19. A ship at sea does not notice the tsunami passing: why?
20. Tide gauges $6000\,\mathrm{km}$ apart record the wave: what the travel time gives, and why the measured speed is slightly less than $\sqrt{gh}$ over an ocean of varying depth.

**Part IV — The spreading packet.**

21. For deep water, compute $\omega''(k) = \dd^2\omega/\dd k^2$ and its value for $\lambda = 100\,\mathrm{m}$ .
22. A packet of $10\,$ such waves ( $\Delta x_0 \approx 1\,\mathrm{km}$ ) is released: spreading time $\Delta x_0^2/|\omega''|$ ; distance travelled by the packet meanwhile.
23. Explain in words why a packet with a narrower spectrum (longer train) spreads more slowly, and why the tsunami of Part III barely spreads at all.
24. The same estimate for an electron of wavelength $0.1\,\mathrm{nm}$ localized within $1\,\mathrm{nm}$ , with $\omega = \hbar k^2/2m$ ( $\hbar =  1.05 \times 10^{-34}\,\mathrm{J}\,\mathrm{s}$ , $m = 9.1 \times 10^{-31}\,\mathrm{kg}$ ): spreading time and the distance travelled — the first contact with the quantum [wave packet](#prop-b2-dispersion-wave-packets-group) of [Chapter 30](https://one-course.com/books/physics/4/en/chapter/30-the-schrodinger-equation-and-wave-functions#ch-b2-schrodinger-wave-functions) .
25. Sum up in a table: for the cable, the swell, the tsunami and the electron, the [dispersion relation](#def-b2-dispersion-wave-packets-relation) , $v_\varphi$ , $v_g$ and whether the medium spreads or merely attenuates.

**Solution of Problem 8.1.**

**1.** $\partial_xv = -\Lambda\partial_ti - Ri$, $\partial_xi = -\Gamma\partial_tv$; with $\eu^{\iu(\omega t - kx)}$: $\iu kv = (\iu\omega\Lambda + R)i$, $\iu ki = \iu\omega\Gamma v$, hence $k^2 = \omega\Gamma(\omega\Lambda - \iu R)$.

**2.** $\omega = R/\Lambda = 4000\,\mathrm{rad}/\mathrm{s}$, $640\,\mathrm{Hz}$: at a few hertz $\Lambda\omega
\ll R$, diffusive.

**3.** $k' = k'' = \sqrt{R\Gamma\omega/2}$ with $R\Gamma = 5 \times 10^{-13}\,\mathrm{s}/\mathrm{m}^{2}$: $1\,\mathrm{Hz}$: $1.25 \times 10^{-6}\,\mathrm{m}^{-1}$, $v_\varphi = 5 \times 10^{6}\,\mathrm{m}/\mathrm{s}$, $\delta = 800\,\mathrm{km}$; $4\,\mathrm{Hz}$: $2.5 \times 10^{-6}\,\mathrm{m}^{-1}$, $1 \times 10^{7}\,\mathrm{m}/\mathrm{s}$, $400\,\mathrm{km}$.

**4.** $8.7L/\delta$: $38\,\mathrm{dB}$ at $1\,\mathrm{Hz}$, $76\,\mathrm{dB}$ at $4\,\mathrm{Hz}$: only the slowest components survive; a sharp dot arrives as a long, low hump.

**5.** $R\Gamma L^2 = 5 \times 10^{-13} \times 1.2 \times 10^{13} = 6\,\mathrm{s}$: a dot every six seconds, a word a minute (the real cable, with sensitive mirror galvanometers, managed a few).

**6.** $\Lambda = R/\omega = 80\,\mathrm{mH}/\mathrm{km}$. Then $k^2 = \Gamma R\omega(1 - \iu)$, $k = \sqrt{\Gamma R\omega}
\,2^{1/4}\eu^{-\iu\pi/8}$: $k' = 3.9 \times 10^{-6}\,\mathrm{m}^{-1}$, $k'' = 1.6 \times 10^{-6}\,\mathrm{m}^{-1}$: $v_\varphi =
6.4 \times 10^{6}\,\mathrm{m}/\mathrm{s}$, $\delta = 620\,\mathrm{km}$, $49\,\mathrm{dB}$ over the cable instead of $76\,\mathrm{dB}$.

**7.** A dot every $6\,\mathrm{s}$ is about $0.2\,\mathrm{bit}/\mathrm{s}$: the fibre is $10^{10}$ to $10^{11}$ times faster.

**8.** $v_\varphi = gT/2\pi$, $v_g = gT/4\pi$.

**9.** $18\,\mathrm{s}$: $v_g = 14.1\,\mathrm{m}/\mathrm{s}$, $4 \times 10^6/14.1 = 3.3\,\mathrm{days}$; $6\,\mathrm{s}$: $4.7\,\mathrm{m}/\mathrm{s}$, $9.9\,\mathrm{days}$: the swell lasts more than six days.

**10.** $t = D/v_g = 4\pi D/gT$, so $1/T = gt/4\pi D$. $\Delta(1/T) = 1/12 - 1/16
= 0.0208\,\mathrm{Hz}$ in $86\,400\,\mathrm{s}$: $g/4\pi D = 2.4 \times 10^{-7}\,\mathrm{s}^{-2}$, $D = 3200\,\mathrm{km}$.

**11.** $\tfrac12\rho gA^2 = 11\,\mathrm{kJ}/\mathrm{m}^{2}$; the energy moves at $v_g = gT/4\pi =
9.4\,\mathrm{m}/\mathrm{s}$: $103\,\mathrm{kW}$ per metre of crest.

**12.** $10^3 \times 103 \times 0.3 = 31\,\mathrm{MW}$: ten wind turbines.

**13.** $\lambda = 2\pi U^2/g = 16\,\mathrm{m}$. The energy of each wave component travels at half its phase speed, so the pattern is confined to a wedge behind the boat (half-angle $19.5{}^{\circ}$, whatever its speed).

**14.** A crest is born at the rear of the group, overtakes it (crests move at $2v_g$ relative to the water, $v_g$ relative to the group) and dies at the front. The group is $120 \times 9.4 = 1.1\,\mathrm{km}$ long, five wavelengths ($\lambda = gT^2/2\pi = 225\,\mathrm{m}$); ten crests pass the buoy, each having lived about two minutes.

**15.** $kh = 0.13$: shallow. $c = 198\,\mathrm{m}/\mathrm{s}$, $T = \lambda/c = 17\,\mathrm{min}$, $8.4\,\mathrm{h}$.

**16.** Non-dispersive: all components at $198\,\mathrm{m}/\mathrm{s}$, arriving together as one pulse (the swell, at $\sim10\,\mathrm{m}/\mathrm{s}$, sorted by period, would take a week).

**17.** $\tfrac12\rho gA^2 = 1.8\,\mathrm{kJ}/\mathrm{m}^{2}$ over $500 \times 200$ km$^2$: $1.8 \times 10^{14}\,\mathrm{J}$, $4\%$ of a megatonne.

**18.** $A = 0.6(4000/20)^{1/4} = 2.3\,\mathrm{m}$; $c = \sqrt{gh} = 14\,\mathrm{m}/\mathrm{s}$, $\lambda =
cT = 14\,\mathrm{km}$. The crest, in deeper water than the trough ahead of it, travels faster and catches up: the front steepens into a bore.

**19.** A rise of $0.6\,\mathrm{m}$ over a quarter of an hour, on a slope of $10^{-5}$: nothing to feel.

**20.** The mean speed, $\sqrt{gh}$ for the mean depth; since $\sqrt h$ is concave, shallow stretches slow the wave more than deep ones speed it up, and the path average is below $\sqrt{g\bar h}$.

**21.** $\omega' = \tfrac12\sqrt{g/k}$, $\omega'' = -\tfrac14\sqrt{g/k^3}$; $k = 0.063\,\mathrm{rad}/\mathrm{m}$: $\omega'' = -50\,\mathrm{m}^{2}/\mathrm{s}$.

**22.** $t_{\text{sp}} = 10^6/50 = 2 \times 10^{4}\,\mathrm{s}$, $5.6\,\mathrm{h}$; the packet has moved $v_gt = 6.25 \times 2 \times 10^4 = 125\,\mathrm{km}$.

**23.** A long train contains a narrow band of $k$, whose group velocities nearly coincide: the components stay together longer. The tsunami’s medium is non-dispersive: equal speeds, no spreading.

**24.** $\omega'' = \hbar/m = 1.2 \times 10^{-4}\,\mathrm{m}^{2}/\mathrm{s}$: $t_{\text{sp}} = 10^{-18}/1.2 \times
10^{-4} = 9 \times 10^{-15}\,\mathrm{s}$; $v_g = \hbar k/m = 7.3 \times 10^{6}\,\mathrm{m}/\mathrm{s}$: $60\,\mathrm{nm}$.

**25.** Cable: $k^2 = -\iu R\Gamma\omega$, $v_\varphi = \sqrt{2\omega/R\Gamma}$, no clean $v_g$, attenuates and smears. Swell: $\omega^2 = gk$, $v_\varphi = 2v_g$, spreads. Tsunami: $\omega = \sqrt{gh}\,k$, $v_\varphi = v_g$, neither. Electron: $\omega = \hbar k^2/2m$, $v_g = 2v_\varphi$, spreads.