# Waves & the Wave Equation
> Before quantum mechanics was mechanics, it was *wave* mechanics — and every wave idea Schrödinger
> needed was already a century old. Superposition, interference, and frequencies quantized by
> boundary conditions all live on an ordinary stretched string. Master the classical wave equation
> now, and the "weird" quantum features of Term 1 turn out to be old friends wearing new collars.
## Learning Objectives
After this lesson you will be able to:
1. Define a **mechanical wave** and derive the **one-dimensional wave equation** from Newton's second law applied to a stretched string.
2. Prove and apply **d'Alembert's solution** $f(x-vt) + g(x+vt)$, reading off propagation speed and direction.
3. Describe **harmonic waves** via amplitude, **wavenumber** $k$, **angular frequency** $\omega$, and the **dispersion relation** $\omega = vk$, in real and **complex-exponential** form.
4. Predict **constructive/destructive interference** from **path difference**, and derive the **beat** envelope of two nearby frequencies.
5. Derive the **normal modes** of a string with fixed ends and explain **quantization by boundary conditions** — the classical ancestor of quantum energy levels.
6. Recognize the same equation in **transverse** and **longitudinal** waves (string, elastic rod, gas column) with $v^2$ = stiffness/inertia, and distinguish the **medium velocity** $\partial_t y$ from the wave speed.
---
## Intuition
Drop a pebble in a pond, or flick the end of a rope: a *pattern* travels while the medium only
jiggles locally. That is a wave — a propagating disturbance carrying energy and momentum **without
transporting matter**. Every water molecule (and every fan in a stadium wave) returns to where it
started; the shape moves on.
Three structural facts about the wave equation matter more than any single solution. It is
**linear**, so solutions add — superposition and interference. It is **second order in time**, so
initial shape *and* initial velocity must be supplied. And **confining** a wave — pinning a string
at both ends — turns a continuum of possible frequencies into a discrete ladder. Replace "string
displacement" by "wavefunction" and these become quantum superposition, quantum dynamics, and
quantized energy levels. This lesson is Term 1 in disguise.
---
## Theory
### The wave equation from Newton's second law
A **mechanical wave** is a disturbance of a deformable medium that propagates through it, driven by
the medium's inertia (it overshoots equilibrium) and its restoring force (it gets pulled back). On
a string the disturbance is the **transverse displacement** $y(x,t)$: each point moves
perpendicular to the propagation direction (sound in air is *longitudinal*; the math is the same).
The state of the medium is a whole *function* $y(x,t)$ — our first infinite-dimensional system.
Take a string of uniform **linear mass density** $\mu$ (kg/m) under uniform **tension** $T$ (N),
displaced slightly from the $x$-axis. The element between $x$ and $x+\Delta x$ has mass
$\mu\,\Delta x$. Tension pulls tangentially at both ends, so with $\theta(x)$ the local angle to
the horizontal, the net transverse force is $F_y = T\sin\theta(x{+}\Delta x) - T\sin\theta(x)$.
For small slopes ($|\partial y/\partial x| \ll 1$) we set
$\sin\theta \approx \tan\theta = \partial y/\partial x$, and the horizontal components cancel to first order (tension stays
uniform; no horizontal acceleration). Then, by the definition of the derivative of
$\partial y/\partial x$,
$$
F_y \approx T\left[\frac{\partial y}{\partial x}\bigg|_{x+\Delta x} -
\frac{\partial y}{\partial x}\bigg|_{x}\right] \approx T\,\frac{\partial^2 y}{\partial x^2}\,\Delta x .
$$
Newton's second law $F_y = (\mu\,\Delta x)\,\partial^2y/\partial t^2$ gives, cancelling $\Delta x$,
$$
\boxed{\;\frac{\partial^2 y}{\partial t^2} = \frac{T}{\mu}\frac{\partial^2 y}{\partial x^2}
= v^2\,\frac{\partial^2 y}{\partial x^2}, \qquad v = \sqrt{T/\mu}\;}
$$
— the **1D wave equation**. The speed is set entirely by the medium: stiffer restoring force
(larger $T$) means faster; more inertia (larger $\mu$) means slower.
### One equation, many media
The string is one instance of a structure that recurs across physics. Waves are first sorted by how
the medium moves relative to the direction of travel. In a *transverse wave* the disturbance is
perpendicular to the propagation direction: the string, shear waves in a solid, and light, whose
fields point across its direction of travel. (Surface waves on water are neither: each parcel of
water moves on a near-circular orbit, mixing both.) In a *longitudinal wave* the disturbance lies
*along* the propagation direction, so the medium is alternately compressed and rarefied: sound in a
gas, compression waves in a rod. The mathematics does not care which. In each case Newton's law on
a small element pits a *stiffness* (the restoring force per unit deformation) against an *inertia*,
and returns the same equation with $v^2 = \text{stiffness}/\text{inertia}$.
**Elastic rod.** Let $\xi(x,t)$ be the longitudinal displacement of the cross-section that sits at
$x$ in equilibrium, in a thin rod of density $\rho$, cross-sectional area $A$ and *Young's modulus*
$Y$. The rod is locally stretched by the strain $\partial\xi/\partial x$, so by Hooke's law the force
transmitted across the section at $x$ is $F = YA\,\partial\xi/\partial x$. The slice between $x$ and
$x+\Delta x$ has mass $\rho A\,\Delta x$ and feels the net force
$F(x+\Delta x) - F(x) \approx YA\,(\partial^2\xi/\partial x^2)\,\Delta x$, hence
$\partial_t^2\xi = (Y/\rho)\,\partial_x^2\xi$: the wave equation again, with
$v_{\text{rod}} = \sqrt{Y/\rho}$. For steel ($Y \approx 2.0\times10^{11}$ Pa,
$\rho \approx 7.85\times10^3$ kg/m³) this is about $5.0$ km/s, fifteen times faster than sound in
air.
**Gas column.** In a pipe of gas with equilibrium pressure $P_0$ and density $\rho_0$, the same
displacement field changes the density by $\delta\rho = -\rho_0\,\partial\xi/\partial x$ and hence the
pressure by $\delta P = \kappa\,\delta\rho/\rho_0 = -\kappa\,\partial\xi/\partial x$, where
$\kappa \equiv \rho_0\,dP/d\rho$ is the gas's *bulk modulus*. The pressure difference across a slab
accelerates it, $\rho_0\,\partial_t^2\xi = -\partial_x(\delta P) = \kappa\,\partial_x^2\xi$,
so $\partial_t^2\xi = (\kappa/\rho_0)\,\partial_x^2\xi$ with $v_{\text{gas}} = \sqrt{\kappa/\rho_0}$.
Differentiating once in $x$ shows that $\delta\rho$ and $\delta P$ obey the same equation, which is
why sound is equally well called a pressure wave or a density wave. The physics hides in $\kappa$.
Newton (1687) took the compressions to be isothermal, $\kappa = P_0$, and got about 290 m/s for
air: some 15% too slow. Laplace (1816) saw that acoustic oscillations are far too fast for heat to
flow between compressed and rarefied regions, so the compressions are *adiabatic*,
$\kappa = \gamma P_0$ with $\gamma = 1.40$ for air. Then
$v = \sqrt{1.40\times1.013\times10^{5}/1.20} \approx 344$ m/s at 20 °C, the measured value.
| Medium | Wave | Stiffness | Inertia | $v$ |
| --- | --- | --- | --- | --- |
| String | transverse | tension $T$ | linear density $\mu$ | $\sqrt{T/\mu}$ |
| Thin rod | longitudinal | Young's modulus $Y$ | density $\rho$ | $\sqrt{Y/\rho}$ |
| Gas | longitudinal | bulk modulus $\kappa = \gamma P_0$ | density $\rho_0$ | $\sqrt{\gamma P_0/\rho_0}$ |
Light belongs on no row of this table: it needs *no medium at all*. Maxwell's equations hand the
same wave equation to the electric and magnetic fields themselves, with $v = c$ in empty space
([P.2.1](../c02-radiation-old-quantum/01-blackbody-radiation.md)). What every wave transports is
energy and momentum; what none transports is the medium.
### d'Alembert's general solution
**Claim:** $y = f(x-vt) + g(x+vt)$ solves the wave equation for *any* twice-differentiable $f,g$,
and every solution has this form. *Direct check:* with $u = x-vt$, the chain rule gives
$\partial_t^2 f(u) = v^2f''(u)$ and $\partial_x^2 f(u) = f''(u)$, so $\partial_t^2 f = v^2\partial_x^2 f$ ✓
(same for $g$; linearity lets us add). *Generality:* in the variables $u = x-vt$, $w = x+vt$ one
has $\partial_x = \partial_u + \partial_w$ and $\partial_t = -v\partial_u + v\partial_w$, so
$$
\partial_t^2 y - v^2\partial_x^2 y
= v^2(\partial_u - \partial_w)^2 y - v^2(\partial_u + \partial_w)^2 y
= -4v^2\,\frac{\partial^2 y}{\partial u\,\partial w} = 0 ,
$$
whence $\partial y/\partial u$ is independent of $w$; integrating, $y = f(u) + g(w)$. The graph of
$f(x-vt)$ is the fixed shape $f(x)$ translated rigidly **rightward** at speed $v$ (its argument is
constant along $x = vt + \text{const}$); $g(x+vt)$ moves left.
The catch is the single $v$: a sum of two movers solves *this* equation only when both travel at
the one speed the equation owns. Give them unequal speeds and the residual
$\partial_z^2\phi - \partial_t^2\phi/v^2$ refuses to vanish for any choice of $v$ (the demo writes the
string coordinate as $z$):
```qm-demo traveling-wave-superposition
```
### Harmonic waves, dispersion, and phase vs group velocity
The workhorse solution is the **harmonic wave** $y = A\cos(kx - \omega t + \varphi)$: **amplitude**
$A$, **wavenumber** $k = 2\pi/\lambda$ (**wavelength** $\lambda$), **angular frequency**
$\omega = 2\pi\nu = 2\pi/\tau$ (**frequency** $\nu$, **period** $\tau$), phase constant $\varphi$.
It has the form $f(x-vt)$ only if $kx - \omega t = k(x - vt)$, i.e. $\omega = vk$, equivalently
$\lambda\nu = v$. A relation $\omega(k)$ is a **dispersion relation**; the string's is *linear*,
so all harmonic waves share one speed and arbitrary shapes propagate undistorted.
Plot $|\omega|$ against $k$ and the string's dispersion relation is a straight line through the
origin with slope $v$, while each harmonic wave is a single point. Two waves obey the same wave
equation exactly when both points land on one line — a condition on the ratio $|\omega|/k$ and
nothing else:
```qm-demo dispersion-line-check
```
In a **dispersive** medium $\omega(k)$ is nonlinear: crests move at the **phase velocity**
$v_p = \omega/k$ while a localized packet (and its energy) moves at the **group velocity**
$v_g = d\omega/dk$, and packets spread. Matter waves will prove *maximally* dispersive
($\omega \propto k^2$) — full treatment in
[P.4.3](../c04-schrodinger-equation/03-matter-waves-uncertainty-momentum.md).
A third velocity is easily confused with these two. The *medium velocity* $u = \partial y/\partial t$
is how fast one bit of the medium moves as the wave passes, not how fast the pattern moves. For the
harmonic wave, $u = \omega A\sin(kx - \omega t + \varphi)$ oscillates with amplitude $\omega A$, set by
the amplitude and frequency and unrelated to $v$. The two are linked only through the slope: for any
right-mover $y = f(x - vt)$ the chain rule gives $\partial_t y = -v f' = -v\,\partial_x y$, so
$u/v = -\partial y/\partial x$. The small-slope assumption behind the wave equation,
$|\partial_x y| \ll 1$, is therefore the statement that the medium moves slowly compared with the wave. A loud 1 kHz
tone in air moves the air at a few mm/s while the sound itself travels at 344 m/s (E6).
### The complex representation
Following [0.3.1](../../term-0-foundations/c03-fourier-methods/01-complex-numbers-functions.md), write
$y = \operatorname{Re}[\tilde A\,e^{i(kx-\omega t)}]$, the complex amplitude $\tilde A = Ae^{i\varphi}$
packaging amplitude *and* phase. Derivatives become multiplications ($\partial_x \to ik$,
$\partial_t \to -i\omega$) and adding waves becomes adding complex numbers. Because the wave
equation is linear with real coefficients, we may compute with $e^{i(kx-\omega t)}$ throughout and
take the real part at the end. For the string this is bookkeeping; in quantum mechanics the complex
exponential becomes the physical object itself — no real part taken (P.4.1).
### Superposition and interference
The operator $\mathcal L = \partial_t^2 - v^2\partial_x^2$ is linear: $\mathcal Ly_1 = \mathcal Ly_2 = 0$
implies $\mathcal L(\alpha y_1 + \beta y_2) = 0$. **Any linear combination of solutions is a
solution** — the superposition principle. Superposing two equal-amplitude waves that differ by a
phase $\delta$ gives, by the sum-to-product identity,
$$
A\cos(kx - \omega t) + A\cos(kx - \omega t + \delta)
= 2A\cos\tfrac{\delta}{2}\,\cos\!\big(kx - \omega t + \tfrac{\delta}{2}\big).
$$
The resultant amplitude $2A|\cos(\delta/2)|$ ranges from $2A$ (**constructive**, $\delta = 2m\pi$)
to $0$ (**destructive**, $\delta = (2m{+}1)\pi$). When the offset comes from two in-phase sources
at different distances, $\delta = k\Delta = 2\pi\Delta/\lambda$ with $\Delta$ the **path
difference**: constructive at $\Delta = m\lambda$, destructive at $\Delta = (m+\tfrac12)\lambda$.
This classical superposition of waves is the direct ancestor of the quantum superposition of
states in [the State Postulate](../../term-1-quantum-mechanics/c01-postulates/01-state-postulate.md):
the same linearity, with amplitudes reinterpreted as probability amplitudes. In three dimensions the
same sum acquires a *direction*, and interference becomes a pattern painted across space rather than
a value at a point — that generalization, and the diffraction and grating physics it unlocks, is
[P.1.2](02-wave-propagation-diffraction.md).
> **Caution.** Superposition adds **amplitudes**, never intensities. Intensity goes as amplitude
> *squared*: two equal in-phase waves give $4I_0$, not $2I_0$; out of phase they give $0$. The
> cross term $2A_1A_2\cos\delta$ *redistributes* energy from dark fringes to bright ones — the
> fringe-averaged intensity is still $2I_0$. The same "add amplitudes, then square" rule governs
> quantum probability amplitudes, where forgetting it erases every interference effect.
### Beats
Superpose equal amplitudes at *slightly different* frequencies, at a fixed point:
$$
y(t) = A\cos\omega_1 t + A\cos\omega_2 t
= 2A\cos\!\Big(\frac{\omega_1 - \omega_2}{2}t\Big)\cos\!\Big(\frac{\omega_1 + \omega_2}{2}t\Big).
$$
The second factor oscillates fast at the mean frequency; the first is a slow **envelope**
$2A\cos(\Delta\omega\,t/2)$. A detector senses intensity $\propto \cos^2(\Delta\omega\,t/2)$,
which peaks *twice* per envelope period — loudness pulses at the **beat frequency**
$\nu_{\text{beat}} = |\nu_1 - \nu_2|$.
### Standing waves: quantization by boundary conditions
Pin the string at both ends: $y(0,t) = y(L,t) = 0$. Seek **normal modes** — motions where every
point oscillates at one common frequency — with the separation ansatz $y = X(x)\phi(t)$.
Substituting and dividing by $X\phi$ gives $\ddot\phi/(v^2\phi) = X''/X$; a function of $t$ alone
equals a function of $x$ alone, so both equal a constant, written $-k^2$ (a positive constant
gives real exponentials that cannot vanish at both ends). Then $X = a\sin kx + b\cos kx$; the
condition $X(0) = 0$ kills $b$, and $X(L) = 0$ forces $\sin kL = 0$:
$$
k_n = \frac{n\pi}{L}, \qquad
\boxed{\;\nu_n = \frac{\omega_n}{2\pi} = \frac{nv}{2L}\;} \qquad n = 1, 2, 3, \dots
$$
Each mode is $y_n = \sin(k_nx)[A_n\cos\omega_nt + B_n\sin\omega_nt]$ with $\omega_n = vk_n$; the
general motion is the superposition $y = \sum_n y_n$, coefficients fixed by initial shape and
velocity via Fourier analysis
([0.3.2](../../term-0-foundations/c03-fourier-methods/02-fourier-series-transforms.md)). Equivalently a
standing wave is two counter-propagating travelers:
$A\sin(kx - \omega t) + A\sin(kx + \omega t) = 2A\sin kx\cos\omega t$.
Drag $k$ and let the boundary conditions do the quantizing. The left end is pinned for free; the
right end lands on its wall only when $kL$ is a whole multiple of $\pi$:
```qm-demo standing-wave-modes
```
This is the most important idea of the lesson: **boundary conditions quantize frequencies** —
nothing quantum happened; a pinned string simply cannot vibrate at arbitrary frequencies. When de
Broglie turns electrons into waves, fitting a whole number of wavelengths around an orbit
quantizes the atom ([P.3.2](../c03-wave-particle-duality/02-matter-as-waves.md)), and confining a
matter wave in a box quantizes its energy — the infinite square well of
[P.5.2](../c05-bound-states/02-step-well-barrier.md) is *this calculation* with a new dispersion
relation. Moreover, each classical normal mode — a frozen spatial shape at a single frequency — is
the direct ancestor of a quantum **stationary state** $\psi_n(x)\,e^{-iE_nt/\hbar}$ with
$E_n = \hbar\omega_n$ ([P.4.2](../c04-schrodinger-equation/02-stationary-states-superposition.md)).
A real pluck is not one mode but a sum of them, $y(x, 0) = \sum_n b_n \sin(n\pi x/L)$, with the
amplitudes $b_n$ set by the starting shape; finding them is a Fourier series
([Fourier Series & Transforms](../../term-0-foundations/c03-fourier-methods/02-fourier-series-transforms.md)).
Below, a string plucked a third of the way along, built mode by mode; the plot shows the string on
the right half and its mirror image on the left, which is how a sine series sees it. Every third
mode is missing: each has a node exactly where the string was pulled.
```qm-demo fourier-synthesis
{ "shape": "plucked", "harmonics": 6 }
```
---
## Worked Examples
### Example 1 — Tuning a guitar string
A guitar's A-string has $L = 0.650$ m and $\mu = 4.00$ g/m $= 4.00\times10^{-3}$ kg/m. What tension
tunes its fundamental to $\nu_1 = 110$ Hz? From $\nu_1 = v/2L$: $v = 2L\nu_1 = 2(0.650)(110) = 143$
m/s, so $T = \mu v^2 = (4.00\times10^{-3})(143)^2 \approx 81.8$ N. The overtones sit at
$\nu_2 = 220$ Hz, $\nu_3 = 330$ Hz, … with $\lambda_n = 2L/n = 1.30,\ 0.650,\ 0.433$ m — an evenly spaced
frequency ladder, courtesy of the linear dispersion relation. ✓
### Example 2 — Two speakers and a path difference
Two speakers driven in phase emit $\nu = 686$ Hz in air ($v = 343$ m/s), so
$\lambda = v/\nu = 0.500$ m. A listener stands $r_1 = 3.00$ m from one speaker, $r_2 = 3.75$ m from the other. Path
difference $\Delta = 0.75$ m $= 1.5\lambda$: **destructive** — a dead spot. Stepping sideways to
$\Delta = 0.50$ m $= 1.0\lambda$: **constructive**, intensity $4I_0$. At an intermediate point
with $\Delta = 0.125$ m, $\delta = 2\pi\Delta/\lambda = \pi/2$, so the amplitude is
$2A\cos(\pi/4) = \sqrt2\,A$ and $I = 2I_0$ — checking the energy bookkeeping of the Caution above.
---
## Hands-on (Python)
```python
import numpy as np
import matplotlib.pyplot as plt
# --- 1. Two-wave interference (snapshots) and beats ---
x = np.linspace(0, 4, 1000) # metres
A, k = 1.0, 2*np.pi # lambda = 1 m
fig, ax = plt.subplots(2, 1, figsize=(8, 6))
for delta in (0.0, np.pi/2, np.pi):
ax[0].plot(x, A*np.cos(k*x) + A*np.cos(k*x + delta), label=f"delta={delta:.2f}")
ax[0].set(title="Two equal waves, t = 0", xlabel="x (m)"); ax[0].legend()
# Expected: amplitudes 2, 1.41, ~0 -> matches 2A|cos(delta/2)|
t = np.linspace(0, 2, 4000) # seconds
nu1, nu2 = 220.0, 224.0 # Hz -> 4 Hz beat
y = np.cos(2*np.pi*nu1*t) + np.cos(2*np.pi*nu2*t)
env = 2*np.cos(np.pi*(nu1 - nu2)*t) # slow envelope
ax[1].plot(t, y, lw=0.5); ax[1].plot(t, env, "r", t, -env, "r")
ax[1].set(title="Beats: 220 Hz + 224 Hz", xlabel="t (s)")
plt.tight_layout(); plt.show()
# Expected: 8 loudness pulses in 2 s = |nu1 - nu2| = 4 Hz beat frequency.
```
```python
import numpy as np
import matplotlib.pyplot as plt
# --- 2. Standing wave from counter-propagating travelers ---
L, v, n = 1.0, 1.0, 3
k = n*np.pi/L; w = v*k; tau = 2*np.pi/w # mode-3 parameters
x = np.linspace(0, L, 500)
for frac in np.arange(0, 1.0, 0.125): # 8 snapshots over one period
t = frac*tau
plt.plot(x, np.sin(k*x - w*t) + np.sin(k*x + w*t), label=f"t={frac:.3f} tau")
plt.title("n = 3 standing wave: nodes never move")
plt.xlabel("x (m)"); plt.legend(fontsize=7); plt.show()
# Expected: every snapshot vanishes at x = 0, 1/3, 2/3, 1 (fixed nodes);
# the profile 2A sin(kx) cos(wt) breathes in place -- no travel survives.
```
```python
import numpy as np
import matplotlib.pyplot as plt
# --- 3. Fourier synthesis: evolving a plucked (triangular) string ---
L, v, a, h, N = 1.0, 1.0, 0.30, 0.05, 60 # pluck at x=a, height h, N modes
x = np.linspace(0, L, 1000)
y0 = np.where(x < a, h*x/a, h*(L - x)/(L - a)) # initial triangle, released at rest
n = np.arange(1, N + 1)
# b_n = (2/L) * integral y0(x) sin(n pi x/L) dx (np.trapz on NumPy < 2.0):
b = np.array([2/L*np.trapezoid(y0*np.sin(m*np.pi*x/L), x) for m in n])
def y(t): # y(x,t) = sum_n b_n sin(k_n x) cos(w_n t)
return np.sin(np.pi*np.outer(x, n)/L) @ (b*np.cos(n*np.pi*v*t/L))
tau = 2*L/v # fundamental period
for frac in (0.0, 0.1, 0.2, 0.3, 0.4, 0.5):
plt.plot(x, y(frac*tau), label=f"t = {frac:.1f} tau")
plt.title("Plucked string via normal-mode expansion")
plt.xlabel("x (m)"); plt.ylabel("y (m)"); plt.legend(fontsize=8); plt.show()
# Expected: the kink splits into two counter-propagating kinks; at t = 0.5 tau
# the shape is the initial triangle inverted & mirrored; it recurs at t = tau.
```
---
## Exercises
**E1 (easy).** A wave on a string is $y(x,t) = 0.020\sin(4\pi x - 200\pi t)$ (SI). Find the
amplitude, wavelength, frequency, speed, and direction of travel.
Solution
$A = 0.020$ m; $k = 4\pi$ m$^{-1}$ so $\lambda = 2\pi/k = 0.50$ m; $\omega = 200\pi$ s$^{-1}$ so
$\nu = 100$ Hz; $v = \omega/k = 50$ m/s; the combination $kx - \omega t$ moves in $+x$.
Solution
With $w = x + vt$: $\partial_t y = vg'(w)$, $\partial_t^2 y = v^2g''(w)$, $\partial_x^2 y = g''(w)$,
so $\partial_t^2 y = v^2\partial_x^2 y$ ✓. The argument is constant along $x = -vt + \text{const}$:
the shape moves in the $-x$ direction at speed $v$.
Solution
Add phasors: $A + Ae^{i\delta} = Ae^{i\delta/2}(e^{-i\delta/2} + e^{i\delta/2}) = 2A\cos(\delta/2)\,e^{i\delta/2}$,
of modulus $2A|\cos(\delta/2)|$ — matching sum-to-product.
Intensity $I = 4I_0\cos^2(\delta/2) = 2I_0(1 + \cos\delta)$; $\cos\delta$ averages to zero over
$[0, 2\pi]$, so $\langle I\rangle = 2I_0$: interference redistributes energy but conserves it.
Solution
$X = a\sin kx$ from $X(0) = 0$; now $X'(L) = ak\cos kL = 0$ requires $kL = (2n-1)\pi/2$, so
$\nu_n = (2n-1)v/4L$: only odd multiples of a (lower) fundamental $v/4L$. Different boundary
conditions, different quantized spectrum — the spectrum *encodes* the boundary conditions.
Solution
$\partial_ty_n = -\omega_nA_n\sin(k_nx)\sin(\omega_nt)$ and
$\partial_xy_n = k_nA_n\cos(k_nx)\cos(\omega_nt)$. Using
$\int_0^L\sin^2(k_nx)\,dx = \int_0^L\cos^2(k_nx)\,dx = L/2$:
$K = \tfrac14\mu L\omega_n^2A_n^2\sin^2\omega_nt$ and $U = \tfrac14TLk_n^2A_n^2\cos^2\omega_nt$.
But $Tk_n^2 = \mu v^2k_n^2 = \mu\omega_n^2$, so
$E_n = K + U = \tfrac14\mu L\omega_n^2A_n^2 = \text{const}$: energy sloshes between kinetic and potential — each normal mode *is* an independent
harmonic oscillator of frequency $\omega_n$. Classical physics then hands every mode an average
thermal energy $k_BT$; counting cavity modes this way leads straight to the ultraviolet
catastrophe of [P.2.1](../c02-radiation-old-quantum/01-blackbody-radiation.md).
Solution
(a) Steel: $v = \sqrt{2.0\times10^{11}/7.85\times10^3} = 5.05\times10^3$ m/s, so
$\lambda = v/\nu = 11.5$ m. Air: $v = \sqrt{1.40\times1.013\times10^5/1.20} = 344$ m/s, so
$\lambda = 0.781$ m. Isothermal ($\kappa = P_0$): $\sqrt{1.013\times10^5/1.20} = 291$ m/s, 15% low,
which is why Laplace's adiabatic correction mattered. The same tone is fifteen times longer in steel
because the frequency is fixed by the source and $\lambda = v/\nu$.
(b) $\omega = 2\pi\times1000 = 6.28\times10^3\ \mathrm{s^{-1}}$, so
$u_{\max} = \omega A = 6.3\times10^{-3}$ m/s, i.e. 6.3 mm/s, and $u_{\max}/v = 1.8\times10^{-5}$.
With $k = \omega/v = 18.3\ \mathrm{m^{-1}}$, $kA = 1.8\times10^{-5}$ ✓, as $u/v = -\partial_x\xi$
requires. (This is a loud tone, roughly 100 dB, yet the collective motion is a drift of millimetres
per second laid over the molecules' random thermal speeds of about 470 m/s.)
Answers
1. Small transverse displacements ($|\partial_xy| \ll 1$), uniform tension and density, purely
transverse motion, no gravity or damping; Newton's law on an element $\mu\,\Delta x$ does the rest.
2. Its argument is constant along $x = vt + \text{const}$, so the graph translates rigidly at $v$.
Rigidity holds because $\omega = vk$ is linear — all Fourier components share one speed.
3. $4I_0$: amplitudes add to $2A$ and intensity is amplitude squared. Averaged over a fringe
pattern the intensity is $2I_0$ — interference only redistributes energy.
4. $y(0) = y(L) = 0$ forces $\sin kL = 0$, so $k_n = n\pi/L$ and $\nu_n = vk_n/2\pi = nv/2L$. For a
quantum particle, the wavefunction vanishing at the box walls quantizes $k$ and hence the
*energy* ([P.5.2](../c05-bound-states/02-step-well-barrier.md)).
5. $v_p = \omega/k$ moves crests; $v_g = d\omega/dk$ moves packets and energy. They coincide
exactly when $\omega \propto k$ — nondispersive media like the ideal string.
6. In each, Newton's law on a small element balances a restoring stiffness against an inertia
linearly in the small displacement, giving $v^2 = \text{stiffness}/\text{inertia}$: $T/\mu$,
$Y/\rho$, $\kappa/\rho_0$. Newton used the isothermal $\kappa = P_0$; acoustic compressions are
too fast for heat to flow, so they are adiabatic, $\kappa = \gamma P_0$.