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.

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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.

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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). 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 $z$):

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:

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.

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, 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: 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.

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). 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 $\pi$:

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), and confining a matter wave in a box quantizes its energy — the infinite square well of P.5.2 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).

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). 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.

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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.

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Hands-on (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.

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.

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.

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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 $+x$.

E2 (easy). Show directly (chain rule) that $y = g(x + vt)$ solves the wave equation for any twice-differentiable $g$, and give its direction of motion.

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 $v$.

E3 (medium). Using complex exponentials, re-derive the resultant amplitude of $A\cos(kx-\omega t) + A\cos(kx-\omega t+\delta)$, then show the intensity averaged over all $\delta$ is $2I_0$.

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.

E4 (medium). A string is fixed at $x = 0$, but its end at $x = L$ rides a frictionless massless ring on a rod, enforcing $\partial y/\partial x\,(L,t) = 0$ (a "free" end). Find the allowed frequencies.

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.

E5 (hard). Compute the total energy of a string vibrating in one mode $y_n = A_n\sin(k_nx)\cos(\omega_nt)$, using kinetic density $\tfrac12\mu(\partial_ty)^2$ and potential density $\tfrac12T(\partial_xy)^2$. Show it is constant in time.

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.

E6 (medium). (a) Find the speed of longitudinal waves in a steel rod ($Y = 2.0\times10^{11}$ Pa, $\rho = 7.85\times10^3$ kg/m³) and of sound in air at 20 °C ($P_0 = 1.013\times10^5$ Pa, $\rho_0 = 1.20$ kg/m³, $\gamma = 1.40$), and the wavelength of a 440 Hz tone in each. What would Newton's isothermal assumption have given for air? (b) A 1.00 kHz sound wave in air has displacement amplitude $1.0\ \mu$m. Find the maximum speed of the air's motion, compare it with the wave speed, and check that the ratio equals the maximum strain $kA$.

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.)

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Checkpoint

1. What physical assumptions produce $\partial_t^2y = (T/\mu)\,\partial_x^2y$ for a string?
2. Why does $f(x - vt)$ describe a rigidly right-moving shape, and what property of the dispersion relation keeps shapes rigid?
3. Two equal in-phase waves overlap. What is the intensity relative to one wave alone, and why is energy still conserved?
4. Why do fixed ends allow only $\nu_n = nv/2L$? What plays the analogous role for a quantum particle in a box?
5. Distinguish phase velocity from group velocity. For which media do they coincide?
6. Why do a string, a rod and a gas column obey the same wave equation, and what sets $v$ in each? Why was Newton's speed of sound wrong?

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).
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, $\kappa = \gamma P_0$.

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Further Reading

• [Gold] Goldstein, Poole & Safko, Ch. 13 — the string's wave equation re-derived from a Lagrangian density; classical field theory in embryo.
• [Gri] Griffiths & Schroeter, §2.4 — wave packets and phase vs group velocity, in their quantum habitat.
• [ER] Eisberg & Resnick, §3.1–3.2 — how the classical wave toolkit is reused for de Broglie's matter waves.

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