# 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. --- ## 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. ### 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. ### 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. 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). ### 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: $A\sin(kx - \omega t) + A\sin(kx + \omega t) = 2A\sin kx\cos\omega t$. 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)). --- ## 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 $+x$.

**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](../c02-radiation-old-quantum/01-blackbody-radiation.md).

--- ## 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?

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.

--- ## 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. --- ← Prev: [Pre-Term Overview](../README.md) · Up: [Pre-Term](../README.md) · Next: [Wave Propagation, Diffraction & Gratings](02-wave-propagation-diffraction.md) →