Waves & Oscillations · Upper Division

The Wave Equation

The wave equation is the PDE that governs sound, light, vibrating strings, water waves, and quantum probability amplitudes. Its solutions — d'Alembert's formula, standing waves, and dispersive packets — appear across all of physics.

PrerequisitesWaveproperties(Ch.8)Fourieranalysis(Ch.F)PartialderivativesWave properties (Ch. 8) \cdot Fourier analysis (Ch. F) \cdot Partial derivatives
Learning Goals
  • Derive the 1D wave equation from Newton's second law applied to a vibrating string.
  • Apply d'Alembert's formula to find the motion of a plucked string from initial conditions.
  • Use separation of variables to find normal modes and their frequencies for a fixed string.
  • Distinguish phase velocity and group velocity and compute them from a dispersion relation.
  • Calculate the power transmitted by a traveling wave from amplitude and wave parameters.

WE.1 Derivation from a Vibrating String

Consider a taut string with tension T and linear mass density μ. For small transverse displacements y(x, t), Newton's second law applied to a small element gives:

2yt2=v22yx2v=Tμ(1D wave equation)\frac{\partial^2y}{\partial t^2}=v^2\frac{\partial^2y}{\partial x^2} \qquad v=\sqrt{\frac{T}{\mu}} \qquad \text{(1D wave equation)}(WE.1)

This is the canonical form of the wave equation with wave speed v. The same equation governs longitudinal sound waves (v = √(B/ρ), B = bulk modulus), EM waves (v = 1/√(με)), and quantum mechanical free particles (v → operator, giving Schrödinger).

WE.2 D'Alembert's Solution

Theorem WE.1D'Alembert's Formula (1747)
Thegeneralsolutiontothe1Dwaveequationwithinitialconditionsy(x,0)=f(x)and\partialy/The general solution to the 1D wave equation with initial conditions y(x,0) = f(x) and \partialy/\partialt(x,0)=g(x)is\partialt(x,0) = g(x) is:y(x,t)=12[f(xvt)+f(x+vt)]+(1/2v)(xvttox+vt)g(s)dsy(x, t) = \frac{1}{2}[f(x-vt) + f(x+vt)] + (1/2v) \int(x-vt to x+vt) g(s) dsThe first term is the superposition of two copies of the initial shape — one traveling right at speed v, one left at speed v. Every solution is a sum of rightward and leftward traveling waves.
Example WE.1Plucked String

AstringoflengthLispluckedatitscentertoheighthandreleasedfromrest(g=0A string of length L is plucked at its center to height h and released from rest (g = 0. Describe the subsequent motion.

Initial shape:f(x)=triangularpeak:2hx/Lfor0xL/2,then2h(Lx)/LforL/2xLf(x) = triangular peak: 2hx/L for 0 \le x \le L/2, then 2h(L-x)/L for L/2 \le x \le L.
D'Alembert:y(x,t)=12[f(xvt)+f(x+vt)],extendedasoddperiodicfunctiony(x,t) = \frac{1}{2}[f(x-vt) + f(x+vt)], extended as odd periodic function.
Motion:The triangle splits into two half-height triangles thattravelinoppositedirections.Theybounceoffthefixedends(boundaryconditiony=0that travel in opposite directions. They bounce off the fixed ends (boundary condition y=0 T=2L/vT = 2L/v
Fourier view:The triangular shape has Fourier series with only odd harmonics.Eachharmonichasfrequencynv/(2L)theharmonicsofaguitarstring.Then=1harmonics. Each harmonic has frequency nv/(2L) — the harmonics of a guitar string. The n=1/(2L).

WE.3 Separation of Variables and Normal Modes

For a string fixed at x=0 and x=L, try y(x,t) = X(x)T(t). Substituting:

T¨T=v2XX=ω2(separation constant)\frac{\ddot T}{T}=v^2\frac{X''}{X}=-\omega^2 \qquad \text{(separation constant)}(WE.2)

The spatial equation X'' + (ω/v)²X = 0 with boundary conditions X(0)=X(L)=0 gives:

Xn(x)=sin(nπxL)ωn=nπvLn=1,2,3,X_n(x)=\sin\left(\frac{n\pi x}{L}\right) \qquad \omega_n=\frac{n\pi v}{L} \qquad n=1,2,3,\ldots(WE.3)

These are the normal modes (standing waves). The general solution is a superposition:

y(x,t)=n=1sin(nπxL)[Ancos(ωnt)+Bnsin(ωnt)]y(x,t)=\sum_{n=1}^{\infty}\sin\left(\frac{n\pi x}{L}\right)\left[A_n\cos(\omega_nt)+B_n\sin(\omega_nt)\right](WE.4)

The coefficients Aₙ and Bₙ are determined by the initial conditions via Fourier sine series — Aₙ = (2/L)∫(0 to L) f(x) sin(nπx/L) dx.

WE.4 Dispersion Relations

Definition WE.1Dispersion Relation
The dispersion relation ω(k)relatesangularfrequencyωtowavenumberkforwavesinamedium.Forthestandardw\omega(k) relates angular frequency \omega to wavenumber k for waves in a medium. For the standard wveequation:ω=vk(linear,nondispersive).Twovelocitiescharacterizedispersivewavesve equation: \omega = vk (linear, non-dispersive). Two velocities characterize dispersive waves:Phase velocity: vp=ω/kspeedofasinglefrequencywavecrestv_{p} = \omega/k — speed of a single-frequency wave crestGroup velocity: vg=dω/dkspeedofawavepacket(carriesenergyandinformationv_{g} = d\omega/dk — speed of a wave packet (carries energy and informationWhenvpvg,themediumisWhen v_{p} ≠ v_{g}, the medium is dispersive: different frequencies travel at different speeds, and pulses spread out over time.

Examples of dispersion relations:

ω2=v2k2+ωp2(plasma, where ωp is the plasma frequency)\omega^2=v^2k^2+\omega_p^2 \qquad \text{(plasma, where }\omega_p\text{ is the plasma frequency)}(WE.5)
ω2=gk+γk3ρ(deep water: gravity + surface tension)\omega^2=gk+\frac{\gamma k^3}{\rho} \qquad \text{(deep water: gravity + surface tension)}(WE.6)
ω=k22m(non-relativistic de Broglie waves)\omega=\frac{\hbar k^2}{2m} \qquad \text{(non-relativistic de Broglie waves)}(WE.7)
Example WE.2Group Velocity of a Wave Packet

Twocosinewaveswithnearlyequalfrequencies:k1=k\deltak,k2=k+\deltak,ω1=ωδω,ω2=ω+δωTwo cosine waves with nearly equal frequencies: k_{1} = k-\deltak, k_{2} = k+\deltak, \omega_{1} = \omega-\delta\omega, \omega_{2} = \omega+\delta\omega. Find the envelope velocity.

Superpose:y=cos(k1xω1t)+cos(k2xω2ty = cos(k_{1}x - \omega_{1}t) + cos(k_{2}x - \omega_{2}t
Sum-to-product:= 2cos((k1k2)x/2(ω1ω2)t/2)cos((k1+k2)x/2(ω1+ω2)t/22 cos((k_{1}-k_{2})x/2 - (\omega_{1}-\omega_{2})t/2) \cdot cos((k_{1}+k_{2})x/2 - (\omega_{1}+\omega_{2})t/2
Result:= 2cos(\deltakxδωt)cos(kx\omegat2 cos(\deltak \cdot x - \delta\omega \cdot t) \cdot cos(kx - \omegat
Interpretation:Fastoscillationat(k,ω)thecarrierwave.Slowenvelopeat(\deltak,δω)thebeat.TheeFast oscillation at (k, \omega) — the carrier wave. Slow envelope at (\deltak, \delta\omega) — the beat. The envelopetravelsatvenvelope=δω/\deltakdω/dkinthelimitnvelope travels at v_{envelope} = \delta\omega/\deltak \to d\omega/dk in the limit.
Conclusion:vgroup=dω/dkthegroupvelocitycarriesthepacketand,inQM,theprobabilitydensitv_{group} = d\omega/dk — the group velocity carries the packet and, in QM, the probability density.

WE.5 Energy and Power in Waves

For a traveling wave on a string y = A sin(kx − ωt), the power transmitted past any point is:

P=12μω2A2v=12TkA2ωP=\frac{1}{2}\mu\omega^2A^2v=\frac{1}{2}TkA^2\omega(WE.8)

More usefully: P = ½μvω²A². The energy per wavelength is E_λ = P · (λ/v) = ½μωA² · λ. The energy density (energy per unit length) is u = ½μω²A² — equally divided between kinetic (½μẏ²) and potential (½T(∂y/∂x)²) at each instant, averaged over time.

Definition WE.2Common Traps
  • Boundary conditions choose the modes: the PDE alone does not decide whether sine, cosine, or mixed modes appear.
  • Phase velocity is not always signal speed: in dispersive media, energy and information travel with group velocity.
  • Fourier coefficients are set by initial conditions: normal modes are the basis, not the full answer by themselves.
  • Average power is cycle-averaged: instantaneous energy density oscillates between kinetic and potential forms.
Exercises — WE.1–WE.5 The Wave Equation
1.
Aguitarstringis65cmlong,under70Ntension,withlineardensityμ=5×104kg/m.FiA guitar string is 65 cm long, under 70 N tension, with linear density \mu = 5\times10^{-4} kg/m. Find the wave speed, fundamental frequency, and first three harmonics.
m/s
Straightforward
2.AGaussianpulsetravelsinamediumwithcubicdispersionω=ck+\alphak3.DescribequalitatA Gaussian pulse travels in a medium with cubic dispersion \omega = ck + \alphak^{3}. Describe qualitatively how the pulse shape changes over time. What is this called in fiber optics?
Intermediate
3.Find the normal modes of a 2D square membrane (fixed boundary, side L). Which modes are degenerate? Sketch the nodal patterns for the lowest modes.
Intermediate
4.ShowthattheSchro¨dingerequationiψ/\partialt=(2/2m)2ψ/\partialx2hasaplanewavesolution.FShow that the Schrödinger equation i\hbar \partial\psi/\partialt = (-\hbar^{2}/2m)\partial^{2}\psi/\partialx^{2} has a plane wave solution. Find the phase and group velocities. Which matches the classical particle velocity?
Challenging
Key Takeaways
  • Waveequation:2y/\partialt2=v22y/\partialx2.Appliestostrings(v=(T/μ)),sound,EM,QMWave equation: \partial^{2}y/\partialt^{2} = v^{2}\partial^{2}y/\partialx^{2}. Applies to strings (v=\sqrt(T/\mu)), sound, EM, QM.
  • DAlembert:generalsolution=f(xvt)+g(x+vt)rightwardplusleftwardwavesD'Alembert: general solution = f(x-vt) + g(x+vt) — rightward plus leftward waves.
  • Normalmodes:sin(n\pix/L)withωn=n\piv/L.SuperposewithFouriercoefficientsNormal modes: sin(n\pix/L) with \omega_{n} = n\piv/L. Superpose with Fourier coefficients.
  • Phasevelocityvp=ω/k(crestspeed);groupvelocityvg=dω/dk(packetspeed,carriesePhase velocity v_{p} = \omega/k (crest speed); group velocity v_{g} = d\omega/dk (packet speed, carries energy).
  • Nondispersive:ω=vkvp=vg.Dispersive:vpvg,pulsesspreadNon-dispersive: \omega = vk \to v_{p} = v_{g}. Dispersive: v_{p} ≠ v_{g}, pulses spread.
  • Energydensity:u=12μω2A2,equallykineticandpotential(averagedovercycleEnergy density: u = \frac{1}{2}\mu\omega^{2}A^{2}, equally kinetic and potential (averaged over cycle.