2D Wave Equation & Chladni Patterns

FDTD drumhead simulator — click the membrane to pluck it, or choose a standing-wave mode

Preset:
Frame: 0
Energy: 0
∂²u/∂t² = c²∇²u

The 2D Wave Equation

A thin elastic membrane stretched over a rigid frame satisfies the 2D wave equation:

∂²u/∂t² = c²(∂²u/∂x² + ∂²u/∂y²) − γ·∂u/∂t where: u(x,y,t) = transverse displacement c = √(T/ρ) wave speed (T = tension, ρ = surface mass density) γ = damping coefficient (energy dissipation) Fixed BCs: u = 0 on all four edges (Dirichlet)
FDTD discretisation: The finite-difference time-domain (FDTD) update rule is: u_new[i,j] = (2·u[i,j] − u_old[i,j]·(1−γΔt) + r²·Lap[i,j]) / (1+γΔt) where r = cΔt/Δx = 0.40 and Lap[i,j] = u[i+1,j]+u[i-1,j]+u[i,j+1]+u[i,j-1]−4u[i,j]. The CFL stability condition r ≤ 1/√2 ≈ 0.707 is satisfied.

Normal Modes of a Rectangular Membrane

For a rectangular membrane of width W and height H with fixed boundaries, the exact normal mode solutions are:

u_{mn}(x,y,t) = A·sin(mπx/W)·sin(nπy/H)·cos(ω_{mn}t + φ) Angular frequencies: ω_{mn} = πc·√(m²/W² + n²/H²) Natural frequencies: f_{mn} = (c/2)·√(m²/W² + n²/H²) For a square membrane (W=H): f_{mn} = (c/2W)·√(m² + n²) Examples (c=1, W=H=1): f₁₁ = c√2 / 2W (fundamental) f₂₁ = f₁₂ = c√5 / 2W (degenerate: two modes at same frequency!) f₂₂ = c√8 / 2W = 2f₁₁ (first overtone at exactly 2×fundamental) f₃₁ = f₁₃ = c√10 / 2W f₃₂ = f₂₃ = c√13 / 2W

Degeneracy: On a square membrane, modes with the same value of m²+n² are degenerate — they share the same frequency but have different spatial patterns. A square drum has many more degenerate pairs than a rectangular drum, making its timbre richer.

Chladni Patterns

Ernst Chladni (1756–1827) famously scattered sand on vibrating metal plates and excited them with a bow. The sand migrated to the nodal lines — curves where u = 0 at all times — forming beautiful geometric patterns. In the simulation the white regions near zero displacement visualise exactly these nodal figures.

Mode (m,n)Frequency ratio to f₁₁Nodal lines
(1,1) Fundamental1.00None (entire membrane moves in phase)
(2,1) / (1,2)√(5/2) ≈ 1.581 vertical or 1 horizontal nodal line
(2,2)2.001 vertical + 1 horizontal (cross pattern)
(3,1) / (1,3)√5 ≈ 1.782 parallel nodal lines
(3,2) / (2,3)√(13/2) ≈ 1.802 + 1 = 3 nodal lines (Chladni figure)
(3,3)3.002 + 2 = 4 nodal lines (checker pattern)

Applications

Frequently Asked Questions

Why does the square membrane have degenerate modes?
Two modes (m,n) and (n,m) have the same frequency ω = πc√(m²+n²)/W on a square membrane whenever m ≠ n. Because both satisfy the wave equation and boundary conditions at the same frequency, any linear combination is also a valid solution. In practice this means the shape of the pattern depends on the symmetry of the driving force — exciting from the corner favours symmetric combinations, from the edge asymmetric ones. On a non-square rectangle there is no such degeneracy (generically), explaining why rectangular drums sound simpler.
What does energy represent in the HUD?
The HUD Energy readout is proportional to the sum of squared displacements Σu² across all grid cells — a proxy for the potential energy of the membrane (true potential energy is ½T·Σ|∇u|² and kinetic energy is ½ρ·Σ(∂u/∂t)²; their sum is conserved in undamped motion). In the Pluck Centre preset with zero damping, watch the energy oscillate as it converts between potential (displacement) and kinetic (velocity) form, just like a spring-mass system.
How is this related to the quantum particle in a 2D box?
Mathematically identical. The Schrödinger equation for a free particle in a 2D infinite square well is −ℏ²/(2m)·∇²ψ = E·ψ, which has eigenfunctions ψ_{mn} = sin(mπx/L)·sin(nπy/L) and energies E_{mn} = ℏ²π²(m²+n²)/(2mL²). These are exactly the spatial factors of the drumhead normal modes, with E_{mn} ∝ ω²_{mn}. The zero-displacement boundary condition (fixed drum edge) maps to the infinite-potential-well boundary condition (ψ=0 at walls). Quantum mechanics and classical wave mechanics share this deep structural similarity.

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