Standing Waves & Normal Modes

Send a sinusoid of amplitude A to the right and an identical one to the left: y = A sin(kx − ωt) + A sin(kx + ωt). The identity sin(P − Q) + sin(P + Q) = 2 sin P cos Q collapses that to y(x, t) = 2A sin(kx) cos(ωt). Look at what the arguments have done. A travelling wave depends on x and t only through the combination x − vt, which is why its whole profile slides along; here x sits inside one factor and t inside another. The shape 2A sin(kx) is frozen in place, and the number cos(ωt) multiplying it merely breathes in and out at angular frequency ω. Every particle is a simple harmonic oscillator of that same ω, but its amplitude 2A|sin kx| is set by where it sits. Check the picture at two instants: at t = 0 the string shows the full envelope and is momentarily at rest; a quarter period later cos(ωt) = 0, the string is instantaneously straight, and all of the mode's energy is transverse motion.

The amplitude 2A|sin kx| vanishes wherever kx = nπ, that is at x = n λ/2 — the nodes. Halfway between them, at x = (2n + 1)λ/4, sin kx = ±1, the amplitude is 2A, and that is the most any point of the medium can reach: twice what either travelling wave carried on its own. So adjacent nodes are half a wavelength apart, and a node and its neighbouring antinode a quarter. That is the measurement this topic buys you: pin two node positions with a ruler and you have λ without timing anything. The phase structure is just as sharp. Inside one loop every particle crosses zero at the same instant and reaches its extreme at the same instant, exactly in phase. Cross a node and sin kx changes sign, so the next loop is a half cycle behind. A travelling wave delays each point smoothly by kΔx; a standing wave has no gradual lag at all, only in step or antiphase.

Superposition alone permits any k whatsoever. Now clamp the string. A rigid end cannot move, so y = 0 there at every instant. Put x = 0 at one clamp and sin(kx) already vanishes; the clamp at x = L demands sin(kL) = 0, which is not automatic. It forces kL = nπ, so kₙ = nπ/L, λₙ = 2L/n and fₙ = v/λₙ = n v/(2L). A continuum of wavelengths has collapsed to a countable list, and the reason is a differential equation plus one condition at each end — the same argument that quantises a particle in a box later in this course. Each allowed k is a normal mode: the whole string moving at a single frequency in a fixed shape. Numbers: clamp a 0.500 m string carrying v = 320 m s⁻¹ and the resonances are 320 Hz, 640 Hz, 960 Hz and nothing else. Drive it at 480 Hz and no large amplitude builds, because the wave returning from the far clamp comes back out of step with the drive.

Not every boundary is a clamp. Tie the far end to a light ring that slides without friction on a transverse rod. The ring is effectively massless, so the transverse force on it, −F ∂y/∂x, must vanish, which sets ∂y/∂x = 0 at x = L. Differentiate 2A sin(kx) cos(ωt) with respect to x and that condition becomes cos(kL) = 0, so kL = (2n − 1)π/2, λₙ = 4L/(2n − 1) and fₙ = (2n − 1) v/(4L). The longest mode now fits a quarter wavelength into L rather than a half, so the fundamental is half the doubly clamped value — an octave lower — and the spectrum keeps only odd multiples of it. Every even harmonic is missing, because an even multiple would demand a node at the free end, and the free end is an antinode. Nothing in the method changed; only the line written at the boundary did.

The transverse power crossing a point of a string is P = −F(∂y/∂x)(∂y/∂t), the transverse force from the tension times the transverse velocity. For y = A_SW sin(kx) cos(ωt) those derivatives are A_SW k cos(kx) cos(ωt) and −A_SW ω sin(kx) sin(ωt), so P = ¼ F A_SW² kω sin(2kx) sin(2ωt). Two results fall out. The factor sin(2ωt) averages to zero over a full cycle, so the mean power is zero at every point: the wave transports nothing, which is what earns it the name. And sin(2kx) is zero at every node and every antinode, so no energy crosses those points at any instant at all. What the instantaneous power does is shuttle energy back and forth inside each quarter wavelength — the stretch from a node to its neighbouring antinode — twice per period, between kinetic energy piled up at the antinodes and elastic energy piled up at the nodes, where the slope is steepest. The mode's total stays fixed at E = ¼ μ ω² A_SW² L.

Three idealisations carry everything above. Linearity: the superposition step assumed slopes small enough to keep the wave equation linear, and a string plucked hard enough to raise its own tension at the extremes sharpens its modes upwards. Ideal ends: a real bridge yields a little, so energy leaks out as sound, which is precisely why you can hear the instrument and why a perfectly standing wave would be silent. Negligible damping: with loss, a driven string still responds at any frequency, but the response is a resonance curve of finite width Δf ≈ fₙ/Q rather than a spike, and free ringing decays as e^(−t/τ). Bending stiffness adds a fourth escape: a real wire resists curvature as well as stretch, so its overtones drift above n f₁ roughly as fₙ ≈ n f₁ √(1 + B n²) with B small. The ideal spectrum is the leading term, not the whole answer.