Standing Waves & Resonance

Write the incident and reflected waves as A sin(kx − ωt) and A sin(kx + ωt). The identity sin α + sin β = 2 sin ½(α + β) cos ½(α − β) turns their sum into y = 2A sin(kx) cos(ωt). Read the two factors separately. The cos(ωt) says every point of the string moves in simple harmonic motion at the same frequency and reaches its extreme at the same instant. The sin(kx) says how far each point goes: nowhere at the nodes, where kx = mπ, and 2A at the antinodes midway between them. For λ = 0.80 m the nodes sit 0.40 m apart, and a point 0.10 m from a node swings with amplitude 2A sin(2π × 0.10/0.80) = 1.41A. Where sin(kx) changes sign the motion reverses, so neighbouring loops are always half a cycle out of step: one is up while the next is down. In this ideal pattern no energy crosses a node; it passes back and forth between kinetic and potential inside each loop.

Nothing so far has fixed λ. The ends do that. A clamped end has y = 0 at every instant, so it must be a node. Put one clamp at x = 0, which the sin(kx) form already satisfies, and the other at x = L: then sin(kL) = 0, so kL = nπ, L = nλ/2 and fₙ = nv/2L. Now replace the far clamp with a light ring sliding on a smooth rod. The rod cannot push the ring along itself, and a ring of negligible mass cannot take a net transverse force, so the string must meet it with zero slope: dy/dx = 0 at x = L, which means cos(kL) = 0 and L = (2n − 1)λ/4. The free end is an antinode. For L = 1.20 m and v = 96 m s⁻¹ the fixed–fixed string resonates at 40, 80 and 120 Hz; with one end freed the list becomes 20, 60 and 100 Hz. The medium is the same. Only the boundary conditions changed, and they moved every frequency.

A normal mode is a motion in which every part of the system oscillates at one frequency with a fixed shape. That makes each mode a single harmonic oscillator, and any free motion of the string is a sum of them. Pluck at the midpoint and you excite the modes with an antinode there, n = 1, 3, 5; rest a fingertip at the midpoint and you remove exactly those, leaving the even ones. Energy follows by adding up slices. A slice of length dx has mass μ dx and swings with amplitude Aₘ sin(kx), so at the instant the string is flat it carries kinetic energy ½(μ dx)ω²Aₘ² sin²(kx) and nothing else. Integrating over whole loops, where sin² averages to ½, gives E = ¼μLω²Aₘ². For μ = 2.0 g m⁻¹, L = 1.00 m, f = 120 Hz and Aₘ = 5.0 mm, ω = 754 rad s⁻¹ and E = 7.1 mJ. Doubling the amplitude quadruples it, as for any harmonic oscillator.

Shake the end at x = 0 through a small amplitude a at frequency f, with the far end clamped. The steady motion that fits both ends is y = a sin(k(L − x)) cos(ωt) / sin(kL): it vanishes at x = L and equals a cos(ωt) at x = 0. Its loops have amplitude a/|sin(kL)|. When kL = (n + ½)π that is just a, and the string barely notices the driver. As kL approaches nπ the denominator goes to zero and the response grows without limit. The reason is timing. A wave launched by the driver returns after 2L/v, having been inverted at each end, and at kL = nπ that delay is exactly n periods, so every round trip adds to the last. With L = 1.00 m and v = 80 m s⁻¹ the round trip takes 25 ms, which is three periods at 120 Hz. At 118.8 Hz, 1 % low, a/|sin(kL)| is already 10.6a; at 119.6 Hz it is 31.8a. Notice where the driver sits: on resonance the driven end is almost a node, because a is tiny beside the loops it sustains.

No real string reaches infinity. The amplitude climbs until the energy dissipated per cycle equals the energy the driver supplies per cycle, and the quality factor measures that balance: Q = 2π × (energy stored)/(energy lost per cycle). For a string shaken at one end the balance gives a loop amplitude of about 2Qa/(nπ) on resonance. The same Q sets the sharpness. Plot amplitude against driving frequency and each mode is a peak of full width Δf = fₙ/Q at half maximum power, so a high-Q string must be tuned with care and a low-Q one responds to almost anything. Take mode 3 at 120 Hz with Q = 150 and a = 0.20 mm: the peak is 0.80 Hz wide and the loops reach 2 × 150 × 0.20/(3π) = 6.4 mm. Switch the driver off and the mode rings down as e^(−πft/Q), falling by the factor e in Q/π = 48 cycles, which is 0.40 s. Width, height and ring-down are three measurements of one number.

Each assumption behind fₙ = nv/2L fails somewhere, and a good report says which one limits it. Nodes are not perfectly still. The driver's energy has to pass through every node to feed the loops beyond, so a driven node keeps a residual motion of order a, smaller than the loops by about nπ/2Q, which is 3 % for the string above. Real wire resists bending as well as stretching, and that pushes high modes sharp: fₙ = nf₀√(1 + Bn²) with f₀ = v/2L, and for B = 4.0 × 10⁻⁴ the sixth mode lies 0.70 % above six times the first. Large amplitudes lengthen the string and raise its tension, so the peak drifts upward as the amplitude grows. Supports that yield slightly make the effective length a little longer than the ruler says. Finally, the peaks have width fₙ/Q while their spacing is f₁; once Q falls towards n the peaks merge, and talking about separate modes stops being useful.