Wave Energy, Power and Intensity

Take a rightward wave y(x, t) = A cos(kx − ωt) on a string of linear density μ under tension F. For small slopes the string on the left of a point pulls the string on the right with a transverse force −F ∂y/∂x, and that force acts through the transverse velocity ∂y/∂t, so the power crossing the point is P(x, t) = −F(∂y/∂x)(∂y/∂t). Put in the derivatives, ∂y/∂x = −kA sin(kx − ωt) and ∂y/∂t = ωA sin(kx − ωt), and you get P = FkωA² sin²(kx − ωt). Use F = μv² and k = ω/v to clear the medium constants: P(x, t) = μvω²A² sin²(kx − ωt). It is never negative, so energy always flows in +x for this wave, and it pulses at twice the wave frequency. Averaging sin² to ½ over a cycle leaves P(av) = ½μvω²A² = ½√(μF) ω²A². Nothing in the argument is special to strings: transverse force times transverse velocity is one case of the general rule that energy crosses a surface at the rate the force there does work.

Both energy densities on the string are quadratic in a first derivative of y: dK/dx = ½μ(∂y/∂t)² and dU/dx = ½F(∂y/∂x)². Every derivative of A cos(kx − ωt) brings down one factor of ω or k, so every energy quantity comes out proportional to A² and to ω². With μ = 0.050 kg m⁻¹ and F = 80 N, v = √(F/μ) = 40 m s⁻¹; at f = 60 Hz, so ω = 377 rad s⁻¹, and A = 5.0 mm, P(av) = ½(0.050)(40)(377)²(5.0 × 10⁻³)² = 3.55 W. Take A to 10 mm and it is 14.2 W. Tension is a weaker lever, since P ∝ √F at fixed A and ω: 80 N to 160 N moves 3.55 W only to 5.02 W. One structural result is worth carrying forward. For any travelling wave ∂y/∂t = −v ∂y/∂x, so the two densities above are equal at every point and every instant, and P(av) = v e(av) with e(av) = ½μω²A². Energy flux is energy density times wave speed — the pattern that reappears for sound and for light.

In two or three dimensions the transported power is spread across a wavefront, so the working quantity is intensity: average power per unit area held perpendicular to the propagation direction, I = P(av)/A(⊥), in W m⁻². The same skeleton builds it. A sound wave with displacement amplitude s₀ has average energy density ½ρω²s₀², so I = ½ρvω²s₀², and since the pressure amplitude is p₀ = ρvωs₀, this is also I = p₀²/(2ρv). The product ρv is the acoustic impedance: for air, 1.20 × 343 = 412 Pa s m⁻¹, and the same pressure amplitude drives less motion, and less power, in a stiffer medium. Ordinary speech at 1 m is about 1.0 × 10⁻⁶ W m⁻², which needs only p₀ = √(2 × 412 × 1.0 × 10⁻⁶) = 0.029 Pa, three parts in ten million of atmospheric pressure, and at 1.0 kHz the air moves through s₀ = p₀/(ρvω) = 1.1 × 10⁻⁸ m. Light obeys the same form: an electromagnetic wave carries I = ½ε₀cE₀².

Let a source emit average power P into a medium that neither absorbs nor scatters. Then the same P crosses every surface enclosing the source, and intensity is just P divided by the area available. A beam held in a pipe, a duct or a waveguide keeps a fixed cross-section, so I stays constant. A source long compared with the distance of interest — a motorway, a strip light, a hot pipe — spreads onto a cylinder of area 2πrL, giving I = P/(2πrL) ∝ 1/r. A source small compared with the distance, radiating equally in all directions, spreads onto a sphere of area 4πr², giving I = P/(4πr²) ∝ 1/r². Amplitude follows as √I in each case: constant, 1/√r, 1/r. On the decibel scale, β = 10 log₁₀(I/I₀) with I₀ = 1.0 × 10⁻¹² W m⁻², each doubling of r costs 10 log₁₀4 = 6.02 dB from a point and 3.01 dB from a line. A 0.50 W isotropic siren gives I = 0.50/(4π × 10.0²) = 4.0 × 10⁻⁴ W m⁻², or 86 dB, at 10.0 m, and 80 dB at 20.0 m. No energy was destroyed on the way.

A real medium converts some wave energy to heat. Each metre of path removes a fixed fraction of whatever is left, so dI/dx = −αI and I(x) = I(0) e^(−αx), with α the absorption coefficient in m⁻¹; after one distance 1/α the intensity is down to 37%. Absorption and spreading multiply rather than replace each other: I(r) = P e^(−αr)/(4πr²). Which one dominates depends on where you stand, and decibels make the comparison clean. Spherical spreading costs 8.686/r dB per metre, a rate that weakens as you move out, while absorption costs a constant 4.343α dB per metre. They are equally expensive at r = 2/α. With α = 0.010 m⁻¹, roughly the upper audio band in air, geometry rules the first 200 m and absorption rules beyond it. Since α climbs steeply with frequency, distant thunder arrives as a rumble: the treble has been absorbed away over kilometres while the bass has only been thinned by spreading.

Test the geometry with a log-log plot, where a power law is a straight line: log I = log P − log 4π − 2 log r for a point source, so the slope is −2, against −1 for a line and 0 for a guided beam. Absorption cannot imitate a change of slope, because e^(−αr) curves the line downward instead of tilting it. Then check the assumptions, because each geometry has a range of validity. Nearer than about one source dimension you are in the near field and no spreading law applies: a 3 m loudspeaker array behaves as a line source at 5 m and as a point source at 200 m. Indoors, reflections build a reverberant field that flattens the curve to nearly constant intensity beyond the critical distance, which is why the 6 dB rule fails in a room. A duct, an optical fibre or an oceanic sound channel removes the spreading in one or two dimensions. A directional source concentrates power into part of the sphere, I = DP/(4πr²) with directivity D, which shifts the intercept but leaves the exponent alone.