Wave Energy, Power & Intensity

Mark one element of a string and watch it while a wave passes. Over one period it rises to +A, falls to −A and returns: its net displacement is exactly zero, and it never advances along the string. With A = 2.5 mm at 100 Hz it traces 4A = 10 mm of path per cycle, an average path speed of 1.0 m s⁻¹, while the pattern itself sweeps past at v = 100 m s⁻¹ — a hundred times faster, and in a direction the element never moves in at all. What actually crosses the room is energy, passed from each element to its neighbour by the transverse component of the tension. That is the whole claim of this topic, and it is why the quantities worth computing here are an energy per unit length, a power, and an intensity, and never a velocity of anything material.

Take y(x, t) = A sin(kx − ωt) and differentiate it two ways. The transverse velocity is ∂y/∂t = −ωA cos(kx − ωt), so a length dx carries kinetic energy ½ μ dx (∂y/∂t)², a kinetic density ½ μ ω² A² cos²(kx − ωt). The elastic energy comes from stretching: a segment of slope ∂y/∂x is longer than dx by about ½ (∂y/∂x)² dx, and pulling that extra length out against the tension F costs ½ F (∂y/∂x)² dx, a potential density ½ F k² A² cos²(kx − ωt). Now use the two facts the wave equation supplied, ω = vk and F = μv²: F k² = μ v² k² = μ ω². The two densities are identical, term for term and instant for instant. Their sum is u = μ ω² A² cos²(kx − ωt). On the numbers above each peaks at 6.17 mJ m⁻¹ and the total at 12.3 mJ m⁻¹.

u(x, t) oscillates, so quote its mean. Hold x fixed and integrate over one period: (1/T) ∫₀^T cos²(kx − ωt) dt = ½, since cos² averages to a half over any whole number of cycles. Hence ū = ½ μ ω² A². The energy lying in a length v dt of string crosses a fixed point in time dt, so the average power is that mean density times the wave speed: Pₐᵥ = ū v = ½ μ v ω² A². The same answer arrives by a route that never mentions energy density. The string on the left of a point pulls the string on the right with a transverse force −F ∂y/∂x, and that force does work at the rate P = −F (∂y/∂x)(∂y/∂t) = F k ω A² cos²(kx − ωt), never negative for a wave moving in +x. Its cycle average is ½ F k ω A², and substituting F = μv² and k = ω/v returns ½ μ v ω² A². Check the dimensions: (M L⁻¹)(L T⁻¹)(T⁻²)(L²) = M L² T⁻³, which is watts.

Algebraically A and ω enter Pₐᵥ = ½ μ v ω² A² identically; in the laboratory they do not. Take μ = 5.0 × 10⁻³ kg m⁻¹ under 50 N, so v = √(F/μ) = 100 m s⁻¹ and ½ μ v = 0.25 kg s⁻¹. At f = 100 Hz and A = 2.5 mm the wave carries 0.25 × (628.3)² × (2.5 × 10⁻³)² = 0.617 W. Double the amplitude and the power quadruples to 2.47 W with the wavelength untouched. Double the frequency instead and it also quadruples, but v is fixed by F and μ alone on a non-dispersive string, so the wavelength halves and it is the transverse speed ωA that has doubled. The difference shows in the quantity that limits the model: the maximum slope is kA = ωA/v, which is 0.0157 here. Turning either dial raises it, and the linear description fails first at whichever dial you turn hardest.

Power is a property of the whole wave; intensity is a property of a place. I = Pₐᵥ/S, where S is the wavefront area the power is crossing, so I is measured in W m⁻². The rest is geometry, provided the medium neither absorbs nor focuses. A point source radiating P uniformly fills nested spheres, and the same P crosses every one, so I(r) = P/(4π r²). A 4.0 W source gives 4.0/(4π × 5.0²) = 0.0127 W m⁻² at 5.0 m and 0.00318 W m⁻² at 10 m, a quarter — while the amplitude, which goes as √I, only halves. A long line source fills nested cylinders of area 2π r ℓ instead: 9.0 W radiated along 60 m of pipeline gives 9.0/(2π × 3.0 × 60) = 7.96 × 10⁻³ W m⁻² at 3.0 m and 3.98 × 10⁻³ W m⁻² at 6.0 m, halved rather than quartered. A plane wave keeps a wavefront of fixed area, so in a lossless medium its intensity does not fall at all.

Linear: every result above used ½(∂y/∂x)² for the stretch and held F constant, which needs a small slope — kA = 0.0157 in the example, so the neglected terms are of order 10⁻⁴. Drive the string hard enough and the tension varies over a cycle, the speed becomes amplitude-dependent, and the power stops going as A². Lossless: real media turn wave energy into heat, so the geometric result acquires a factor e^(−αr) for an absorption coefficient α in m⁻¹. At α = 0.02 m⁻¹ and r = 20 m that factor is 0.670, and the lossless prediction is half as large again as the truth. Far field: I = P/(4π r²) also assumes the source radiates equally in all directions and is small compared with r, so within a wavelength or two of a real transducer it simply does not hold. Two limiting checks keep the algebra honest: A → 0 or ω → 0 must send Pₐᵥ → 0, and over a patch small compared with r a sphere is a plane, so the spherical result must flatten into the plane-wave one.