University Physics I · Optional extension · Institutional Extension · Waves & Sound · WS.4
Wave Energy, Power & Intensity
A wave carries energy through a medium whose particles go nowhere. Here is the power a string wave delivers, why doubling the amplitude quadruples it, and how that power thins as the wavefront spreads.
Build the model
Connect the measurement to the mechanism.
A travelling wave moves energy, not matter. Every element of the medium oscillates about a fixed point, and while it does so it does work on the element ahead: on a string the transverse component of the tension, −F ∂y/∂x, acts through the transverse velocity ∂y/∂t, so power flows along the string. For y = A sin(kx − ωt) that instantaneous power is μvω²A²cos²(kx − ωt), averaging over a cycle to P = ½μvω²A².
Two features of that result travel everywhere: the energy goes as the square of the amplitude and as the square of the frequency, so a wave that merely looks twice as tall delivers four times the power. Spread the same power over a growing wavefront and you have intensity, I = P/area — falling as 1/r² from a point source, as 1/r from a line source, not at all for a plane wave. Every step assumes a linear, lossless medium carrying one wave in one direction.
- Simple definition
- A travelling wave transports energy without transporting matter. On a string its average power is P = ½μvω²A², and its intensity is that power divided by the wavefront area it crosses, measured in W m⁻².
- Example
- A 60.0 Hz wave of amplitude 5.00 mm on a string with μ = 0.0500 kg m⁻¹ under 80.0 N of tension (v = 40.0 m s⁻¹) carries 3.55 W. Ask for twice the amplitude and the shaker must supply 14.2 W.
The transverse pull of the string behind acting through the velocity of the element ahead.
F in N; slope and transverse velocity from the wave function
Kinetic and potential shares are equal at every point and at every instant.
Average ½μω²A², in J m⁻¹; μ in kg m⁻¹, ω in rad s⁻¹, A in m
Energy per unit length times the speed at which it is handed along: P = v × dE/dx.
P in W; v = √(F/μ) in m s⁻¹
Power per unit area. On a string, where the wave runs along one line, power is the whole story.
W m⁻²; S is the wavefront area normal to the propagation direction
The same ½ × inertia × speed × ω² × amplitude², with density in place of linear density.
ρ in kg m⁻³; s is displacement amplitude in m, Δp in Pa
Fixed power over a growing wavefront. A plane wavefront never grows, so a lossless plane wave never weakens.
Lossless medium; amplitude then goes as 1/r, 1/√r, and 1/1
What actually moves
Shake one end of a taut string and the disturbance runs its whole length while every particle of string stays where it started, moving only across the line of travel. What propagates is energy, delivered element to element. Take the element at x. The string on its left pulls along its own direction with tension F, and for small slopes the transverse component of that pull is −F ∂y/∂x. That force acts on an end which is itself moving transversely at ∂y/∂t, so the left does work on the right at a rate P(x, t) = −F (∂y/∂x)(∂y/∂t). Nothing there is special to strings: a wave carries power because the stress across any surface acts through the velocity of that surface. Substitute y = A sin(kx − ωt), so ∂y/∂x = kA cos(kx − ωt) and ∂y/∂t = −ωA cos(kx − ωt), and P(x, t) = Fkω A² cos²(kx − ωt). With v = ω/k and F = μv² this is μvω²A² cos²(kx − ωt) — never negative, so for a wave travelling towards +x the energy always flows towards +x.
Kinetic and potential energy arrive together
A length dx of string has mass μ dx and transverse speed ∂y/∂t, so dK = ½μ(∂y/∂t)² dx = ½μω²A² cos²(kx − ωt) dx. The potential energy is stored in stretching: a segment of horizontal extent dx is drawn out to √(dx² + dy²) ≈ dx[1 + ½(∂y/∂x)²], and the tension does work F times that extra length, giving dU = ½F(∂y/∂x)² dx, which is the same ½μω²A² cos²(kx − ωt) dx once F = μv² and ω = vk are used. The two are equal everywhere and always. That deserves a pause, because a mass on a spring does the opposite: its K peaks where its U vanishes. On a travelling wave the element at a crest is instantaneously at rest and the string there is instantaneously unstretched, so it holds neither; the element crossing y = 0 is moving fastest and sits where the string is steepest, so it holds the most of both. Total energy per unit length is μω²A² cos²(kx − ωt), averaging to ½μω²A².
Average power, and what it scales with
cos² averages to ½ over a cycle, so the average power is P = ½μvω²A². Read it as energy density times transport speed: the string holds ½μω²A² joules in every metre and hands that content along at v. Substituting v = √(F/μ) gives P = ½√(μF) ω²A², which separates what the medium supplies, μ and F, from what the source chooses, ω and A. Both source choices enter squared. Take μ = 0.0500 kg m⁻¹ and F = 80.0 N, so v = √(80.0/0.0500) = 40.0 m s⁻¹, and drive it at 60.0 Hz with A = 5.00 mm, ω = 377 rad s⁻¹. Then ½μω²A² = 0.0888 J m⁻¹ and P = (0.0888)(40.0) = 3.55 W. Ask for 10.0 mm at the same frequency and the shaker must supply 14.2 W; drop to 30.0 Hz at the original amplitude and it needs 0.888 W. Amplitude is what the eye reads off a photograph, but frequency counts exactly as heavily.
From power to intensity
A wave spreading through a volume needs power per unit area rather than power: the intensity I = P/S, in W m⁻², across a surface held normal to the propagation direction. The structure of the string result carries straight across. For a sound wave of displacement amplitude s in a fluid of density ρ, I = ½ρvω²s² — the same ½ × inertia × speed × ω² × amplitude², with mass per unit volume replacing mass per unit length. Sound is usually described by its pressure amplitude instead, Δp = ρvωs, which recasts the result as I = Δp²/(2ρv). Air at ρ = 1.20 kg m⁻³ and v = 343 m s⁻¹ has ρv = 412 kg m⁻² s⁻¹, so a painfully loud I = 1.00 × 10⁻² W m⁻² means Δp = √(2 × 412 × 0.0100) = 2.87 Pa, twenty-eight parts per million of atmospheric pressure, and a displacement amplitude at 1.00 kHz of only 1.11 μm. Sound is a very small perturbation, which is why air behaves as a linear medium.
Geometry sets the falloff
Once the power has left the source, geometry alone decides how the intensity falls — provided the medium neither absorbs nor scatters. Draw a closed surface round a point source radiating P uniformly: all of that power crosses every sphere, so I(r) = P/(4πr²). Intensity goes as 1/r², and because I ∝ A², the amplitude goes as 1/r. A 1.00 W source gives I = 1.00/(4π(2.00)²) = 1.99 × 10⁻² W m⁻² at 2.00 m and 4.97 × 10⁻³ W m⁻² at 4.00 m: a quarter of the intensity, half the amplitude. Change the geometry and the exponent changes with it. A long line source — a busy road, a vibrating wire — spreads its power over a cylinder of area 2πrL, so I = P′/(2πr) with P′ the power per unit length, giving intensity as 1/r and amplitude as 1/√r. A plane wavefront never grows at all. None of this is dissipation; it is the same energy, spread thinner.
Where the ideal model stops
Every result above assumes a linear, lossless, unbounded medium carrying one wave in one direction, and each assumption is a place to check. Real media absorb, turning wave energy into heat, so intensity picks up an extra factor exp(−αr) on top of the spreading — and α climbs steeply with frequency, which is why distant thunder rumbles instead of cracking. Inverse-square needs the far field: within roughly one source dimension the radiation has not yet organised into spherical wavefronts, and indoors reflections build a reverberant field whose intensity stops falling with distance at all. Few sources are isotropic; a loudspeaker beams. The single-wave assumption bites hardest — add an equal wave running the other way and you have a standing wave, whose average power flow is exactly zero, energy sloshing between neighbouring quarter-wavelengths without advancing. Comparing intensities on a logarithmic scale is the next topic's business.
Change one variable at a time
Make the relationship visible.
Take the amplitude from 4.0 to 8.0 mm and watch the power readout quadruple, then leave it and raise the frequency to see it climb just as steeply — while the crest dot always lands where the energy curve reads zero.
WAVE SPEED v40.0 m s⁻¹
WAVELENGTH λ0.667 m
MEAN ENERGY / METRE0.0888 J m⁻¹
AVERAGE POWER P3.55 W
Live interpretationWAVE SPEED v: 40.0 m s⁻¹. WAVELENGTH λ: 0.667 m. MEAN ENERGY / METRE: 0.0888 J m⁻¹. AVERAGE POWER P: 3.55 W
Catch the common trap
Explain before calculating.
A small loudspeaker radiates uniformly in every direction into a lossless medium, and a microphone 2.0 m away records pressure amplitude Δp. The speaker is then driven so that its radiated power is four times larger, and the microphone is moved to 4.0 m. What pressure amplitude does it record now?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyA string of linear density μ = 0.0100 kg m⁻¹ is held at 64.0 N tension. A shaker drives one end at 40.0 Hz with amplitude 4.00 mm. What average power does the wave carry along the string?
- The medium fixes the speed: v = √(F/μ) = √(64.0 N ÷ 0.0100 kg m⁻¹) = √6400 = 80.0 m s⁻¹.
- The source fixes the angular frequency: ω = 2πf = 2π(40.0 Hz) = 251.3 rad s⁻¹, so ω² = 6.3166 × 10⁴ rad² s⁻².
- Put both into P = ½μvω²A², with A in metres: ½(0.0100 kg m⁻¹)(80.0 m s⁻¹) = 0.400 kg s⁻¹.
- P = 0.400 × 6.3166 × 10⁴ × (4.00 × 10⁻³ m)² = 0.400 × 6.3166 × 10⁴ × 1.60 × 10⁻⁵ = 0.404 W.
AnswerP = 0.404 W
MediumThe same string, tension and frequency, but the shaker cannot deliver more than 1.00 W. What is the largest amplitude it can drive?
- With μ, v and ω all held, P = ½μvω²A² leaves A as the only free quantity, so P ∝ A².
- Scale from the known point: A(max) = A √(P(max)/P) = 4.00 mm × √(1.00 W ÷ 0.404 W) = 4.00 mm × √2.474.
- √2.474 = 1.573, so A(max) = 4.00 × 1.573 = 6.29 mm.
- Check it straight from the formula: A² = 2P/(μvω²) = 2(1.00) ÷ (0.0100 × 80.0 × 6.3166 × 10⁴) = 3.958 × 10⁻⁵ m², so A = 6.29 × 10⁻³ m.
AnswerA(max) = 6.29 mm — 2.47 times the power buys only 57% more amplitude
HardA small loudspeaker radiates 25.0 mW uniformly in every direction into still air (ρ = 1.20 kg m⁻³, v = 343 m s⁻¹, no absorption). Find the intensity and the pressure amplitude 3.00 m away, and the distance at which the intensity has fallen to a quarter of that.
- All of the power crosses every sphere drawn round the source, so I = P/(4πr²) = 0.0250 W ÷ (4π(3.00 m)²) = 0.0250 ÷ 113.1 = 2.21 × 10⁻⁴ W m⁻².
- Invert I = Δp(max)²/(2ρv): Δp(max) = √(2ρvI), and 2ρv = 2(1.20)(343) = 823 kg m⁻² s⁻¹.
- Δp(max) = √(823 × 2.21 × 10⁻⁴) = √0.182 = 0.427 Pa — about four parts per million of atmospheric pressure.
- Nothing absorbs, so the fall is pure geometry: I ∝ 1/r², and a quarter of the intensity needs twice the distance, r = 2 × 3.00 = 6.00 m. The amplitude there is halved, not quartered.
AnswerI = 2.21 × 10⁻⁴ W m⁻²; Δp(max) = 0.427 Pa; quarter intensity at r = 6.00 m