University Physics II · Optional extension · Unnumbered preflight · Waves for E&M and Optics · W.4
Wave Energy, Power and Intensity
A wave delivers energy by having each part of the medium do work on the next. Power goes as amplitude squared, intensity is that power per unit area of wavefront, and how intensity falls with distance is a statement about geometry.
Build the model
Connect the measurement to the mechanism.
A wave transports energy without transporting the medium. On a string each element does work on the one ahead through the transverse pull of the tension, P = −F(∂y/∂x)(∂y/∂t), which for y = A cos(kx − ωt) averages to P(av) = ½μvω²A². Two features carry into every medium.
Power goes as the square of the amplitude and the square of the frequency, because each energy density is quadratic in a first derivative of y. And energy flux equals energy density times wave speed, which is what turns power into intensity, I = P(av)/A(⊥) in W m⁻², once the wave spreads in two or three dimensions. With the source power fixed, geometry alone sets how I falls: constant in a guide, 1/r from a line, 1/r² from a point, with amplitude going as √I in every case.
That is bookkeeping, not loss. Absorption is the separate mechanism, exponential in path length, and the two have different signatures in data.
- Simple definition
- A wave carries energy at the wave speed. Average power is the energy crossing a point per second, ½μvω²A² on a string; intensity is that power per unit area of wavefront, I = P(av)/A(⊥), and both scale as amplitude squared.
- Example
- A 60 Hz wave of amplitude 5.0 mm on a string with μ = 0.050 kg m⁻¹ under 80 N carries 3.55 W. Double the amplitude and it carries 14.2 W — same string, same tension, four times the power.
The transverse pull of the tension acting through the transverse velocity.
F in N, μ in kg m⁻¹; never negative for a wave moving in +x
⟨sin²⟩ = ½ over a cycle. Double A, or double f, and the power quadruples.
kg m⁻¹ × m s⁻¹ × s⁻² × m² = kg m² s⁻³ = W
Kinetic and potential energy peak together, unlike a mass on a spring.
e(av) = ½μω²A² in J m⁻¹; the two halves are equal at every point
Energy density times wave speed again, now counted per unit area.
I in W m⁻²; ρv is the impedance, 412 Pa s m⁻¹ in air
The same power through a larger surface. Nothing has been absorbed.
Amplitude ∝ √I: 1/r, 1/√r, constant · −6.02 and −3.01 dB per doubling
Absorption costs 4.343α dB m⁻¹, spreading 8.686/r dB m⁻¹; equal at r = 2/α.
α in m⁻¹; I₀ = 1.0 × 10⁻¹² W m⁻² for sound level in dB
From work on the next element to average power
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. It is the general statement that a wave delivers energy by having each part of the medium do work on the next.
Why the squares, and why the flux pattern transfers
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.
Intensity, and what the medium contributes
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.
Spreading is geometry, not loss
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.
Absorption is exponential, and it competes with spreading
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) = I₀e(−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.
Checking the model across geometries
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.
Change one variable at a time
Make the relationship visible.
Set absorption to zero and step n from 2 to 1 to 0: the line tilts but stays dead straight. Then raise α — the tilt is untouched while the curve bows below the line and keeps steepening, so a new straight slope means new geometry and curvature means absorption.
LEVEL AT r55.7 dB
LOST TO SPREADING40.0 dB
LOST TO ABSORPTION4.3 dB
AMPLITUDE A(r)/A(1 m)0.0061
Live interpretationLEVEL AT r: 55.7 dB. LOST TO SPREADING: 40.0 dB. LOST TO ABSORPTION: 4.3 dB. AMPLITUDE A(r)/A(1 m): 0.0061
Catch the common trap
Explain before calculating.
In still air, well away from reflecting surfaces, a microphone measures intensity from 2.0 m to 20 m from a steady source. A plot of log₁₀I against log₁₀r is straight, with slope −1.00 across the whole range. Which conclusion does that best support?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyA string of linear density μ = 0.040 kg m⁻¹ is held under a tension of 90 N. A travelling wave of amplitude 4.0 mm and frequency 50 Hz runs along it. Find the wave speed and the average power the wave carries. Then find the power when the amplitude is raised to 6.0 mm at the same frequency.
- Wave speed from the medium alone: v = √(F/μ) = √(90/0.040) = √2250 = 47.4 m s⁻¹. Angular frequency ω = 2πf = 2π(50) = 314 rad s⁻¹, so ω² = 9.8696 × 10⁴ s⁻².
- The medium factor in P(av) = ½μvω²A² is μv = √(μF) = √(0.040 × 90) = √3.6 = 1.897 kg s⁻¹.
- P(av) = ½(1.897)(9.8696 × 10⁴)(4.0 × 10⁻³)² = ½(1.897)(9.8696 × 10⁴)(1.6 × 10⁻⁵) = 1.50 W. Units: kg s⁻¹ × s⁻² × m² = kg m² s⁻³ = W.
- Amplitude enters squared, so raising it from 4.0 mm to 6.0 mm multiplies the power by (6.0/4.0)² = 2.25: P(av) = (1.4981)(2.25) = 3.37 W. The string, the tension and the frequency were never touched.
Answerv = 47.4 m s⁻¹ and P(av) = 1.50 W; at 6.0 mm amplitude, 3.37 W.
MediumA small loudspeaker radiates 2.5 mW of acoustic power uniformly in all directions in open air, far from any reflecting surface. Take ρ = 1.20 kg m⁻³, v = 343 m s⁻¹ and I₀ = 1.0 × 10⁻¹² W m⁻², and ignore absorption. Find the intensity and the sound level at 4.0 m, the pressure amplitude there, and the level at 16 m.
- Nothing absorbs, so all 2.5 mW crosses every sphere: I = P/(4πr²) = 2.5 × 10⁻³/[4π(4.0)²] = 2.5 × 10⁻³/201.06 = 1.243 × 10⁻⁵ W m⁻².
- β = 10 log₁₀(I/I₀) = 10 log₁₀(1.243 × 10⁻⁵/1.0 × 10⁻¹²) = 10 log₁₀(1.243 × 10⁷) = 70.9 dB.
- Pressure amplitude from I = p₀²/(2ρv), with the impedance ρv = 1.20 × 343 = 411.6 Pa s m⁻¹: p₀ = √(2 × 411.6 × 1.243 × 10⁻⁵) = √(1.0236 × 10⁻²) = 0.101 Pa — one part in a million of atmospheric pressure.
- From 4.0 m to 16 m is two doublings of r, each costing 10 log₁₀4 = 6.02 dB, so β = 70.9 − 12.0 = 58.9 dB. Direct check: I = 2.5 × 10⁻³/[4π(16)²] = 7.77 × 10⁻⁷ W m⁻², which is 10 log₁₀(7.77 × 10⁵) = 58.9 dB.
- The source power never changed. The same 2.5 mW is thinned over 16 times the area, so I is 16 times smaller and the amplitude 4 times smaller. No energy was destroyed between 4.0 m and 16 m.
AnswerI = 1.24 × 10⁻⁵ W m⁻² and β = 70.9 dB at 4.0 m, with p₀ = 0.101 Pa; β = 58.9 dB at 16 m.
HardAn ultrasonic transducer, small compared with the distances involved, runs at steady power in a tank of liquid whose absorption coefficient is α = 0.080 m⁻¹. The intensity 0.50 m from it is 4.0 × 10⁻² W m⁻². Find the intensity at 2.0 m, split the total drop in decibels into the spreading part and the absorption part, and find the distance at which the two mechanisms cost the same number of decibels per metre.
- Both mechanisms act at once and multiply: I(r) = I₁(r₁/r)² e(−α(r − r₁)), a point source spreading onto spheres inside an absorbing liquid.
- Spreading factor: (0.50/2.0)² = 0.0625. Absorption factor: e(−0.080 × 1.50) = e(−0.120) = 0.8869.
- I(2.0) = (4.0 × 10⁻² W m⁻²)(0.0625)(0.8869) = 2.22 × 10⁻³ W m⁻².
- In decibels the two costs add: spreading 20 log₁₀(2.0/0.50) = 12.04 dB, absorption 4.343(0.080)(1.50) = 0.52 dB, total 12.56 dB. Check against the intensities: 10 log₁₀(4.0 × 10⁻²/2.217 × 10⁻³) = 10 log₁₀(18.04) = 12.56 dB.
- Rates per metre: spreading costs 8.686/r dB m⁻¹, which weakens as you move out, while absorption costs a constant 4.343α = 4.343(0.080) = 0.347 dB m⁻¹. Setting 8.686/r = 0.347 gives r = 2/α = 2/0.080 = 25 m.
- Equal rates are not equal totals. Out to 25 m spreading has already taken 20 log₁₀(25/0.50) = 34.0 dB while absorption has taken only 4.343(0.080)(24.5) = 8.5 dB — geometry has done four times the work and is only now handing over.
AnswerI(2.0 m) = 2.2 × 10⁻³ W m⁻², from 12.04 dB of spreading and 0.52 dB of absorption; the two cost the same per metre at r = 25 m.