Energy Density and Power Transport

Derivation

Energy Density and Power Transport

D-068 Home PU-103 Threads energy · waves Depends on The Wave Equation as a Continuum Limit, Energy Exchange and Conservation in SHM

Statement

For a small-amplitude transverse wave y(x,t) = A cos(kx − ωt) on a uniform string of linear mass density μ under tension T, the kinetic and potential energy densities are equal at every point, the total energy per unit length is μω²A² sin²(kx − ωt), and the energy flux (power) travels at the wave speed v with time average ⟨P⟩ = ½ μvω²A², so that transported power scales as amplitude squared times frequency squared.

Why it matters

The dependence P ∝ A²ω² is the single most reused fact about waves. It fixes why loudness goes as the square of pressure amplitude, why the intensity of light depends on the field squared, and why high-frequency modes dominate radiative losses. Every "intensity" you will meet later — acoustic, electromagnetic, gravitational — inherits this quadratic-in-amplitude, quadratic-in-frequency structure from exactly this calculation.

It also delivers the first clean example of a flux: a locally conserved quantity (energy) obeying a continuity equation, with the flux equal to density times a transport velocity. That template — density, current, continuity — is the backbone of thermodynamics, electromagnetism, and fluid mechanics.

Assumptions

Small transverse displacements.If the slope ∂y/∂x is not small, the potential energy is no longer ½T(∂y/∂x)² per length; the exact arc-length stretch adds higher-order terms and tension itself varies along the string, so the neat equality of energies breaks.

Uniform, lossless string.If μ or T varies with position, v varies and part of the wave reflects; if there is internal friction, energy leaks to heat and the flux is no longer conserved along the string.

Tension constant in time and unaffected by the wave.If the displacement changes the tension appreciably (large amplitude), T couples to y and the wave equation becomes nonlinear, so a single travelling A cos(kx−ωt) is no longer an exact solution.

A single travelling harmonic, not a standing wave.For a superposition of oppositely moving waves the cross terms give a net flux that oscillates and can time-average to zero (a pure standing wave transports no energy on average), so ⟨P⟩ = ½μvω²A² is specific to one-way propagation.

Derivation

v = √(T/μ), ω = vk

Imported from wave-equation-from-loaded-string: the string obeys ∂²y/∂t² = v² ∂²y/∂x², and A cos(kx−ωt) solves it only when ω=vk. A

∂y/∂t = Aω sin(kx−ωt), ∂y/∂x = −Ak sin(kx−ωt)

Differentiate the given waveform. Both derivatives carry the same phase argument φ = kx−ωt, which is what makes the two energy densities track each other. A

dK/dx = ½ μ (∂y/∂t)² = ½ μA²ω² sin²φ

A length dx has mass μ dx moving transversely at speed ∂y/∂t; its kinetic energy is ½(μdx)(∂y/∂t)². Divide by dx. A

dU/dx = ½ T (∂y/∂x)² = ½ TA²k² sin²φ

Stretching a segment against tension stores work; to leading order the extra length is ½(∂y/∂x)²dx, and multiplying by T gives the elastic energy per length. This is where small slope is used. B t3

Tk² = μv²k² = μω² ⇒ dU/dx = dK/dx

Substitute T = μv² and vk = ω from step 1. The potential density collapses onto the kinetic density: equipartition holds pointwise, not merely on average. B t3

dE/dx = dK/dx + dU/dx = μA²ω² sin²φ

Add the two equal densities. The energy is bunched where sin²φ = 1 (steepest, fastest points) and vanishes at the crests and troughs, contrary to naive intuition. A

P(x,t) = −T (∂y/∂x)(∂y/∂t)

The material to the left pulls the material to the right with transverse force −T ∂y/∂x; the rate it does work on the right is that force times the transverse velocity ∂y/∂t. That work rate is the energy crossing the point per unit time. C t3

P = −T(−Ak sinφ)(Aω sinφ) = TA²kω sin²φ

Insert the derivatives from step 2. P ≥ 0 everywhere for a wave moving in +x: energy always flows in the propagation direction. B

P = TA²kω sin²φ = v · μA²ω² sin²φ = v (dE/dx)

Using Tk = μv²k = μvω. Flux equals energy density times wave speed — the continuity statement ∂(dE/dx)/∂t + ∂P/∂x = 0 with transport velocity v. C t3

⟨sin²φ⟩ = ½ ⇒ ⟨P⟩ = ½ μvω²A², ⟨dE/dx⟩ = ½ μω²A²

Average sin² over a full cycle. The frequency-squared and amplitude-squared laws now stand out explicitly. A

Result

⟨P⟩ = ½ μvω²A² = v ⟨dE/dx⟩, dK/dx = dU/dx

Reading. A one-way harmonic wave carries, on average, an energy per unit length ½μω²A², split equally and instant-by-instant between kinetic and potential. That energy streams past any point at the wave speed, giving an average power that quadruples when you double either the amplitude or the frequency. Double the frequency and the same-sized wiggle delivers four times the power; halve the amplitude and power drops to a quarter.

Units check. [μ][v][ω]²[A]² = (kg·m⁻¹)(m·s⁻¹)(s⁻²)(m²) = kg·m²·s⁻³ = W. Energy density: (kg·m⁻¹)(s⁻²)(m²) = kg·m·s⁻² = J·m⁻¹. Both correct.

Limiting cases

ω → 0 (static displacement): ⟨P⟩ → 0. A frozen shape stores energy but transports none — flux needs motion.
A → 0: power vanishes as A², so the small-amplitude regime is exactly where the formula is trustworthy.
Fix A, raise ω: power grows as ω² — high notes and blue light carry disproportionately more energy at equal amplitude.
Fix input power: required amplitude scales as 1/ω, so a heavier, slower wave needs bigger displacement to move the same watts.
Standing wave (equal counter-propagating pair): the two fluxes cancel on average, ⟨P⟩ → 0, though the energy density is nonzero — the formula for one-way flux no longer applies.

Breaks when

Large amplitude / steep slope. Once (∂y/∂x)² is not small, the leading-order stretch energy in step 4 is wrong, tension varies along the string, and the wave equation turns nonlinear — the pointwise equality of K and U and the clean A² law both fail.
Dispersive or dissipative medium. If v depends on ω (stiffness, boundary effects) energy travels at the group velocity, not v = ω/k; if the string has damping the flux decays exponentially and is not conserved along x.
Non-uniform string / interfaces. A change in μ or T partially reflects the wave; the transmitted power is reduced and ⟨P⟩ = ½μvω²A² must be applied piecewise with the local values.
Non-sinusoidal or transient pulses. For a pulse there is no single ω; one must integrate the energy densities over the actual profile rather than use ⟨sin²⟩ = ½.

Failure modes

Peak-vs-average slip. Quoting P = μvω²A² (the peak) where ⟨P⟩ = ½μvω²A² (the average) is meant — a factor-of-two error from dropping ⟨sin²⟩ = ½.
Energy "at the crests." Believing the energy sits at the crests. It is largest at the zero-crossings where the string moves fastest and is most stretched (sin²φ = 1), and zero at the crests.
f vs ω confusion. Writing f² where ω² = (2πf)² belongs — a missing factor 4π² ≈ 39.5.
Treating K and U like a pendulum. Assuming kinetic and potential are 90° out of phase (max K at zero U). In a travelling wave they are in phase and equal everywhere; the out-of-phase picture belongs to a single oscillator or a standing wave.
Forgetting the sign / direction of flux. Dropping the minus sign in P = −T(∂y/∂x)(∂y/∂t) and concluding energy can flow backward for a +x wave.
Using string speed for energy speed. Confusing the transverse particle speed ∂y/∂t (max Aω) with the propagation speed v that actually carries the energy.

Discussion

The result is really two statements bolted together. First, energy is locally conserved: the density dE/dx and the flux P satisfy a continuity equation ∂_t(dE/dx) + ∂_xP = 0, exactly the form ∂ρ/∂t + ∂j/∂x = 0 you will re-meet for charge and mass. Second, the flux is the density carried at the propagation speed, P = v(dE/dx) — the mechanical ancestor of the Poynting vector S = E×H and of acoustic intensity I = p u.

The pointwise equality dK/dx = dU/dx is special to a travelling wave and is often misremembered. For a single oscillator, kinetic and potential energy trade off a quarter-period out of phase; a mass at the extreme of its swing has all potential, none kinetic. In a travelling string wave every element behaves differently: at the moment it flashes through y = 0 it has both maximum speed and maximum slope, so K and U peak together. The energy therefore travels in packets located at the steep zero-crossings, not at the visually obvious crests. This connects forward to shm-energy-conservation: the total per particle is the same ½mω²A²-type expression, but its time-partition is what distinguishes a standing oscillation from a running wave.

The amplitude-squared law is why "intensity" is defined the way it is throughout physics. Because detectors and ears respond to time-averaged power, and that average is quadratic in amplitude, we build logarithmic scales (decibels, magnitudes) on top of it and habitually work with amplitude² as the physically meaningful "strength." The frequency-squared factor is equally consequential: it is behind the strong ω-dependence of radiated power (dipole radiation goes as ω⁴ once you also let the wave spread), the preferential loss of high harmonics, and the reason a stiff, high-frequency mode is an efficient energy carrier.

More carefully, "energy travels at v" is a statement about a single Fourier component; for a wave packet built from a band of frequencies the energy propagates at the group velocity v_g = dω/dk, which coincides with the phase velocity v = ω/k only in a non-dispersive medium such as the ideal string. The averaged energy density and flux can be derived in one stroke from the second-order term in the wave's Lagrangian density ℒ = ½μ(∂_ty)² − ½T(∂_xy)²: the canonical stress-energy tensor of this field theory has T⁰⁰ = dE/dx and T⁰¹ = P, and its conservation ∂_μT^μν = 0 is the continuity equation of step 9 — the same machinery that yields energy–momentum conservation for the electromagnetic and other classical fields.

Common misconceptions. (i) The energy does not oscillate between "the whole string kinetic" and "the whole string potential" — at every instant half is kinetic and half potential. (ii) A bigger amplitude does not raise the wave speed; v depends only on T and μ, while amplitude sets only how much energy rides along. (iii) A standing wave stores energy but, averaged over a cycle, transports none — zero net flux is not zero energy.

Worked examples

Guitar-like string. μ = 5.0×10⁻³ kg·m⁻¹, T = 80 N, A = 5.0 mm, f = 60 Hz. Find v, ⟨dE/dx⟩, and ⟨P⟩.

A driven string with a one-way harmonic wave; apply the boxed results directly. A

v = √(T/μ) = √(80 / 5.0×10⁻³) = √(1.6×10⁴) = 126.5 m·s⁻¹

Wave speed from tension and density. A

ω = 2πf = 2π(60) = 377 rad·s⁻¹, ω² = 1.421×10⁵ s⁻²

Convert frequency to angular frequency before squaring — this is where the 4π² hides. A

⟨dE/dx⟩ = ½μω²A² = ½(5.0×10⁻³)(1.421×10⁵)(5.0×10⁻³)²

Insert numbers; A² = 2.5×10⁻⁵ m². A

⟨dE/dx⟩ = 8.9×10⁻³ J·m⁻¹, ⟨P⟩ = v⟨dE/dx⟩ = 126.5 × 8.9×10⁻³

Flux equals density times speed. A

⟨dE/dx⟩ ≈ 8.9 mJ·m⁻¹, ⟨P⟩ ≈ 1.1 W

Reading. Only about a watt flows even at a healthy 5 mm swing — strings are gentle energy carriers, which is why an unamplified instrument is quiet. Quadrupling to 120 Hz at the same amplitude would push this to about 4.5 W.

Heavy rope, slow shake. μ = 0.020 kg·m⁻¹, T = 50 N, A = 2.0 cm, f = 4.0 Hz. Find ⟨P⟩, then the amplitude needed to double it at fixed f.

Same formula, larger displacement and lower frequency; then invert for A. B

v = √(50 / 0.020) = √2500 = 50 m·s⁻¹, ω = 2π(4.0) = 25.13 rad·s⁻¹

Speed and angular frequency. A

⟨P⟩ = ½μvω²A² = ½(0.020)(50)(631.5)(2.0×10⁻²)²

ω² = 631.5 s⁻², A² = 4.0×10⁻⁴ m². A

⟨P⟩ = 0.126 W ≈ 0.13 W

Collect the factors: ½·0.020·50 = 0.50; ×631.5 = 315.8; ×4.0×10⁻⁴ = 0.126 W. A

⟨P⟩ ∝ A² ⇒ A_new = A√2 = (2.0 cm)(1.414) = 2.83 cm

To double the power at fixed ω and v, scale amplitude by √2, not 2. B

⟨P⟩ ≈ 0.13 W; doubling power needs A = 2.83 cm

Reading. Despite a swing four times larger than Example 1, the low frequency keeps the power an order of magnitude smaller — the ω² factor dominates over the A² factor here. And because power is quadratic in amplitude, a mere 41% bigger shake, not a doubled one, doubles the delivered watts.

Problems

A string has μ = 8.0×10⁻³ kg·m⁻¹ and T = 200 N. A 50 Hz wave of amplitude 4.0 mm travels along it. Find the average power transmitted.
Solution
v = √(200/8.0×10⁻³) = √25000 = 158.1 m·s⁻¹. ω = 2π(50) = 314.2 rad·s⁻¹, ω² = 9.87×10⁴ s⁻². A² = 1.6×10⁻⁵ m². ⟨P⟩ = ½(8.0×10⁻³)(158.1)(9.87×10⁴)(1.6×10⁻⁵) = ½×8.0×10⁻³×158.1 = 0.632; ×9.87×10⁴ = 6.24×10⁴; ×1.6×10⁻⁵ = 1.0 W.
A wave transports 2.0 W. Without changing the string or the frequency, by what factor must the amplitude change to transmit 8.0 W? What if instead only the frequency is changed?
Solution
At fixed ω, v: ⟨P⟩ ∝ A², so A_new/A = √(8.0/2.0) = √4 = 2.0×. At fixed A: ⟨P⟩ ∝ ω² ∝ f², so f_new/f = √4 = 2.0× as well. Either doubling works because both enter squared.
Show from the pointwise densities that, for a travelling wave, the ratio of the maximum energy density to its time-averaged value is exactly 2. Where along the wave does the maximum occur?
Solution
dE/dx = μω²A² sin²φ. Maximum is at sin²φ = 1, giving (dE/dx)_max = μω²A². The average uses ⟨sin²φ⟩ = ½, giving ½μω²A². Ratio = 2. Since φ = kx − ωt, sin²φ = 1 where sinφ = ±1, i.e. where cosφ = 0 — the zero-crossings (y = 0), not the crests. There the transverse speed and the slope are both maximal.
Two identical waves of the same amplitude and frequency travel in opposite directions on the same string, forming a standing wave. Using P = −T(∂y/∂x)(∂y/∂t), show that the time-averaged flux is zero.
Solution
Superpose y = Acos(kx−ωt) + Acos(kx+ωt) = 2A coskx cosωt. Then ∂y/∂x = −2Ak sinkx cosωt and ∂y/∂t = −2Aω coskx sinωt. So P = −T(−2Ak sinkx cosωt)(−2Aω coskx sinωt) = −4TA²kω sinkx coskx sinωt cosωt = −TA²kω sin(2kx) sin(2ωt). The time average of sin(2ωt) over a cycle is 0, so ⟨P⟩ = 0 everywhere: a standing wave carries no net energy, though its energy density is nonzero.
A motor drives one end of a rope (μ = 0.015 kg·m⁻¹, v = 40 m·s⁻¹) with a 6.0 mm amplitude harmonic wave and must deliver 0.50 W. What driving frequency is required? What transverse peak force does the motor exert? (Peak transverse driving force magnitude = T k A.)
Solution
From ⟨P⟩ = ½μvω²A²: ω² = 2⟨P⟩/(μvA²) = 2(0.50)/[(0.015)(40)(6.0×10⁻³)²] = 1.0/[0.015×40×3.6×10⁻⁵] = 1.0/(2.16×10⁻⁵) = 4.63×10⁴ s⁻². ω = 215.2 rad·s⁻¹, so f = ω/2π = 34.2 Hz. Tension: T = μv² = 0.015×1600 = 24 N; k = ω/v = 215.2/40 = 5.38 m⁻¹. Peak force = TkA = 24×5.38×6.0×10⁻³ = 0.77 N.