Wave Energy and Intensity
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Direct answer
A wave is a moving patch of stressed, moving medium: both potential and kinetic energy travel with it, and both scale as the square of amplitude, giving average string power P = ½μvω²A² and three-dimensional intensity I = ½ρvω²A² (μ mass per unit length, ρ density, v wave speed, ω angular frequency). Doubling amplitude quadruples the intensity; doubling frequency quadruples it too. From a point source the power spreads over spheres, so I = P/4πr² falls as 1/r² while the amplitude itself falls only as 1/r — a distinction JEE tests relentlessly. Sound levels compress this range logarithmically: β = 10 log10(I/I0) with the threshold of hearing I0 = 10⁻¹² W/m², so every ten-fold intensity rise adds 10 dB, every doubling adds 3 dB, and doubling distance from a point source subtracts 6 dB.
What you must remember
- Squares law: intensity ∝ A² and ∝ ω² (i.e. f²); quadrupling amplitude raises intensity sixteen-fold — check exponents before anything else in the question.
- String wave power: P = ½μvω²A², with v = √(T/μ) the wave speed; energy transport needs the medium to move, and the power flows at the wave speed.
- Three-dimensional intensity: I = ½ρvω²A²; for a point source radiating power P, I = P/4πr² at distance r.
- Inverse-square consequences: I ∝ 1/r² but A ∝ 1/r; doubling distance quarters intensity (−6 dB) and halves amplitude.
- Decibel scale: β = 10 log10(I/I0), I0 = 10⁻¹² W/m²; ×10 intensity = +10 dB, ×2 = +3 dB; normal conversation near 60 dB (10⁻⁶ W/m²), pain threshold 120 dB (1 W/m²).
- Standing waves carry no net energy: energy oscillates locally between antinodes instead of propagating — superposition redistributes energy, it never destroys it.
- Beat and interference link: at destructive interference the missing energy has relocated to constructive regions; total energy remains conserved across the pattern.
Worked comparison: two sources
A small source radiates 0.5 W of sound. At 10 m, I = P/4πr² = 0.5/1257 ≈ 4 × 10⁻⁴ W/m², a level of 10 log10(4 × 10⁸) ≈ 86 dB. Walk to 20 m and intensity falls to 10⁻⁴ W/m² — 80 dB: doubling distance removed 6 dB exactly as the inverse-square law demands. Now keep the position fixed and double the source's amplitude instead: intensity quadruples to 1.6 × 10⁻³ W/m², a jump of +6 dB from the original — amplitude doubling and distance doubling produce equal-magnitude, opposite-sign level changes, a symmetry worth internalising because options are set to catch it.
Second comparison, on a string: two waves of the same frequency, amplitudes A and 3A, travel on the same wire. Their power ratio is 1:9 (amplitude squared), so if the smaller wave carries 2 W, the larger carries 18 W. Now let them interfere: at a constructive point the resultant amplitude is 4A, and the local intensity is sixteen times that of the first wave alone — yet summed over the whole pattern the total power is just 20 W. The arithmetic makes the conservation point unmissable: interference piles energy into antinodes by draining the nodes, and any question implying energy destruction at destructive interference has a wrong premise built in.
Exam angles on intensity
The classic slip is assigning amplitude a 1/r² fall; that belongs to intensity, amplitude falls as 1/r for spherical waves. Second, decibel arithmetic inverts under subtraction: going from 80 dB to 60 dB is a hundred-fold intensity drop, not a quarter, and linear instincts about "dB difference" misfire. Third, frequency dependence is forgotten: at the same amplitude, a 2 kHz wave carries four times the intensity of a 1 kHz wave, a fact questions exploit through "identical amplitude" phrasing. Fourth, the string formula's v is the wave speed √(T/μ), not the particle speed ωA — a dimensional slip that produces wildly wrong answers and a favourite of diagnostic multiple-correct items. Main tests plug-in intensity and dB numbers; Advanced prefers reasoning — sketch the 1/r versus 1/r² behaviour, compare two interfering sources' power budgets, or connect wave energy to why the second harmonic contributes disproportionately to a plucked guitar string's brightness.
Frequently asked questions
How does the energy carried by a wave depend on amplitude and frequency?
Intensity scales with the square of both: I ∝ A²ω², so doubling either amplitude or frequency quadruples the intensity.
Why does amplitude fall as 1/r but intensity as 1/r² from a point source?
Power spreads over a sphere of area 4πr², so intensity must fall as 1/r²; since intensity goes as amplitude squared, amplitude itself falls only as 1/r.
What intensity corresponds to the threshold of hearing and of pain?
10⁻¹² W/m² defines 0 dB at the hearing threshold, and 1 W/m² corresponds to 120 dB, the pain threshold — a factor of 10¹² spanned by the decibel scale.
How many decibels does doubling the intensity add?
3 dB, since 10 log10(2) ≈ 3; a ten-fold intensity increase adds exactly 10 dB.
Do standing waves transport energy?
No — in a standing wave energy shuttles between adjacent antinodes locally rather than propagating, which is why the net power flow past any point averages to zero.