Energy & Intensity of Waves

A wave's whole purpose is to carry energy from one place to another without carrying any matter. So the honest questions to ask of any wave are: how much energy does it carry, and how is that energy spread out when it arrives? Those two questions — energy and intensity — are what this article answers, and the answers run every loudspeaker, every radio link and every safety limit on a laser. Start from a fact you already own. Each particle of the medium is doing simple harmonic motion, and an SHM oscillator of mass $m$ , amplitude $A$ and angular frequency $\omega$ stores energy $E = \tfrac12 m\omega^2 A^2$ (from Energy in SHM). A travelling wave is just a bucket-chain passing that energy along, element to element, so the power it delivers and the intensity it produces both inherit the same shape:

I \propto A^2 f^2

Two consequences fall straight out, and both are examined relentlessly. Double the amplitude and you quadruple the energy — because energy goes as $A^2$ , not $A$ . And double the frequency and you also quadruple the energy — the $f^2$ that students forget. A wave twice as tall, or twice as shrill, is four times as energetic. Then let the wave spread. A small source radiating into open space sends its power out through ever-larger spheres, and the same energy thinned over a sphere of area $4\pi r^2$ gives the famous inverse-square law, $I = P/4\pi r^2$ — the reason a sound, a light or a radio signal fades as the square of the distance, and its amplitude as $1/r$ . By the end of this page you will be able to say why standing twice as far from a speaker makes it a quarter as intense, put a real wattage on a vibrating string, and see where the decibel scale — the topic of the sound chapter ahead — comes from.

Key takeaways

Each element of the medium does SHM, storing on average $\tfrac12(\mathrm{d}m)\omega^2 A^2$ ; a wave passes that energy along.
Power carried by a wave on a string: $P = \tfrac12\mu v\omega^2 A^2$ . Intensity of a wave spreading through a volume: $I = \tfrac12\rho v\omega^2 A^2 = uv$ , where $u = \tfrac12\rho\omega^2 A^2$ is the energy density.
The headline both share: $I \propto A^2 f^2$ . Double the amplitude → 4× the energy; double the frequency → 4× the energy.
Intensity is power per unit area (W m⁻²). A point source radiating power $P$ into open space gives $I = P/4\pi r^2$ — the inverse-square law — so amplitude falls as $A \propto 1/r$ .
Anchors: a string ( $\mu = 2.0\times10^{-3}$ kg m⁻¹, $v = 200$ m s⁻¹, $f = 50$ Hz, $A = 5.0$ mm) carries $P = 0.493$ W. A 1.0 W point source gives $I = 7.96\times10^{-4}$ W m⁻² at 10 m and exactly a quarter of that, $1.99\times10^{-4}$ W m⁻², at 20 m.
Absorption (energy soaked up by the medium) is a separate effect from spreading — do not confuse the two.
Key idea: a wave's energy comes from the SHM of its particles, so its intensity goes as $A^2 f^2$ ; and when a point source's power spreads over a sphere of area $4\pi r^2$ , the intensity obeys the inverse-square law and the amplitude falls as $1/r$ .

The core insight

Because every particle in a wave carries SHM energy $\tfrac12 m\omega^2 A^2$ , the wave's power and intensity both scale as $A^2 f^2$ — double either and the energy quadruples — and a point source spreading power $P$ over a sphere of area $4\pi r^2$ produces intensity $I = P/4\pi r^2$ , an inverse-square fall with amplitude $\propto 1/r$ .

Where this lives today

The $A^2 f^2$ law and the inverse-square law together set the rules for every technology that moves information or energy by waves. Mobile networks and Wi-Fi live and die by $I = P/4\pi r^2$ : a signal's intensity collapses as the square of the distance to the tower, which is why coverage maps are drawn as filled circles and why moving a router a few metres can rescue a connection. Stage and concert engineering budgets loudspeaker power against the same law — doubling the throw distance demands four times the power for the same intensity — while the $A^2$ dependence is why a modest rise in a driver's excursion yields a large jump in output. Laser safety limits are intensity limits: the same beam power spread over a smaller spot is more intense and more dangerous, so the standards are written in W m⁻². Ultrasound and sonar read intensities to form images and gauge distance, and astronomy turns the inverse-square law around — measuring a star's apparent intensity and, knowing its true power, deducing how far away it is, the method that first put a scale on the galaxy. In every case the two facts of this page are doing the work: energy scales as amplitude and frequency squared, and it thins as the square of the distance.

The physical picture

A wave is a moving parcel of energy. To see how much, look at what each bit of the medium is doing: oscillating in simple harmonic motion about its rest position. From Energy in SHM, a mass $m$ in SHM of amplitude $A$ and angular frequency $\omega$ carries total mechanical energy $E = \tfrac12 m\omega^2 A^2$ — kinetic plus potential, forever swapping between the two but always summing to that. A wave sets one stretch of the medium after another into exactly this motion, so the energy stored per unit length (or per unit volume) is built from that single-oscillator result, and the wave then transports it at the wave speed. Two features of $E = \tfrac12 m\omega^2 A^2$ decide everything. First, the energy depends on the square of the amplitude. A wave twice as tall does not carry twice the energy — it carries four times as much, because $A$ is squared. This is why a modest increase in the height of a wave can be so destructive, and why "twice as loud" is nowhere near "twice the amplitude". Second, the energy depends on the square of the frequency (through $\omega^2 = (2\pi f)^2$ ). A wave of the same amplitude but twice the frequency carries four times the energy, because its particles, completing twice as many cycles a second, are moving twice as fast and so have four times the kinetic energy. Amplitude and frequency enter symmetrically, each squared — the source of the headline $I \propto A^2 f^2$ .

Figure

Wave energy grows as the square of both amplitude and frequency. Left: keeping frequency fixed, doubling the amplitude from A to 2A raises the energy fourfold, a bar four times as tall. Right: keeping amplitude fixed, doubling the frequency from f to 2f likewise raises the energy fourfold. Energy is proportional to A squared times f squared.

Energy density, power and intensity

Three closely related quantities describe a wave's energy, and keeping them straight is half the battle. The energy density $u$ is the energy stored per unit volume of the medium; averaged over a cycle it is

u = \tfrac12\,\rho\,\omega^2 A^2 \qquad (1)

where: $\rho$ is the medium's mass density (kg m⁻³), $\omega = 2\pi f$ the angular frequency, $A$ the amplitude. This is just the SHM energy $\tfrac12 m\omega^2 A^2$ divided by volume, since $\rho$ is mass per volume. The power a wave transports past a point is that energy travelling at the wave speed. For a wave on a string, whose inertia is the linear density $\mu$ , the average power is

P = \tfrac12\,\mu\,v\,\omega^2 A^2 \qquad (2)

where: $v = \sqrt{T/\mu}$ is the wave speed on the string (Speed of a Wave on a Stretched String). For a wave spreading through a three-dimensional medium, the corresponding quantity is the intensity $I$ — the power crossing unit area held square to the wave:

I = \frac{P}{\text{area}} = \tfrac12\,\rho\,v\,\omega^2 A^2 = u\,v \qquad (3)

where: $I$ is in watts per square metre (W m⁻²). Read equation (3) as "energy density carried along at the wave speed": the intensity is the energy packed into each cubic metre, $u$ , swept past at $v$ metres per second. All three — $u$ , $P$ , $I$ — carry the same $\omega^2 A^2$ , which is why they all obey the one headline below.

The headline: I ∝ A² f²

Strip the constants from equation (3) and what remains is the fact worth memorising:

I \propto A^2 f^2 \qquad (4)

since $\omega^2 = (2\pi f)^2 \propto f^2$ . Two independent handles, each squared. Amplitude: double $A$ and the intensity goes up by $2^2 = 4$ ; treble it and by $9$ . Frequency: double $f$ and, at the same amplitude, the intensity again goes up by $4$ — the factor students most often drop, and the one competitive papers most often test ("the amplitude is unchanged but the frequency is doubled; by what factor does the intensity change?" — answer, four). Both together: a wave with twice the amplitude and twice the frequency is $4\times4 = 16$ times as intense. Amplitude and frequency are equal partners here, and both are squared.

Spreading out — the inverse-square law

A wave from a small source does not keep its intensity as it travels, even through a perfectly transparent medium, because the same energy is forced to cover an ever-growing area. Picture a point source radiating power $P$ equally in all directions (isotropically). At distance $r$ , all that power is streaming outward through an imaginary sphere of radius $r$ and surface area $4\pi r^2$ , so the intensity there is the power divided by that area:

I = \frac{P}{4\pi r^2} \qquad (5)

where: $P$ is the source's total radiated power (W) and $r$ the distance from it (m). This is the inverse-square law: the intensity falls as $1/r^2$ . Move twice as far away and the intensity drops to a quarter, not a half; ten times as far, to a hundredth. And since $I \propto A^2$ , the amplitude itself falls only as $A \propto 1/r$ — halving each time you double the distance. (A line source, like a long straight road of traffic, spreads its energy over a cylinder of area $\propto r$ , so its intensity falls more gently as $1/r$ and its amplitude as $1/\sqrt r$ — one honest sentence, no more.) It is vital to keep this geometric thinning apart from absorption, in which the medium actually soaks up wave energy and turns it to heat. The inverse-square law describes energy merely spreading thinner while being fully conserved; absorption removes energy from the wave altogether. Fog does not dim a distant lamp by the inverse-square law — that is absorption and scattering; the inverse-square law is what dims it even on a perfectly clear night.

Figure

A point source radiates power P equally in all directions. The same power flows through every sphere drawn around it, but each larger sphere has a bigger area, so the intensity — power per unit area — falls as one over r squared. At radius r the intensity is I; at 2r it is I over 4; at 3r it is I over 9. The amplitude falls only as one over r.

Intensity, loudness and the decibel

Intensity is a physical quantity in watts per square metre; loudness is the sensation your ear makes of it, and the two are not the same. The quietest sound a healthy young ear can detect, the threshold of hearing, is an intensity of about $I_0 = 1\times10^{-12}$ W m⁻² — an astonishingly tiny power. Because the ear responds to enormous ratios of intensity (from that threshold up to a pain-inducing $1$ W m⁻², a factor of $10^{12}$ ), sound is measured on a logarithmic scale, the decibel, defined by comparing $I$ with $I_0$ . The intensity our 1.0 W source produces at 10 m, $7.96\times10^{-4}$ W m⁻², sits about 89 decibels above threshold — roughly the level of heavy traffic. We quote the number here only to place the physics; the decibel scale itself, and how loudness really tracks it, belong to the Sound chapter ahead and are not developed on this page.

Two derivations

Derivation AThe power carried by a wave on a string

We want to put an actual wattage on a travelling wave — to answer "how many joules per second does this wave deliver?" — and to do it we only need one earlier result: the energy of a single SHM oscillator. The goal is to build $P = \tfrac12\mu v\omega^2 A^2$ from the ground up, by finding the energy stored in a length of string and asking how fast that energy is carried past a point. The whole derivation is one idea: a wave is energy on the move, so its power is "energy per unit length" times "length per unit time".

Step 1. Take a short length $\ell$ of the vibrating string. Its mass is $\mathrm{d}m = \mu\ell$ , and each part of it is doing SHM of amplitude $A$ and angular frequency $\omega$ . From Energy in SHM, the average mechanical energy of that mass is $\Delta E = \tfrac12(\mathrm{d}m)\omega^2 A^2 = \tfrac12\mu\ell\,\omega^2 A^2$ .

Step 2. So the average energy stored per unit length of the string is $\dfrac{\Delta E}{\ell} = \tfrac12\mu\,\omega^2 A^2$ . This is the string's linear energy density — how many joules sit in each metre of the oscillating string.

Step 3. That energy is not sitting still; it travels along at the wave speed $v$ . In a time $\Delta t$ the wave advances a distance $v\,\Delta t$ , so the energy that sweeps past a fixed point in that time is the energy in a length $v\,\Delta t$ : $\ \Delta E_{\text{past}} = \left(\tfrac12\mu\omega^2 A^2\right)(v\,\Delta t)$ .

Step 4. Power is energy delivered per unit time: $P = \dfrac{\Delta E_{\text{past}}}{\Delta t} = \tfrac12\mu v\,\omega^2 A^2$ — equation (2). Notice the shape it hands you: $P \propto \omega^2 A^2 \propto f^2 A^2$ , the headline. Put the anchor numbers in ( $\mu = 2.0\times10^{-3}$ kg m⁻¹, $v = 200$ m s⁻¹, $f = 50$ Hz so $\omega = 314.2$ rad s⁻¹, $A = 5.0$ mm $= 5.0\times10^{-3}$ m): $P = \tfrac12(2.0\times10^{-3})(200)(2\pi\times50)^2(5.0\times10^{-3})^2 = 0.493$ W. Half a watt of mechanical power, flowing along a vibrating string. Verified: P = ½ μ v ω² A² = 0.493 W for the anchor string, recomputed in verify_hs2-wave-energy-intensity.py.

Go deeper · Derivation B · 1 minThe inverse-square law from energy conservation

Derivation BThe inverse-square law from energy conservation

Why should intensity fall as $1/r^2$ exactly — why the square? The goal here is to derive $I = P/4\pi r^2$ from a single unbreakable principle, the conservation of energy, with no other physics at all, and then to read off how the amplitude must behave. The argument is geometric and it is airtight: if the medium neither creates nor destroys energy, then whatever power leaves the source must cross every sphere around it, and a sphere's area is what carries the $r^2$ .

Step 1. Let a point source radiate a steady total power $P$ equally in all directions, into a medium that absorbs nothing. Draw an imaginary sphere of radius $r$ centred on the source.

Step 2. Every joule the source emits must pass outward through that sphere — there is nowhere else for the energy to go, and none is lost on the way. So the total power crossing the sphere equals the source power $P$ , whatever the radius $r$ : the power through the sphere at $2r$ is the same $P$ as through the sphere at $r$ .

Step 3. Intensity is power per unit area, and the sphere's area is $4\pi r^2$ , so $I = P/(4\pi r^2)$ — equation (5). The $r^2$ is nothing but the surface area of a sphere; energy conservation supplies the constant power, and geometry supplies the spreading. Double the radius and the same power covers $2^2 = 4$ times the area, so the intensity drops to a quarter.

Step 4. Translate to amplitude. Since intensity goes as the square of amplitude, $I \propto A^2$ , and $I \propto 1/r^2$ , we have $A^2 \propto 1/r^2$ , hence $A \propto 1/r$ . Check on the anchor: a 1.0 W source gives $I(10\,\mathrm{m}) = 1.0/(4\pi\times10^2) = 7.96\times10^{-4}$ W m⁻² and $I(20\,\mathrm{m}) = 1.0/(4\pi\times20^2) = 1.99\times10^{-4}$ W m⁻² — a ratio of exactly 4 for a doubling of distance, while the amplitude merely halves. (A line source spreads over a cylinder of area $2\pi r L \propto r$ , giving $I \propto 1/r$ and $A \propto 1/\sqrt r$ instead.) Verified: I = 7.96×10⁻⁴ W m⁻² at 10 m, 1.99×10⁻⁴ at 20 m (ratio 4), for a 1.0 W source, in verify_hs2-wave-energy-intensity.py.

Limiting cases

Case	Result	Meaning
amplitude doubled ( $A \to 2A$ )	intensity $\times 4$	$I \propto A^2$ — the squared amplitude
frequency doubled ( $f \to 2f$ )	intensity $\times 4$	$I \propto f^2$ — the factor students drop
both doubled	intensity $\times 16$	$A^2 f^2$ , each squared
point source, distance doubled	intensity $\times \tfrac14$ ; amplitude $\times\tfrac12$	$I = P/4\pi r^2$ , $A \propto 1/r$
line source, distance doubled	intensity $\times\tfrac12$ ; amplitude $\times 1/\sqrt2$	cylindrical spreading, $I \propto 1/r$
medium absorbs energy	extra fall on top of spreading	absorption is a separate mechanism
$A \to 0$ or $f \to 0$	$I \to 0$	no oscillation, no energy transported

Common trapIntensity goes as amplitude SQUARED — and do not forget the frequency

The two errors that sink intensity questions are both about the exponents. First: $I \propto A^2$ , not $A$ . Doubling a wave's amplitude quadruples its intensity and its power; trebling it multiplies them by nine. If a problem says "the amplitude is halved", the intensity drops to a quarter, not a half. Second: the intensity also goes as $f^2$ , and this is the one that is silently dropped. Two waves of the same amplitude but frequencies in the ratio 1 : 2 carry energy in the ratio 1 : 4 — a favourite NEET and JEE trick, "same amplitude, double the frequency, find the ratio of intensities" (the answer is 4, not 2). Keep both handles, each squared: $I \propto A^2 f^2$ . Two housekeeping traps ride alongside: put the amplitude in metres (a wave written as " $A = 5$ mm" must become $5\times10^{-3}$ m before it goes into any SI energy formula), and use $r^2$ , not $r$ , in the inverse-square law.

Worked examples

Board level — the power in a wave on a string (CBSE/ICSE/A-Level pattern)

A transverse wave travels on a string of linear density $\mu = 2.0\times10^{-3}$ kg m⁻¹ at speed $v = 200$ m s⁻¹. The wave has frequency 50 Hz and amplitude 5.0 mm. (a) Find the average power it transmits. (b) If the amplitude is doubled, what is the new power? (c) If instead the frequency is doubled, what is the new power?

Step 1. Convert and gather, in SI: $\omega = 2\pi f = 2\pi\times50 = 314.2$ rad s⁻¹, and $A = 5.0$ mm $= 5.0\times10^{-3}$ m (the conversion that must happen first).

Step 2. Part (a): $P = \tfrac12\mu v\omega^2 A^2 = \tfrac12(2.0\times10^{-3})(200)(2\pi\times50)^2(5.0\times10^{-3})^2 = \boxed{0.493\ \mathrm{W}}$ . About half a watt of mechanical power flowing along the string.

Step 3. Part (b): doubling the amplitude multiplies $P$ by $2^2 = 4$ , so $P = 4\times0.493 = \boxed{1.97\ \mathrm{W}}$ . Nothing else changed; the $A^2$ did the work.

Step 4. Part (c): doubling the frequency also multiplies $P$ by $2^2 = 4$ (through $\omega^2 \propto f^2$ ), so again $P = 4\times0.493 = \boxed{1.97\ \mathrm{W}}$ . The symmetry of $A^2 f^2$ made parts (b) and (c) give the identical factor — the point of the question.

Competitive level — a point source and the inverse-square law (JEE-Main/NEET pattern)

A small source radiates sound uniformly in all directions with a total power of 1.0 W. (a) Find the intensity at 10 m and at 20 m. (b) By what factor does the intensity change, and what happens to the amplitude? (c) Express the intensity at 10 m in decibels above the threshold $I_0 = 1\times10^{-12}$ W m⁻² — and say which chapter owns that scale.

Step 1. Part (a): use $I = P/4\pi r^2$ . At 10 m, $I = 1.0/(4\pi\times10^2) = 1.0/1256.6 = \boxed{7.96\times10^{-4}\ \mathrm{W\,m^{-2}}}$ . At 20 m, $I = 1.0/(4\pi\times20^2) = 1.0/5026.5 = \boxed{1.99\times10^{-4}\ \mathrm{W\,m^{-2}}}$ .

Step 2. Part (b): the distance doubled, so the intensity fell by $2^2 = 4$ — and indeed $7.96\times10^{-4}/1.99\times10^{-4} = 4.0$ . Because $I \propto A^2$ , the amplitude fell only by $\sqrt4 = 2$ : the wave at 20 m is a quarter as intense but half as tall as at 10 m.

Step 3. Part (c): the level in decibels is $10\log_{10}(I/I_0) = 10\log_{10}(7.96\times10^{-4}/10^{-12}) = 10\log_{10}(7.96\times10^{8}) = \boxed{89\ \mathrm{dB}}$ — about the loudness of heavy traffic. The decibel scale itself belongs to the Sound chapter ahead; here it is only a way to size the intensity we computed. Timing note: about 90 seconds. The trap is to use $r$ instead of $r^2$ (giving a fall of 2 instead of 4), or to confuse the intensity ratio (4) with the amplitude ratio (2). Write $I \propto 1/r^2$ and $A \propto 1/r$ side by side before you answer.

Exam traps

$I \propto A$ instead of $A^2$ . Intensity, power and energy all go as the square of the amplitude. Halving the amplitude quarters the intensity; doubling it quadruples it. Never leave the amplitude to the first power.

Forgetting the $f^2$ . Intensity also goes as frequency squared. "Same amplitude, double the frequency" multiplies the intensity by four, not two — the single most common competitive-exam trick in this topic. Carry both $A^2$ and $f^2$ .

Using $r$ instead of $r^2$ . The inverse-square law is $I = P/4\pi r^2$ . Doubling the distance quarters the intensity. Writing $I \propto 1/r$ (the line-source law) for a point source loses the question.

Confusing intensity with loudness. Intensity is watts per square metre, a physical quantity; loudness is the ear's logarithmic response to it, measured in decibels. They are related but not equal — a topic developed in the Sound chapter, not here.

Amplitude in millimetres inside an SI formula. A wave given as " $A = 5$ mm" must be converted to $5\times10^{-3}$ m before it enters $P = \tfrac12\mu v\omega^2 A^2$ or any energy formula, or the $A^2$ makes the error a factor of a million.

Board scope — IGCSE. The intensity formulae $I = \tfrac12\rho v\omega^2 A^2$ and the inverse-square law are not on the IGCSE 0625 syllabus (IGCSE stops at the wave model and $v = f\lambda$ ). If you are an IGCSE candidate, this page is enrichment — but A-Level, IB (SL and HL) and AP Physics 2 all do carry intensity $\propto$ amplitude² and the inverse-square law, so for those boards it is core.

Origins · 1 minLord Rayleigh and how this was discovered

Lord Rayleigh

1842–1919 · England

John William Strutt — Lord Rayleigh from 1873 — succeeded Maxwell as Cavendish Professor at Cambridge and became the great systematiser of classical wave physics. His two-volume The Theory of Sound (1877–78) is the book that put wave energy and intensity on a rigorous footing: it is there that the amplitude-squared law, the treatment of how sound spreads and fades, and the analysis of the ear's threshold of hearing were assembled into a single theory, and acousticians still reach for it. The same eye for the intensity of waves led him to explain why the sky is blue — the intensity of scattered light rising as the inverse fourth power of wavelength — and to predict the surface elastic wave that bears his name and carries much of an earthquake's energy. He won the 1904 Nobel Prize (for the discovery of argon). Nearly every quantity on this page — energy density, intensity, the $A^2$ law — is set out in his Sound.

History

1760

— Johann Heinrich Lambert, in Photometria, states the inverse-square law of intensity for a point source and founds the quantitative measurement of light.

1871

— Lord Rayleigh shows the intensity of scattered light rises as the inverse fourth power of wavelength, explaining the blue of the sky.

1877

— Rayleigh publishes the first volume of The Theory of Sound, systematising wave energy, intensity and the amplitude-squared law.

1878

— The second volume completes the treatise, still a foundation of modern acoustics.

1885

— Rayleigh predicts the surface wave on an elastic solid that now bears his name and carries much of an earthquake's destructive energy.

Going further

Texts. NCERT Physics Class 11, Ch. 15 "Waves" (energy transport by a wave) · H. C. Verma, Concepts of Physics Vol. 1, "Wave Motion and Waves on a String" — power and intensity problems · Halliday, Resnick & Walker, "Waves — I", rate of energy transmission · Cambridge A-Level 9702 topic 7.3 (intensity $\propto$ amplitude²) and IB Physics C.2 (intensity and the inverse-square law).

Provenance. Standard results. Recompute in `verify_hs2-wave-energy-intensity.py`: the string power $P = \tfrac12(2.0\times10^{-3})(200)(2\pi\times50)^2(5.0\times10^{-3})^2 = 0.493$ W, with $\omega = 314.2$ rad s⁻¹, and its $\times4$ under a doubled amplitude or frequency ( $1.97$ W); the point-source intensities $I(10) = 1.0/(4\pi\times100) = 7.96\times10^{-4}$ W m⁻² and $I(20) = 1.99\times10^{-4}$ W m⁻², ratio exactly 4; the amplitude ratio 2; the level $10\log_{10}(7.96\times10^{-4}/10^{-12}) = 89$ dB; and the quiz distance (one-quarter intensity at twice the distance, $5 \to 10$ m).

Prerequisites. Energy in SHM — the $\tfrac12 m\omega^2 A^2$ that every result here is built on · Wave Speed: v = fλ — the speed $v$ that carries the energy along. Leads to. Superposition & Interference of Waves — where two waves' intensities combine (and $I \propto A^2$ decides the bright and dark fringes). The decibel scale and loudness are taken up in the Sound chapter that follows this one.

Go deeper (UG). Why intensity is properly a time-averaged quantity, and how the averaging is done carefully for waves whose energy density oscillates, is developed in Intensity and Time-Averaged Quantities in the university electromagnetism track. It is enrichment; the averaged results used here are complete for this course.

Why does intensity depend on the square of the amplitude, not just the amplitude?

Because energy in any oscillation goes as the square of the amplitude, and a wave is a train of oscillations. A mass in SHM stores energy $\tfrac12 m\omega^2 A^2$ : doubling the amplitude doubles both the maximum speed (so the kinetic energy quadruples) and the maximum displacement (so the potential energy quadruples), and the total quadruples. A wave inherits this exactly — its energy density, power and intensity all carry the factor $A^2$ . So the honest relationship is $I \propto A^2$ : a wave twice as tall is four times as energetic. This is also why "twice as loud" is not "twice the amplitude" — the ear's response to intensity is another layer on top, but the physics underneath is the squared amplitude.

Why does doubling the frequency also quadruple the energy?

Because the particles' speed depends on the frequency. In SHM the maximum speed of a particle is $v_{p,\max} = A\omega = 2\pi A f$ : at twice the frequency, each particle whips back and forth twice as fast, so its kinetic energy — proportional to speed squared — is four times as large. Since the wave's energy is built from these oscillating particles, doubling the frequency (at fixed amplitude) quadruples the energy the wave carries, exactly as doubling the amplitude does. That is the symmetry in $I \propto A^2 f^2$ : amplitude and frequency are equal partners, each entering squared. Forgetting the $f^2$ is the classic slip, because problems that hold the amplitude fixed and change only the frequency look at first as though nothing energetic has happened.

Is the inverse-square law caused by the medium absorbing energy?

No — and keeping these two apart is important. The inverse-square law is pure geometry: a point source's energy is conserved, but it spreads over spheres whose area grows as $r^2$ , so the energy per unit area (the intensity) falls as $1/r^2$ even in a perfectly transparent, loss-free medium. Absorption is a different mechanism — the medium actually converting wave energy into heat as the wave passes — and it removes energy from the wave altogether, on top of the geometric thinning. A distant streetlight looks dim on a clear night entirely because of the inverse-square law; on a foggy night it looks dimmer still, and that extra dimming is absorption and scattering. The inverse-square law would hold even in a vacuum; absorption needs a medium to do the absorbing.

What is the difference between intensity and loudness?

Intensity is a physical quantity — the power a wave delivers per unit area, in watts per square metre — and it is what every formula on this page computes. Loudness is a sensation, the impression your ear and brain form of that intensity, and it does not track intensity in a simple straight line. The ear responds to huge ratios of intensity (the loudest bearable sound is about a million million times as intense as the faintest audible one), so loudness follows a roughly logarithmic scale, which is why sound is measured in decibels. A tenfold rise in intensity does not feel ten times as loud — it feels like a fixed step up (10 dB, and roughly "twice as loud"). Intensity is the physics; loudness is the perception; the decibel bridges them, and it is developed properly in the Sound chapter, not here.

Why is there a factor of one half in P = ½ μ v ω² A²?

It comes straight from the energy of simple harmonic motion, whose total is $\tfrac12 m\omega^2 A^2$ — the same one-half you have met in $\tfrac12 kx^2$ for a spring and $\tfrac12 mv^2$ for kinetic energy. When you build the wave's energy from its oscillating elements and work out the average energy stored per unit length, that $\tfrac12$ carries through, and it survives into the power and the intensity. Physically it is an average: as the wave passes, each element's energy sloshes between kinetic and potential, and the mean of the oscillating energy density is half of its peak value. So the one-half is not decoration — it is the time-average built into every SHM energy, and dropping it doubles your answer.

Does a bigger wave always carry more energy than a smaller one?

Not necessarily — because energy depends on frequency as well as amplitude, and on the medium too. A tall, slow ocean swell and a short, fast ripple can carry very different energies even at the same height, because intensity goes as $A^2 f^2$ : the higher-frequency wave, whose particles move faster, packs more energy for the same amplitude. And the medium's properties matter: the intensity $\tfrac12\rho v\omega^2 A^2$ also carries the density $\rho$ and speed $v$ , so the "same" wave in a denser or faster medium transports more power. The safe statement is not "taller means more energetic" but " $I \propto A^2 f^2$ (with the medium's $\rho$ and $v$ fixed)" — compare like with like, and remember that frequency is an equal partner to amplitude.