University Physics I · Optional extension · Institutional Extension · Waves & Sound · WS.8

Sound Intensity & Decibels

The quietest sound you can hear and the loudest you can tolerate differ in intensity by a factor of a million million. The decibel scale compresses that range into two digits — and changes how every measurement must be combined.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Sound intensity is the average power a wave carries through unit area, I = P/A, and for a travelling sound wave the pressure amplitude fixes it: I = p₀²/(2ρv). A small source radiating into open space spreads one power over a growing sphere, so I = P/(4πr²) and intensity falls as 1/r². The range the ear covers, roughly 10⁻¹² to 1 W m⁻², is twelve orders of magnitude, so acoustics reports a level rather than an intensity: β = 10 log₁₀(I/I₀), with I₀ = 1.00 × 10⁻¹² W m⁻² fixed by definition rather than measured.

The logarithm compresses the range and then dictates the arithmetic. Levels never add; the intensities behind them do. Doubling the intensity adds 3.01 dB, doubling the distance from a point source subtracts 6.02 dB, and a source 10 dB below another lifts the total by only 0.41 dB.

So read a meter in two steps: convert back to an intensity, then check that the geometry the inverse-square law assumes actually holds.

Simple definition: Sound intensity is the average acoustic power crossing unit area, in W m⁻². The sound-intensity level is its logarithm against a fixed reference: β = 10 log₁₀(I/I₀) with I₀ = 1.00 × 10⁻¹² W m⁻², reported in decibels.
Example: A 0.50 W source radiating uniformly gives I = 0.50/(4π × 10²) = 4.0 × 10⁻⁴ W m⁻² at 10 m, which is 86.0 dB. Walk out to 20 m and the intensity quarters, but the level falls only to 80.0 dB.

Intensity of a sound waveI = P/A = ½ρvω²s₀² = p₀²/(2ρv)

Average power per unit area. It goes as amplitude squared, whether you track pressure or displacement.

I in W m⁻²; ρv is the acoustic impedance, 412 Pa s m⁻¹ for air at 20 °C

Inverse-square spreadingI = P/(4πr²)

One power spread over a growing sphere. Doubling r quarters I; ten times r cuts it by 100.

Point source, free field, no reflection or absorption; P in W, r in m

Sound-intensity levelβ = 10 log₁₀(I/I₀), I₀ = 1.00 × 10⁻¹² W m⁻²

A ratio made dimensionless, then compressed. 0 dB means I = I₀, not silence.

β in dB; I₀ is fixed by definition, not measured

Sound pressure levelL(p) = 20 log₁₀(p(rms)/p(ref)), p(ref) = 20 μPa

What a microphone actually measures. Pressure, not power, so the multiplier doubles.

Factor 20 because I ∝ p²; in air the two levels agree to 0.12 dB

Difference between two levelsβ₂ − β₁ = 10 log₁₀(I₂/I₁) = 20 log₁₀(r₁/r₂)

Twice the intensity is +3.01 dB; twice the distance is −6.02 dB; ten times the intensity, +10 dB.

Second form for a point source in a free field

Combining incoherent sourcesβ(total) = 10 log₁₀(Σ 10^(βᵢ/10))

Convert to intensity, add, convert back. Two equal sources give +3.01 dB, not double.

Uncorrelated sources only; coherent sources add pressures instead

What intensity is, and what sets it

Intensity is the average power a wave carries through unit area held perpendicular to the propagation direction, I = P/A, in W m⁻². For a travelling sound wave the general wave result specialises to I = ½ρvω²s₀², where s₀ is the displacement amplitude of the oscillating air and ρv is the medium's acoustic impedance — 1.20 × 343 = 412 Pa s m⁻¹ for air at 20 °C. Pressure and displacement amplitudes are linked by p₀ = ρvωs₀, so the same intensity reads I = p₀²/(2ρv), and that form is the working one because microphones respond to pressure. Both say amplitude squared. Put the threshold of hearing through them: I = 1.0 × 10⁻¹² W m⁻² needs only p₀ = √(2Iρv) = 2.9 × 10⁻⁵ Pa, under three parts in ten thousand million of atmospheric pressure, and at 1.0 kHz a displacement amplitude s₀ = p₀/(ρvω) = 1.1 × 10⁻¹¹ m, roughly a tenth of an atomic diameter. A detector that sensitive spans an enormous range, which is exactly why a linear scale is unusable.

Spreading: why 1/r², and when it is not

A small source radiating steadily into open air sends the same average power P through every sphere centred on it, because nothing creates or destroys energy on the way out. That sphere has area 4πr², so I = P/(4πr²) and intensity falls as the inverse square of distance; the pressure amplitude, going as √I, falls as 1/r. The law is geometry, not a property of sound, and it carries four assumptions: the source is small compared with r, it radiates equally in all directions, nothing reflects the sound back, and the air absorbs none of it. Break one and the exponent changes. A long line of traffic spreads its power over a cylinder of area 2πrL, so I ∝ 1/r and each doubling of distance costs 3.01 dB rather than 6.02. Indoors, reflected energy fills the room, and beyond the critical distance the level stops falling at all. Over hundreds of metres, molecular absorption strips high frequencies faster than spreading alone — a few dB per 100 m at 4 kHz — which is why near thunder cracks and distant thunder rumbles.

The decibel and its reference

From the quietest audible sound to the onset of pain the intensity climbs by a factor of about 10¹², and the ear responds to ratios of intensity far more nearly than to differences. Both facts point to a logarithm. Define the sound-intensity level β = 10 log₁₀(I/I₀), with I₀ = 1.00 × 10⁻¹² W m⁻² adopted by convention as the nominal threshold of hearing at 1 kHz. Three consequences follow. β is dimensionless: the decibel names a logarithmic ratio, not a physical unit, so a level is meaningless without its reference, and 0 dB means I = I₀ rather than silence. Negative levels are ordinary — a good anechoic chamber sits near −10 dB. And because I₀ is exact by definition, all the uncertainty in a level comes from the measured intensity. Running the definition backwards is a routine step: I = I₀ × 10^(β/10), so a reading of 86.0 dB is I = 10⁻¹² × 10^8.60 = 4.0 × 10⁻⁴ W m⁻².

The arithmetic a logarithm forces

Because β is a logarithm, levels do not add — the intensities behind them do. To combine uncorrelated sources, convert each level to an intensity, sum, and convert back: β(total) = 10 log₁₀(Σ 10^(βᵢ/10)). Two identical machines each giving 70 dB alone give 10 log₁₀(2 × 10⁷) = 73.0 dB together, not 140. Eight of them sit 10 log₁₀ 8 = 9.0 dB above one. A source 10 dB below another barely registers: 70 dB and 60 dB combine to 10 log₁₀(1.1 × 10⁷) = 70.4 dB. The same arithmetic runs the other way. Multiply intensity by 2 and add 3.01 dB; by 10 and add exactly 10 dB. Move a point source twice as far off and the intensity quarters, so the level drops 10 log₁₀ 4 = 6.02 dB — the familiar 6 dB per doubling of distance. All of this assumes incoherent sources, with random relative phase. Two coherent sources in phase at a point double the pressure amplitude, quadruple the intensity and add 6.02 dB; out of phase they cancel.

Reading a meter: what the number does not say

A sound level meter reports sound pressure level, L(p) = 20 log₁₀(p(rms)/p(ref)) with p(ref) = 20 μPa. The 20 appears because intensity goes as pressure squared, and the reference was chosen so the two scales nearly coincide: 20 μPa in air is I = p(rms)²/(ρv) = 9.7 × 10⁻¹³ W m⁻², which is 0.12 dB below I₀. Treat β and L(p) as the same number in air — but not in water, where p(ref) = 1 μPa and the impedance is 3600 times larger, putting levels about 62 dB above the air figure for the same intensity. Three qualifiers travel with every reading. Frequency weighting: dBA discounts low frequencies to imitate the ear, so a rumbling plant room reads far lower in dBA than unweighted. Averaging time: a peak, a one-second reading and an eight-hour L(eq) are three different quantities. And background: a machine reading 62 dB against a 58 dB background is itself producing 10 log₁₀(10^6.2 − 10^5.8) = 59.8 dB, not 62.

Where the model stops

A level is a physical quantity; loudness is a perception, and they are not proportional. Sensitivity depends strongly on frequency — a 1 kHz tone at 60 dB is conversational while a 50 Hz tone at 60 dB is barely audible — which is what equal-loudness contours and the A weighting try to encode. As a rough rule a sound must gain about 10 dB, a factor of 10 in intensity, before most listeners call it twice as loud; the 3 dB from a second identical machine is hard to notice and still doubles the energy delivered. That gap governs exposure, because averaging levels arithmetically understates the dose. Seven hours at 80 dB plus one hour at 95 dB give an eight-hour L(eq) of 10 log₁₀[(7 × 10⁸ + 10^9.5)/8] = 86.8 dB, not the 81.9 dB an arithmetic mean suggests. Geometry matters too: a source on hard ground radiates into a hemisphere, so I = P/(2πr²) and the level runs 3.01 dB above free field. Close in, the far-field assumption fails and no simple power of r applies.

Change one variable at a time

Make the relationship visible.

Interactive model

Meter distance r10 m

Incoherent sources n4

Power per source P0.20 W

Drag the meter from 2 m to 4 m and watch the level fall 6.0 dB; then drag the source count from 1 to 8 and watch eight sources add just 9.0 dB, not eight times the level.

LEVEL AT METER88.0 dB

LOSS FROM 1 m20.00 dB

SOURCES ADD6.02 dB

RMS PRESSURE0.512 Pa

Live interpretationLEVEL AT METER: 88.0 dB. LOSS FROM 1 m: 20.00 dB. SOURCES ADD: 6.02 dB. RMS PRESSURE: 0.512 Pa

Catch the common trap

Explain before calculating.

A small source radiates uniformly into a free field, and a meter 5.0 m away reads 88.0 dB. The meter is then moved out to 40 m and, at the same moment, seven more identical sources are switched on at the original position. The eight sources are mutually incoherent. What does the meter now read?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyA meter records an intensity of 3.2 × 10⁻⁵ W m⁻². Find the sound-intensity level, then find it again after the intensity doubles.

β = 10 log₁₀(I/I₀) with I₀ = 1.00 × 10⁻¹² W m⁻², so form the ratio first: I/I₀ = 3.2 × 10⁻⁵ / 1.00 × 10⁻¹² = 3.2 × 10⁷.
log₁₀(3.2 × 10⁷) = 7.505, so β = 10 × 7.505 = 75.05 dB → 75.1 dB.
Doubling I to 6.4 × 10⁻⁵ W m⁻² doubles the ratio, and 10 log₁₀ 2 = 3.01 dB, so β = 75.05 + 3.01 = 78.06 dB → 78.1 dB. The level gains 3 dB, not 100%.

Answerβ = 75.1 dB; after doubling, 78.1 dB

MediumA small loudspeaker radiates 25 mW of acoustic power uniformly into a free field. Find the level at 3.0 m, then at 12 m.

The power spreads over a sphere: I = P/(4πr²). At r = 3.0 m, I = 0.025/(4π × 3.0²) = 0.025/113.1 = 2.21 × 10⁻⁴ W m⁻².
β = 10 log₁₀(2.21 × 10⁻⁴ / 10⁻¹²) = 10 log₁₀(2.21 × 10⁸) = 83.4 dB.
Moving to 12 m multiplies r by 4, so I falls by 4² = 16: Δβ = −10 log₁₀ 16 = −12.04 dB.
β(12 m) = 83.44 − 12.04 = 71.40 dB → 71.4 dB. Check straight from the definition: I = 0.025/(4π × 144) = 1.38 × 10⁻⁵ W m⁻², which is 71.4 dB.

Answer83.4 dB at 3.0 m; 71.4 dB at 12 m

HardA machine under test reads 74.0 dB at the operator's position; switched off, the background alone reads 68.0 dB. (a) What level does the machine alone produce? (b) A second identical, incoherent machine is switched on beside it — what does the meter read then?

Levels never subtract. Convert both to intensities in units of I₀: total 10^(74.0/10) = 2.512 × 10⁷, background 10^(68.0/10) = 6.310 × 10⁶.
Machine alone: (2.512 − 0.631) × 10⁷ = 1.881 × 10⁷ I₀, so β = 10 log₁₀(1.881 × 10⁷) = 72.7 dB — 1.3 dB below the raw reading.
Two identical incoherent machines double that intensity: 2 × 1.881 × 10⁷ = 3.762 × 10⁷ I₀, which is +3.01 dB.
Add the background back in intensity: 3.762 × 10⁷ + 0.631 × 10⁷ = 4.393 × 10⁷ I₀.
β(total) = 10 log₁₀(4.393 × 10⁷) = 76.4 dB. The meter moved 2.4 dB, not the 3.01 dB the machinery gained, because the background did not double with it: the total intensity rose only by 4.393/2.512 = 1.75, and 10 log₁₀ 1.75 = 2.43 dB.

Answer(a) 72.7 dB from the machine alone; (b) 76.4 dB with both machines and the background