
Physics III course sequence · Interactive lesson
Sound Intensity & the Decibel Scale
Microphones read pressure, noise limits are written in decibels, and the two are joined by an intensity you rarely measure directly. This topic builds that chain in both directions, then shows why silencing the quieter of two machines almost never helps.
UPIII-02.07 · Lesson 7 of 9 in this unit. Position is not a completion score.Build the model
A level is a compressed ratio, so levels never simply add.
The key idea
The sound intensity level of a sound is ten times the base-ten logarithm of its intensity divided by a reference intensity fixed by definition at 1.00 × 10⁻¹² W m⁻², and it is reported in decibels.
Example. An intensity of 1.00 × 10⁻³ W m⁻² is 10⁹ times the reference, so its level is 10 log₁₀(10⁹) = 90.0 dB — and an intensity twice as large is 93.0 dB, not 180 dB.
The complete physical model and assumptions
Sound carries energy, and the quantity that matters at a listening point is intensity: the time-averaged power crossing unit area, I = pₘₐₓ²/(2ρv), where the ½ is the cycle average of a squared sinusoid and ρv is the medium's characteristic impedance, 412 Pa s m⁻¹ for air at 20 °C. Two things follow. Geometry alone spreads a point source's power P over a sphere, so I = P/(4πr²) and the pressure amplitude falls as 1/r. And because hearing spans twelve decades of intensity, from about 10⁻¹² to 1 W m⁻², levels are quoted logarithmically against a reference fixed by definition at I₀ = 1.00 × 10⁻¹² W m⁻²: β = 10 log₁₀(I/I₀). A decibel is therefore a compressed ratio, not a unit of sound, and that compression is what makes it treacherous — intensities add when sources are uncorrelated, logarithms do not, so every combination must be converted back to W m⁻², summed, and converted forward again. The model assumes a free non-reverberant field, a source small compared with r, and no absorption along the path; indoors the reflected field eventually dominates and the level stops falling at all.
1Intensity is a squared pressure divided by the medium
A sound wave transports energy, and the quantity a listener meets is the intensity I: the time-averaged power per unit area crossing a surface that faces the source, in W m⁻². Instantaneously the power flux is the excess pressure p multiplied by the particle velocity u, and for a travelling plane wave in a non-dispersive fluid those two are in phase with p = ρv u, so the flux is p²/(ρv). Averaging a squared sinusoid over one period, (1/T)∫₀ᵀ sin²(ωt) dt = ½, leaves I = pₘₐₓ²/(2ρv). The product z = ρv is the medium's characteristic impedance; for air at 20 °C, z = 1.20 × 343 = 412 Pa s m⁻¹. Feed the reference intensity into that relation and the pressure amplitude at the threshold of hearing is √(2 × 412 × 1.00 × 10⁻¹²) = 2.87 × 10⁻⁵ Pa, about 20 μPa rms — two parts in ten thousand million of atmospheric pressure, and a displacement amplitude at 1 kHz of only 1.1 × 10⁻¹¹ m.
2Geometry alone gives the inverse-square law
Surround a small source with a sphere of radius r. If the medium neither absorbs nor reflects, every watt radiated must cross every such sphere, so P = I × 4πr² and I = P/(4πr²). Nothing about sound entered that argument: it is conservation of energy and the area of a sphere. Since I ∝ pₘₐₓ², the pressure amplitude falls only as 1/r. A 0.25 W source gives I = 0.25 ÷ (4π × 16.0) = 1.24 × 10⁻³ W m⁻² at 4.0 m, and exactly a quarter of that at 8.0 m. Three assumptions are buried here: the source is small compared with r, so the field is spherical; the field is free, so no reflected energy returns; and the air does not absorb, which fails above a few kilohertz over tens of metres. A line source such as a motorway spreads over a cylinder instead, so I ∝ 1/r and the fall per doubling is halved.
3The scale is logarithmic because the range is
From the faintest audible 1 kHz tone to the onset of pain, intensity spans twelve decades, 1.00 × 10⁻¹² to about 1 W m⁻². No linear axis can show that, and the ear's response to a change is closer to a ratio than to a difference. So levels are defined as β = 10 log₁₀(I/I₀), with the reference fixed by definition at I₀ = 1.00 × 10⁻¹² W m⁻² rather than measured. A decibel is therefore a compressed ratio, and a level quoted without its reference means nothing. Two consequences are worth committing to memory. Doubling the intensity adds 10 log₁₀ 2 = 3.01 dB, and a factor of ten adds exactly 10 dB. Doubling the distance from a point source quarters I, so it subtracts 10 log₁₀ 4 = 6.02 dB — the six decibels per doubling an acoustician quotes without calculating.
4Pressure levels use 20 log, and the two references agree
Microphones are calibrated in pascals, so field data are usually reported as a sound pressure level, Lₚ = 20 log₁₀(pᵣₘₛ/p₀) with p₀ = 20 μPa. The 20 is not a rival convention. Because I ∝ p², taking the logarithm brings the square down as a factor of two: 10 log₁₀(p²/p₀²) = 20 log₁₀(p/p₀). The two references were chosen to agree, since p₀²/(ρv) = I₀ requires ρv = 400 Pa s m⁻¹ exactly; air at 20 °C offers 412, so the two levels differ by 10 log₁₀(412/400) = 0.13 dB and the difference is ignored in practice. The factor 20 also fixes the pressure arithmetic: doubling the pressure amplitude adds 6.02 dB, and halving the distance from a point source, which doubles p, does the same.
5Never add levels: convert, add, convert back
Two uncorrelated sources — machines whose vibrations hold no fixed phase relation — deliver intensities that add, because the cross term in (p₁ + p₂)² averages to zero over a cycle. Levels do not add, because the logarithm of a sum is not the sum of the logarithms. The procedure is therefore fixed: convert each level with I = I₀ × 10(β/10), add the intensities, convert back with β = 10 log₁₀(I/I₀). Take 86.0 dB and 80.0 dB at one point. The intensities are 3.98 × 10⁻⁴ and 1.00 × 10⁻⁴ W m⁻², their sum is 4.98 × 10⁻⁴ W m⁻², and the level is 87.0 dB. Silencing the quieter machine completely would save 1.0 dB; silencing the louder would save 7.0 dB. That asymmetry is the whole of noise control: the loudest source sets the answer, and anything more than about 10 dB below it is already irrelevant.
6Where the free-field model stops
Indoors only part of the energy reaches you directly; the rest arrives after reflections, and the reverberant field is roughly uniform through the room. Near the source the direct field wins and the level falls at 6.02 dB per doubling, but beyond a critical distance the reverberant field dominates and the level stops falling, so walking further away buys nothing. Real sources also radiate unevenly: a directivity factor Q multiplies the free-field intensity, and a source sitting on a hard floor radiates into a hemisphere, giving Q = 2 and a level 3.01 dB above the free-field value. Finally, intensities add only for uncorrelated sources. Two loudspeakers driven by one signal are coherent, so where their pressures arrive in phase the amplitudes add, the intensity quadruples and the level rises 6.02 dB, while in antiphase it collapses.
Key equations
Read the units, assumptions and limits before substituting.
Intensity from pressure amplitude
I = pₘₐₓ²/(2ρv)
A microphone reads pressure; this turns that reading into a power flux.
pₘₐₓ in Pa, ρ in kg m⁻³, v in m s⁻¹, I in W m⁻²Free-field spreading from a point source
I = P/(4πr²)
The same power crosses every sphere, so doubling r quarters the intensity.
P is the radiated acoustic power in W, r the distance in mSound intensity level
β = 10 log₁₀(I/I₀)
Twelve decades of intensity compress into a 0 dB to 120 dB span.
β in dB, I and I₀ in W m⁻², with I₀ = 1.00 × 10⁻¹² W m⁻² by definitionRecovering an intensity from a level
I = I₀ × 10(β/10)
The step that must come first whenever levels are combined or compared.
I in W m⁻²; β = 90.0 dB returns I = 1.00 × 10⁻³ W m⁻²Combining uncorrelated sources
βₜₒₜₐₗ = 10 log₁₀(Σ 10(βᵢ/10))
Two equal machines add 3.01 dB and ten add 10.0 dB, never double the level.
each βᵢ is that source's own level in dB at the same pointSound pressure level
Lₚ = 20 log₁₀(pᵣₘₛ/p₀)
Matches the intensity level whenever ρv = 400 Pa s m⁻¹, and air is close.
pᵣₘₛ in Pa, with p₀ = 20 μPa by definition; the 20 appears because I ∝ p²Interactive lab
Predict first. Explore the model. Compare the result.
Explore the model
Try this. Take the distance from 20.0 m in to 5.0 m: the dashed source curve climbs 12.0 dB, two doublings at 6.02 dB each, but the solid total climbs only 8.8 dB, because at 20.0 m the background already carried most of the sound. Raise the background and the total flattens.
Live results
- DIRECT LEVEL β
- 77.0 dB
- TOTAL LEVEL
- 80.5 dB
- BACKGROUND ADDS
- 3.56 dB
- PRESSURE AMPLITUDE
- 0.202 Pa
Results follow this lesson’s model and assumptions. Displayed values are rounded.
Predict → change one variable → check
Plan a change before running it. Every other input stays at the saved baseline.
Saved baseline. Source acoustic power P: 0.25 W; Listening distance r: 20.0 m; Background level: 78 dB
This experiment is kept only while this page is open. It does not change saved course progress.
Catch the common trap
The usual error is to treat a level as though it were the quantity itself, and so to add, average or subtract decibels directly. Two 80.0 dB machines are neither 160 dB nor still 80.0 dB: their intensities add, giving 10 log₁₀(2 × 10⁸) = 83.0 dB. Nor is 86.0 dB combined with 80.0 dB their average of 83.0 dB — it is 87.0 dB. The reason is the logarithm: β is a compressed ratio, and log(I₁ + I₂) is not log I₁ + log I₂. Adding two levels multiplies the intensities instead of adding them, which is what a 160 dB answer really claims: 10¹⁶ times the reference, a pressure amplitude near 2.9 kPa, far past the point where the small-amplitude linear model still holds. Convert to W m⁻² first, every time.
Practice & worked examples
3 authored examples. Make an attempt, then reveal the reasoning one step at a time.
EasyExample 1
A small loudspeaker radiates 0.25 W of acoustic power uniformly in every direction in an open field. Find the sound intensity and the sound intensity level 4.0 m away, then state the level at 8.0 m without recomputing the intensity.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 4 steps revealed.
- Free field, point source: every watt crosses the sphere of radius r, so I = P/(4πr²).
- At r = 4.0 m: I = 0.25 W ÷ (4π × 16.0 m²) = 0.25 ÷ 201.06 = 1.24 × 10⁻³ W m⁻².
- Level: β = 10 log₁₀(1.24 × 10⁻³ ÷ 1.00 × 10⁻¹²) = 10 log₁₀(1.24 × 10⁹) = 10 × 9.0946 = 90.9 dB.
- Doubling r to 8.0 m quarters I, and 10 log₁₀ 4 = 6.02 dB, so the level falls to 90.9 − 6.0 = 84.9 dB. Check: 0.25 ÷ (4π × 64.0) = 3.11 × 10⁻⁴ W m⁻² gives 84.9 dB directly.
Answer. I = 1.24 × 10⁻³ W m⁻² and β = 90.9 dB at 4.0 m; β = 84.9 dB at 8.0 m.
MediumExample 2
Three machines run in an open yard. Measured one at a time at the operator's position, they read 85.0 dB, 82.0 dB and 79.0 dB. Find the level with all three running, then decide whether silencing any single machine brings the position below an 85.0 dB limit.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 5 steps revealed.
- Levels cannot be added, so convert each one with I = I₀ × 10(β/10), taking I₀ = 1.00 × 10⁻¹² W m⁻².
- The three intensities are 10⁻¹² × 10(8.50) = 3.16 × 10⁻⁴ W m⁻², 10⁻¹² × 10(8.20) = 1.58 × 10⁻⁴ W m⁻², and 10⁻¹² × 10(7.90) = 7.94 × 10⁻⁵ W m⁻².
- The machines are uncorrelated, so the intensities add, which is the Σ of the combining rule written out: Iₜₒₜₐₗ = 5.54 × 10⁻⁴ W m⁻².
- Convert back: βₜₒₜₐₗ = 10 log₁₀(5.54 × 10⁻⁴ ÷ 1.00 × 10⁻¹²) = 10 log₁₀(5.54 × 10⁸) = 87.4 dB.
- Now drop one machine at a time. Without the 79.0 dB machine, 4.75 × 10⁻⁴ W m⁻² gives 86.8 dB, a saving of 0.7 dB; without the 82.0 dB machine, 3.96 × 10⁻⁴ W m⁻² gives 86.0 dB, a saving of 1.5 dB; without the 85.0 dB machine, 2.38 × 10⁻⁴ W m⁻² gives 83.8 dB, a saving of 3.7 dB. Only the last clears the limit, and silencing both quieter machines instead leaves the loudest alone at 85.0 dB — exactly at the limit, not below it.
Answer. βₜₒₜₐₗ = 87.4 dB, and only the 85.0 dB machine is worth treating: silencing it leaves 83.8 dB, while the 82.0 dB machine is worth 1.5 dB and the 79.0 dB machine only 0.7 dB.
HardExample 3
A ventilation fan behaves as a point source in a free field. A meter 1.5 m away reads a sound intensity level of 94.0 dB. Take air at 20 °C, with ρ = 1.20 kg m⁻³ and v = 343 m s⁻¹. Find the intensity, the acoustic power radiated, the pressure amplitude, and the distance at which the level has fallen to 65.0 dB.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 5 steps revealed.
- Intensity from the level: I = I₀ × 10(β/10) = 1.00 × 10⁻¹² × 10(9.40) = 2.51 × 10⁻³ W m⁻².
- Power through the sphere of radius 1.5 m: P = I × 4πr² = 2.51 × 10⁻³ × 4π × 2.25 = 2.51 × 10⁻³ × 28.27 = 7.10 × 10⁻² W.
- Pressure amplitude from I = pₘₐₓ²/(2ρv), with ρv = 1.20 × 343 = 412 Pa s m⁻¹: pₘₐₓ = √(2 × 412 × 2.51 × 10⁻³) = √2.068 = 1.44 Pa.
- Distance for 65.0 dB: the level must drop 29.0 dB, so the intensity must drop by 10(2.90) = 794, and because I ∝ 1/r² the distance grows by √794 = 28.2.
- r = 1.5 m × 28.2 = 42.3 m. Check independently: I₆₅ = 1.00 × 10⁻¹² × 10(6.50) = 3.16 × 10⁻⁶ W m⁻², so r = √(7.10 × 10⁻² ÷ (4π × 3.16 × 10⁻⁶)) = √1787 = 42.3 m.
Answer. I = 2.51 × 10⁻³ W m⁻², P = 71.0 mW, pₘₐₓ = 1.44 Pa, and the level reaches 65.0 dB at r = 42.3 m — only 71 mW of radiated sound is already uncomfortably loud at arm's length.
Quick check
Select an answer and explain it before checking.
Review & continue
The sound intensity level of a sound is ten times the base-ten logarithm of its intensity divided by a reference intensity fixed by definition at 1.00 × 10⁻¹² W m⁻², and it is reported in decibels.
- Explain the relationship without looking at the equation.
- Check a sign, a unit and a model assumption in your working.
- Return to the lab or worked example if a step is still unclear.