Sound intensity — decibels (logarithmic scale): Worked examples

Definition The one formula everything on this page uses (recap)

Sound intensity $I$ is power per unit area, $I = P / A$ , measured in $W/m^{2}$ . The decibel level compares $I$ to a fixed reference: $β = 10 lo g_{10} (\frac{I}{I _{0}}) dB, I_{0} = 1 0^{- 12} W/m^{2}$

Where $I_{0}$ comes from: it is the threshold of hearing — the quietest sound a human can detect at 1 kHz. We measure every sound relative to this floor, so the quietest audible sound reads $lo g_{10} (1) = 0 dB$ .
Why the log: the audible range spans a factor of $\sim 1 0^{12}$ ; a logarithm counts factors of ten and compresses that into $0$ – $120$ . (See Logarithms and exponentials.)
What a "bel" is: $lo g_{10} (I / I_{0})$ by itself is the level in bels — one bel = one whole factor of ten in intensity. A bel is a coarse step, so we work in decibels = tenths of a bel, which is where the $\times 10$ out front comes from. Hence "deci-bel" (dB).
Difference rule (used constantly below): $Δ β = β_{2} - β_{1} = 10 lo g_{10} (I_{2} / I_{1})$ ; the $I_{0}$ cancels, so you only need the ratio.

The scenario matrix

Every decibel problem is one (or a combination) of these case classes. The right-hand column names the example that covers it.

Cell	Case class	What makes it tricky	Covered by
A	Intensity → dB (forward)	just plug in, watch the exponents	Ex 1
B	dB → intensity (invert the log)	undo ×10, then undo $lo g$	Ex 2
C	Ratio only — sources add	$I_{0}$ cancels; intensities add, not dB	Ex 3
D	Distance / inverse-square	ratio comes from $(r_{1} / r_{2})^{2}$	Ex 4
E	Degenerate: $I = I_{0}$ and $I < I_{0}$	zero dB, then negative dB	Ex 5
F	Pressure given, not intensity	use $20 lo g_{10}$ because $I \propto p^{2}$	Ex 6
G	Word problem: combine distance and counting	chain two ratios	Ex 7
H	Exam twist: "how many quiet sources = one loud one?"	solve for a count inside the log	Ex 8
I	Limiting behaviour: $I \to 0$	$β \to - \infty$ ; what silence "means"	Ex 9

Notice the three signs a $β$ can take: positive (louder than threshold, the usual case), zero (exactly threshold), and negative (quieter than the reference — real, just below $I_{0}$ ). Ex 5 and Ex 9 make sure you've seen all three.

Example 1 — Cell A · Intensity → dB

Worked example A quiet library reads

$I = 3.2 \times 1 0^{- 9} W/m^{2}$ . Find $β$ . Forecast: guess the dB before reading on. Threshold is $1 0^{- 12}$ , this is a few thousand times bigger, so expect something in the 30s.

Step 1. Form the ratio $\frac{I}{I _{0}} = \frac{3.2 \times 1 0 ^{- 9}}{1 0 ^{- 12}} = 3.2 \times 1 0^{3} = 3200$ . Why this step? $β$ only ever cares about how many times above threshold you are — a pure number. Dividing $1 0^{- 9}$ by $1 0^{- 12}$ means subtracting exponents: $- 9 - (- 12) = + 3$ .

Step 2. Take the log: $lo g_{10} (3200) = lo g_{10} (3.2) + 3 = 0.505 + 3 = 3.505$ . Why this step? The log counts factors of ten. We split $3200 = 3.2 \times 1 0^{3}$ so the $1 0^{3}$ contributes exactly $3$ and only the leading $3.2$ needs a lookup.

Step 3. Multiply by 10: $β = 10 \times 3.505 = 35.05 \approx 35 dB$ . Why this step? Decibels are tenths of a bel, so the formula carries the ×10.

Verify: $35 dB$ sits between whisper (20) and conversation (60) — spot on for a library, and it matched our forecast of "mid-30s". ✓

Example 2 — Cell B · dB → intensity

Example 3 — Cell C · Sources add (ratio only)

Worked example A single cricket chirps at

$42 dB$ . A field has five identical crickets. Combined level? Forecast: Not $42 \times 5$ . Five times the intensity is less than one factor of ten, so expect roughly $+ 7 dB$ → about $49 dB$ .

Step 1. Five identical sources → intensities add → $I_{new} = 5 I_{one}$ , so the ratio is $5$ . Why this step? Energy flows add up (see Wave energy and power); dB do not add, because they're logarithms.

Step 2. Use the difference rule — $I_{0}$ cancels: $Δ β = 10 lo g_{10} (5) = 10 \times 0.699 = 6.99 dB$ . Why this step? We only know the ratio $5$ , not each intensity. The difference rule is built exactly for "ratio in, dB-change out".

Step 3. Add to the original: $β = 42 + 6.99 \approx 49.0 dB$ .

Verify: Sanity — doubling gives $+ 3$ , ten-fold gives $+ 10$ ; five (between $2$ and $10$ ) gives $+ 7$ , which is between. ✓ And far from the "wrong" answer $210 dB$ warned about in the parent note's mistake box.

Example 4 — Cell D · Inverse-square with distance

Figure — Sound intensity — decibels (logarithmic scale)

Worked example A siren reads

$95 dB$ at $r_{1} = 10 m$ . What does it read at $r_{2} = 30 m$ (open field, no echoes)? Forecast: Distance ×3 → intensity ÷9. That's about one factor of ten, so roughly $- 9$ or $- 10 dB$ → mid-80s.

Step 1. Intensity ratio from geometry: $\frac{I _{2}}{I _{1}} = (\frac{r _{1}}{r _{2}})^{2} = (\frac{10}{30})^{2} = \frac{1}{9}$ . Why this step? All the source's power $P$ crosses the sphere surrounding it. A sphere of radius $r$ has area $A = 4 π r^{2}$ , so the intensity is $I = P / A = P / (4 π r^{2})$ — the area grows as $r^{2}$ , hence $I$ falls as $1/ r^{2}$ (see Inverse-square law for radiation). Triple $r$ , area ×9, intensity ÷9. Look at the pink arc in the figure: three times the radius, nine times the surface.

Step 2. $Δ β = 10 lo g_{10} (\frac{1}{9}) = - 10 lo g_{10} (9) = - 10 \times 0.954 = - 9.54 dB$ . Why this step? Ratio below $1$ → log is negative → the level drops, exactly as physics demands.

Step 3. $β_{2} = 95 - 9.54 \approx 85.5 dB$ .

Verify: Sign check — farther away is quieter, and we got a decrease. Magnitude check — ÷9 is nearly ÷10 ( $- 10$ ), so $- 9.5$ is right. Matches forecast. ✓

Example 5 — Cell E · Degenerate inputs ( $I = I_{0}$ and $I < I_{0}$ )

Worked example (i) A sound has exactly

$I = I_{0} = 1 0^{- 12} W/m^{2}$ . (ii) A sound has $I = 4 \times 1 0^{- 13} W/m^{2}$ . Find both levels. Forecast: (i) should be exactly $0 dB$ . (ii) is below threshold, so brace for a negative number.

Step 1 (i). $β = 10 lo g_{10} (\frac{I _{0}}{I _{0}}) = 10 lo g_{10} (1) = 10 \times 0 = 0 dB$ . Why this step? $lo g_{10} (1) = 0$ because $1 0^{0} = 1$ : zero factors of ten above threshold. This is why the scale was anchored at $I_{0}$ — the quietest audible sound reads $0$ .

Step 2 (ii). Ratio $= \frac{4 \times 1 0 ^{- 13}}{1 0 ^{- 12}} = 0.4$ , which is less than 1. Why this step? Below the reference the ratio drops under $1$ , and the log of a number under $1$ is negative.

Step 3 (ii). $β = 10 lo g_{10} (0.4) = 10 \times (- 0.398) = - 3.98 \approx - 4.0 dB$ . Why this step? Negative dB is perfectly legal — it just means "quieter than the reference". Your ear wouldn't catch it, but the maths is defined for all $I > 0$ .

Verify: (i) exactly $0$ ✓. (ii) $- 4 dB$ is negative as forecast; and $0.4 \approx \frac{1}{2.5}$ , roughly half, and half is $- 3 dB$ , so $- 4$ (a bit less than half) is consistent. ✓

Example 6 — Cell F · Pressure given (use $20 lo g_{10}$ )

Worked example A microphone measures the sound

pressure doubling, $p_{2} = 2 p_{1}$ . By how many dB does the level rise? Forecast: Doubling pressure is not $+ 3 dB$ (that was intensity). Pressure hits intensity twice over, so expect $+ 6 dB$ .

Step 1. Recall the link $I \propto p^{2}$ from Sound waves — pressure & displacement. So the intensity ratio is $\frac{I _{2}}{I _{1}} = (\frac{p _{2}}{p _{1}})^{2} = 2^{2} = 4$ . Why this step? Intensity is proportional to pressure squared, so doubling $p$ quadruples $I$ . The decibel definition is about intensity, so we must convert first.

Step 2. $Δ β = 10 lo g_{10} (4) = 10 \times 0.602 = 6.02 dB$ . Why this step? Once we have the intensity ratio ( $4$ ), the ordinary $10 lo g_{10}$ rule applies.

Shortcut (same answer). $Δ β = 10 lo g_{10} (p_{2} / p_{1})^{2} = 20 lo g_{10} (p_{2} / p_{1}) = 20 lo g_{10} (2) = 6.02 dB$ . Why this step? Pulling the exponent $2$ out front of the log turns $10$ into $20$ — that's exactly where the " $20 lo g$ for pressure" rule comes from.

Verify: Both routes give $6.02 dB$ . ✓ And this is the parent note's third mistake, now solved cleanly.

Example 7 — Cell G · Combine distance and counting

Worked example A single machine reads

$88 dB$ at $2 m$ . You install four identical machines and stand at $6 m$ from the group. Estimate the level (treat them as one point source, free field). Forecast: More machines push up; more distance pulls down. Expect the two effects to partly cancel — somewhere near the original.

Step 1. Counting factor: $4$ machines → intensity ×4. Why this step? Intensities add, so four equal sources give four times the power flowing.

Step 2. Distance factor: $r$ goes $2 \to 6$ , that's ×3, so intensity $\times (\frac{2}{6})^{2} = \times \frac{1}{9}$ . Why this step? Inverse-square again — the ratio is $(r_{1} / r_{2})^{2}$ .

Step 3. Chain the two ratios (they multiply): $\frac{I _{2}}{I _{1}} = 4 \times \frac{1}{9} = \frac{4}{9}$ . Why this step? Independent effects on intensity multiply; inside the log, multiplication becomes a sum of two dB changes.

Step 4. $Δ β = 10 lo g_{10} (\frac{4}{9}) = 10 (lo g_{10} 4 - lo g_{10} 9) = 10 (0.602 - 0.954) = - 3.52 dB$ . Then $β_{2} = 88 - 3.52 \approx 84.5 dB$ .

Verify: Split-check — counting alone was $+ 10 lo g_{10} 4 = + 6.0$ , distance alone was $- 10 lo g_{10} 9 = - 9.5$ ; sum $= - 3.5$ . ✓ Ends up near the original as forecast.

Example 8 — Cell H · Exam twist: solve for a count

Worked example A concert must reach

$118 dB$ . Each speaker delivers $100 dB$ (alone, at the audience). How many identical speakers $N$ are needed? Forecast: We need $+ 18 dB$ . That's between $+ 10$ (×10) and $+ 20$ (×100), so guess between $10$ and $100$ speakers.

Step 1. The needed gain: $Δ β = 118 - 100 = 18 dB$ . Why this step? $N$ speakers give $N \times$ the intensity of one; that ratio is what raises the dB.

Step 2. Set the difference rule equal: $18 = 10 lo g_{10} (N)$ , so $lo g_{10} (N) = 1.8$ . Why this step? The intensity ratio is exactly the count $N$ , so it sits inside the log.

Step 3. Invert: $N = 1 0^{1.8} = 63.1$ . Why this step? Undo the log by raising 10 to the power. You can't buy $63.1$ speakers, so round up to $N = 64$ to be sure you clear $118$ .

Verify: $N = 63.1$ gives exactly $118 dB$ ; with $64$ , $β = 100 + 10 lo g_{10} 64 = 100 + 18.06 = 118.06 dB \geq 118$ . ✓ And $63.1$ lies between $10$ and $100$ as forecast.

Example 9 — Cell I · Limiting behaviour ( $I \to 0$ )

Worked example What happens to

$β$ as the intensity fades toward perfect silence, $I \to 0$ ? Forecast: Silence should be "infinitely below" the reference. Expect $β \to - \infty$ , not some finite floor.

Step 1. Watch the ratio: as $I \to 0^{+}$ , $\frac{I}{I _{0}} \to 0^{+}$ (a positive number shrinking toward zero). Why this step? $I_{0}$ is fixed; only the top shrinks.

Step 2. Track the log: $lo g_{10} (x) \to - \infty$ as $x \to 0^{+}$ — see the graph in Logarithms and exponentials. So $β = 10 lo g_{10} (I / I_{0}) \to - \infty$ . Why this step? Each further ÷10 of intensity subtracts another $10 dB$ , forever — there's no smallest value.

Step 3. Interpretation: there is no finite dB for true silence; $β = 0$ is only the hearing threshold, not the bottom of the scale.

Verify: Numerically, $I = 1 0^{- 13}$ gives $- 10 dB$ ; $1 0^{- 15}$ gives $- 30$ ; $1 0^{- 20}$ gives $- 80$ — marching down with no floor. ✓ Confirms the $- \infty$ limit and the forecast.

Recall Which cell is this? (self-test)

"A speaker reads 90 dB at 4 m; find the level at 12 m." — which case class? ::: Cell D (inverse-square distance), like Ex 4. "Pressure triples; find the dB rise." — which case class? ::: Cell F (use $20 lo g_{10}$ ), like Ex 6. "How many 70 dB violins to reach 80 dB?" — which case class? ::: Cell H (solve for a count), like Ex 8. Can $β$ be negative, and what does it mean? ::: Yes — it means $I < I_{0}$ , the sound is below the hearing reference (Ex 5). Why do you round the speaker count up in Ex 8? ::: $63.1$ speakers is impossible and would fall short; $64$ guarantees you clear the target. What is a "bel", and why do we use decibels? ::: One bel = $lo g_{10} (I / I_{0})$ , a whole factor of ten in intensity — too coarse; a decibel is a tenth of a bel, hence the ×10.