Sound intensity — decibels (logarithmic scale): Exercises

Intuition How to use this page

Work each problem before opening its solution. The problems climb a ladder: first just recognising the formula, then applying it, then analysing multi-step situations, then synthesising several ideas at once, and finally mastery problems that mix everything. After each level there is a [!mistake] callout for the exact trap that catches most people at that rung.

Everything you need is in the parent note: Sound intensity — decibels (logarithmic scale). Keep these three facts in your hand at all times — the level formula, the difference rule, and the reference intensity: $β = 10 lo g_{10} (\frac{I}{I _{0}}) dB, Δ β = 10 lo g_{10} (\frac{I _{2}}{I _{1}}), I_{0} = 1 0^{- 12} W/m^{2} .$

Useful log values (memorise the first two): $lo g_{10} 2 \approx 0.301$ , $lo g_{10} 3 \approx 0.477$ , $lo g_{10} 5 \approx 0.699$ , $lo g_{10} 16 \approx 1.204$ .

Intuition Edge case — what does a

negative dB mean? The formula does not stop working when a sound is quieter than the reference $I_{0}$ . If $I < I_{0}$ , the ratio $I / I_{0}$ is a number less than 1, and the logarithm of a number below 1 is negative — so $β$ comes out negative.

$I = I_{0} \Rightarrow β = 10 lo g_{10} (1) = 0 dB$ (the threshold, the natural "zero").
$I = \frac{1}{10} I_{0} = 1 0^{- 13} W/m^{2} \Rightarrow β = 10 lo g_{10} (1 0^{- 1}) = - 10 dB$ .

So $0$ dB is not silence and not the floor of the scale — it is just the marker "as loud as the quietest sound a typical person can hear." A $- 10$ dB sound is a real, physical sound; it is simply too faint for most ears (a very sensitive microphone or a keen listener in a soundproof room could still register it). Negative dB = below the reference, not "no sound." True silence ( $I = 0$ ) has $β = 10 lo g_{10} (0) = - \infty$ , off the bottom of the scale entirely.

Prerequisites you may want open: Logarithms and exponentials, Wave energy and power, Inverse-square law for radiation, Sound waves — pressure & displacement, Loudness vs intensity — psychoacoustics, Doppler effect.

Level 1 — Recognition

Goal: plug into the formula and read it back. No traps yet — just get fluent.

Recall Solution 1.1

WHAT we do: put $I$ and $I_{0}$ into $β = 10 lo g_{10} (I / I_{0})$ . WHY: $β$ asks "how many factors of ten above threshold?", and the log counts them. $\frac{I}{I _{0}} = \frac{1 0 ^{- 9}}{1 0 ^{- 12}} = 1 0^{- 9 - (- 12)} = 1 0^{3} .$ Subtracting exponents ( $- 9 - (- 12) = 3$ ) tells us the library is $3$ factors of ten above the threshold. $β = 10 lo g_{10} (1 0^{3}) = 10 \times 3 = 30 dB .$

Recall Solution 1.2

WHAT we do: invert the formula step by step. WHY: every operation that built $β$ we now undo in reverse order. $70 = 10 lo g_{10} \frac{I}{I _{0}} .$ Divide by 10 (undo the "deci" ×10): $lo g_{10} \frac{I}{I _{0}} = 7.$ Raise 10 to both sides (undo the log — it asks "10 to what gives this ratio?"): $\frac{I}{I _{0}} = 1 0^{7} \Rightarrow I = 1 0^{7} \times 1 0^{- 12} = 1 0^{- 5} W/m^{2} .$

Recall Solution 1.3

WHAT we do: use the difference rule $Δ β = 10 lo g_{10} (I_{2} / I_{1})$ with ratio $10$ . WHY the difference rule: we only care about the change, and $I_{0}$ cancels when we subtract two levels. $Δ β = 10 lo g_{10} (10) = 10 \times 1 = + 10 dB .$ This is the headline fact: every ×10 in intensity is exactly +10 dB.

Level 2 — Application

Goal: combine the formula with one physical idea (doubling, distance, ratios).

Recall Solution 2.1

WHAT we do: intensities add, so two equal fans give $2 I$ . Then apply the difference rule. WHY intensities add and not levels: intensity is energy-flow density (a physical amount of power per area), and power adds. Levels are logarithms — logs of a sum are not the sum of logs. $Δ β = 10 lo g_{10} (2) = 10 \times 0.301 = 3.01 \approx + 3 dB .$ $β_{new} = 45 + 3 = 48 dB .$

Recall Solution 2.2

WHAT we do: use $I \propto 1/ r^{2}$ (from Inverse-square law for radiation) to get the intensity ratio, then convert to dB. WHY: a point source spreads its power over a sphere of area $4 π r^{2}$ ; halving $r$ shrinks that area by $4$ , so intensity quadruples. $\frac{I _{2}}{I _{1}} = (\frac{r _{1}}{r _{2}})^{2} = (\frac{10}{5})^{2} = 2^{2} = 4.$ $Δ β = 10 lo g_{10} (4) = 10 \times (2 lo g_{10} 2) = 10 \times 0.602 = + 6.0 dB .$ $β_{new} = 58 + 6 = 64 dB .$

The figure below shows the two arcs the sound crosses. Look at the green arc at $5 m$ : the same power is packed onto one-quarter of the area of the blue arc at $10 m$ , so the intensity there is four times larger — and four times is exactly the $+ 6 dB$ we computed.

Figure — Sound intensity — decibels (logarithmic scale)

Recall Solution 2.3

WHAT we do: invert the difference rule to solve for the ratio. WHY: the difference in dB is $10 lo g_{10}$ of the intensity ratio, so we undo the log. $Δ β = 95 - 65 = 30 dB = 10 lo g_{10} \frac{I _{2}}{I _{1}} .$ $lo g_{10} \frac{I _{2}}{I _{1}} = 3 \Rightarrow \frac{I _{2}}{I _{1}} = 1 0^{3} = 1000 .$ A 30 dB gap is a thousand times the intensity — not "about 1.5×".

Level 3 — Analysis

Goal: several steps, or a non-power-of-ten ratio that needs real log values.

Recall Solution 3.1

WHAT we do: distance goes ×4, so by inverse-square intensity goes ÷ $4^{2} = 16$ . WHY the square: $I \propto 1/ r^{2}$ , and $(r_{1} / r_{2})^{2} = (2/8)^{2} = (1/4)^{2} = 1/16$ . $Δ β = 10 lo g_{10} \frac{1}{16} = - 10 lo g_{10} 16 = - 10 \times 1.204 = - 12.0 dB .$ The negative sign says the level drops (we moved away). $β_{new} = 80 - 12 = 68 dB .$

The figure below plots the level as a smooth curve against distance. Follow the blue curve from the green dot ( $80 dB$ at $2 m$ ) down to the red dot ( $68 dB$ at $8 m$ ): the gray double-arrow marks the $- 12 dB$ drop, and notice how the curve falls steeply near the source but flattens far away — that is the $1/ r^{2}$ law losing its bite at large $r$ .

Recall Solution 3.2

WHAT we do: the rise is $80 - 60 = 20 dB$ ; find the intensity ratio, which equals $N$ . WHY $N$ : $N$ identical sources give $N$ times one machine's intensity. $20 = 10 lo g_{10} (N) \Rightarrow lo g_{10} (N) = 2 \Rightarrow N = 1 0^{2} = 100 .$ Sanity check the compression: one hundred machines make only 20 dB more than one — that is the whole point of the log scale.

Recall Solution 3.3

WHAT we do: " $10%$ gets through" means $I_{2} / I_{1} = 0.10 = 1 0^{- 1}$ . WHY read it as a ratio: dB only ever sees ratios; convert the percentage to a fraction first. $Δ β = 10 lo g_{10} (1 0^{- 1}) = 10 \times (- 1) = - 10 dB .$ So cutting 90% of the energy is only a 10 dB drop — a caution about how much energy a "small" dB change hides.

Level 4 — Synthesis

Goal: stitch two independent effects together in one problem.

Recall Solution 4.1

WHAT we do: handle the two effects as two intensity ratios and multiply them, because both scale the same intensity. WHY multiply ratios: intensity is proportional to (number of sources) and to $1/ r^{2}$ , so $I_{final} / I_{ref} = N \times (r_{1} / r_{2})^{2}$ .

More sources: $\times 4$ .
Farther away: $(4/12)^{2} = (1/3)^{2} = 1/9$ . $\frac{I _{2}}{I _{1}} = 4 \times \frac{1}{9} = \frac{4}{9} .$ Now convert to dB: $Δ β = 10 lo g_{10} \frac{4}{9} = 10 (lo g_{10} 4 - lo g_{10} 9) = 10 (0.602 - 0.954) = 10 (- 0.352) = - 3.52 dB .$ $β_{final} = 100 - 3.5 = 96.5 dB (\approx 96 - 97 dB) .$ The 4 sirens add +6 dB, but the extra distance takes away about −9.5 dB, netting a small drop.

Recall Solution 4.2

WHAT we do: convert each level to an intensity ratio (relative to $I_{0}$ ), add the intensities, convert back. WHY: you cannot add 70 and 74 dB directly — only physical intensities add.

Work in units of $I_{0}$ so we never touch $1 0^{- 12}$ : $\frac{I _{A}}{I _{0}} = 1 0^{70/10} = 1 0^{7}, \frac{I _{B}}{I _{0}} = 1 0^{74/10} = 1 0^{7.4} .$ Add: $\frac{I _{A} + I _{B}}{I _{0}} = 1 0^{7} + 1 0^{7.4} = 1 0^{7} (1 + 1 0^{0.4}) = 1 0^{7} (1 + 2.512) = 1 0^{7} \times 3.512.$ Convert the total back to dB: $β = 10 lo g_{10} (1 0^{7} \times 3.512) = 10 (7 + lo g_{10} 3.512) = 10 (7 + 0.5457) = 75.5 dB .$ Note it sits just above the louder source (74 dB), exactly as expected: adding a slightly quieter sound nudges the total up only a little.

Level 5 — Mastery

Goal: pressure–vs–intensity subtlety, inversion, and a Doppler-flavoured mix.

Recall Solution 5.1

WHAT we do: turn the pressure factor into an intensity factor first, then into dB. WHY $I \propto p^{2}$ : a sound wave's energy density goes as the square of the pressure amplitude (like kinetic energy $\propto v^{2}$ ). So multiplying $p$ by 10 multiplies $I$ by $1 0^{2} = 100$ . $\frac{I _{2}}{I _{1}} = (\frac{p _{2}}{p _{1}})^{2} = 1 0^{2} = 100.$ $Δ β = 10 lo g_{10} (100) = 10 \times 2 = + 20 dB .$ Equivalently, $Δ β = 20 lo g_{10} (p_{2} / p_{1}) = 20 lo g_{10} 10 = 20$ dB — that is why pressure ratios carry a factor of 20 instead of 10.

Recall Solution 5.2

WHAT we do: find the current level, compare to 80 dB, then convert any excess back to an intensity factor. WHY both directions: we go intensity → dB to judge, then dB → ratio to prescribe the fix. $β = 10 lo g_{10} \frac{3.2 \times 1 0 ^{- 4}}{1 0 ^{- 12}} = 10 lo g_{10} (3.2 \times 1 0^{8}) = 10 (8 + lo g_{10} 3.2) .$ Since $lo g_{10} 3.2 = lo g_{10} (2^{5} /10) = 5 (0.301) - 1 = 0.505$ : $β = 10 (8.505) = 85.05 dB .$ That is above the 80 dB cap by $5.05 dB$ — not compliant. To remove $5.05$ dB we need an intensity factor $f$ with $10 lo g_{10} f = - 5.05$ : $lo g_{10} f = - 0.505 \Rightarrow f = 1 0^{- 0.505} = 0.313.$ They must reduce intensity to about $31%$ of the current value — i.e. cut it by roughly a factor of $3.2$ . (Pleasingly, the cut factor equals the leading $3.2$ , because they were exactly $lo g_{10} 3.2$ dB over.)

Recall Solution 5.3

WHAT we do: separate frequency (what Doppler shifts) from intensity (power per area), then read off the dB change. WHY they are independent: the decibel level depends only on $I = P / (4 π r^{2})$ — the radiated power and the distance. Doppler changes the frequency the listener detects, but it does not change how much power crosses each square metre at that instant. Pitch and intensity are two different attributes of the same wave; the dB scale measures energy flow, not pitch.

If the radiated power $P$ is unchanged and the distance $r$ is momentarily fixed, then $I$ is unchanged, so the intensity ratio is exactly $1$ : $Δ β = 10 lo g_{10} (1) = 0 dB .$
So why does an approaching train seem to get louder in real life? Two reasons, neither of them the Doppler pitch shift itself:
1. The distance $r$ is actually shrinking as the train nears, and $I \propto 1/ r^{2}$ makes it genuinely more intense — a pure distance effect (see Problem 2.2 / 3.1).
2. A psychoacoustic bias: at equal intensity, higher-pitched tones can feel louder to the human ear (see Loudness vs intensity — psychoacoustics). This is perception, not physics of the wave. The Doppler frequency shift, considered on its own at fixed distance and power, contributes exactly $0$ dB.

Recall One-line answer key (peek only to check)

1.1 → 30 dB · 1.2 → $1 0^{- 5}$ W/m² · 1.3 → +10 dB 2.1 → 48 dB · 2.2 → 64 dB · 2.3 → 1000× 3.1 → 68 dB · 3.2 → 100 machines · 3.3 → −10 dB 4.1 → ≈96.5 dB · 4.2 → ≈75.5 dB 5.1 → +20 dB · 5.2 → 85.05 dB, cut to ≈31% (factor ≈3.2) · 5.3 → 0 dB from Doppler alone