1.6.23 · D5Oscillations & Waves
Question bank — Sound intensity — decibels (logarithmic scale)
True or false — justify
Two 60 dB speakers together produce 120 dB.
False. Levels are logarithmic and never add directly; the intensities add, doubling , which adds only dB → 63 dB.
A 0 dB sound means "no sound at all".
False. 0 dB is the threshold of hearing — a real (barely audible) intensity, not zero. Silence would be dB because diverges.
Negative decibel levels are impossible.
False. Any intensity below gives ; e.g. gives dB. Negative just means "quieter than the reference".
Doubling the intensity always adds the same +3 dB, no matter what level you start at.
True. The difference rule depends only on the ratio; a ratio of 2 always gives dB whether you start at 20 dB or 100 dB.
Because , doubling your distance from a point source halves the decibel reading.
False. Doubling distance quarters intensity, and dB — a subtraction, not a halving of the dB number. See Inverse-square law for radiation.
The decibel is a unit like the metre or the watt.
False. dB is dimensionless — it labels a ratio . We write "dB" only to record which reference and which prefactor (10 vs 20) we used.
If you know a sound is 40 dB you can compute its intensity exactly.
True, provided you also know the reference is ; then . Without the reference the dB number alone is meaningless.
A sound at 100 dB is delivering ten times the intensity of one at 10 dB.
False. The difference is 90 dB factors of ten, so is times larger, not 10 times. dB differences are exponents, not multipliers.
Spot the error
"Ten identical drummers each at 70 dB give dB."
You multiplied levels. Ten sources give ten times the intensity, so dB → 80 dB, not 700.
"The problem gives me a pressure ratio of 100, so the level is dB."
Wrong prefactor. For pressure/amplitude use because ; here it's dB. See Sound waves — pressure & displacement.
" and , so ."
Units mismatch: is in microwatts, in watts. Convert first () so the ratio is a pure number before taking the log.
"Going from 60 dB to 63 dB is only a 3-unit change, so basically nothing happened to the intensity."
A 3 dB rise means the intensity doubled. Small dB steps hide large physical changes — that compression is the whole point of the scale.
"Since , a source of zero intensity reads 0 dB."
is not 0 — it is . Zero intensity has no finite decibel value; 0 dB corresponds to , not .
" dB is twice as intense as dB because ."
The scale is logarithmic, so you subtract, not divide: dB , i.e. a million times more intense.
Why questions
Why do we take a logarithm of intensity instead of just quoting intensity in ?
Audible intensities span ; a log turns "multiply by 10" into "add 1", compressing that trillion-fold range into a readable 0–120 scale — see Logarithms and exponentials.
Why is the prefactor 10 (deci-bel) and not 1 (bel)?
One bel is a coarse step (a full factor of ten). Multiplying by 10 gives tenths-of-a-bel resolution, so a barely-noticeable change registers as roughly 1 dB instead of a fraction of a bel.
Why does the reference disappear when we compare two sounds?
Subtracting gives ; the terms cancel, so ratio problems never need it.
Why do intensities add but decibel levels do not?
Intensity is energy flow per area — physically additive. The decibel is a log of intensity, and , so you must return to intensity, sum, then re-log. See Wave energy and power.
Why does the decibel scale match how loud things feel better than raw intensity does?
Human loudness perception is roughly logarithmic (Weber–Fechner), so equal ratio increases feel like equal steps — exactly what a log encodes. Detailed in Loudness vs intensity — psychoacoustics.
Why is used for pressure but for intensity?
Because intensity is proportional to pressure squared, ; substituting into pulls the 2 out front, giving .
Edge cases
What decibel level corresponds to exactly?
dB — the definition anchor. This is why the threshold of hearing is the natural zero, not silence.
What happens to as ?
, since of a number approaching zero grows unboundedly negative. Real silence has no finite decibel reading.
If a point source's power doubles but you also move twice as far away, what is the net dB change?
Intensity scales as : power ×2 gives dB, distance ×2 gives dB, net dB — the ratios combine additively in dB.
Can two different-frequency tones at 50 dB each be added as intensities the same way as identical sources?
Yes for total energy: uncorrelated intensities add, giving dB → 53 dB. (Only coherent same-phase waves could interfere; independent tones do not.)
A moving siren appears to shift pitch via the Doppler effect — does that Doppler shift by itself change its decibel level?
No. Doppler changes the observed frequency, not the intensity, so the level is unchanged by the pitch shift; the level changes only through the accompanying distance (inverse-square) variation.
If you halve the distance to a point source in a free field, what is the exact dB change?
Distance ÷2 → intensity ×4 → dB. Every halving of distance adds about 6 dB.
Recall One-line trap detector
Before answering any dB question, ask two things ::: (1) Is this an absolute level (needs ) or a difference (needs only the ratio)? (2) Is the given quantity intensity () or pressure/amplitude ()?