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Acoustics and decibels: calculations, SPL versus HL versus SL, resonance, filters and room acoustics

Most acoustics questions are a short calculation: the decibel is a logarithmic ratio, so you add intensities, not decibels. A handful of rules is enough — 10log for intensity, 20log for pressure, the inverse square law, converting between SPL, HL and SL, tube resonance and the Sabine equation.

Why it matters in the exam

The rules that recur in the bank: doubling pressure adds 6 dB and doubling intensity 3 dB; ten times the intensity is 10 dB and ten times the pressure 20 dB; 1 pascal is 94 dB SPL, and 0.2 pascal is 80 dB SPL (reference 20 micropascals). Two equal incoherent sources add 3 dB; N identical sources add 10log(N); a 6 dB gap between two sources adds about 1 dB to the louder; and to remove background noise you subtract intensities. In a free field each doubling of distance takes off 6 dB, but in a room, beyond the critical distance, the reflected field is nearly uniform and the level stops falling.

The scale questions: the HL scale exists because normal threshold in SPL differs from frequency to frequency, so the audiometer removes a reference value (RETSPL) at each frequency; dB SPL = dB HL + RETSPL. Sensation level (SL) is measured above the person's own threshold — 40 dB SL above an SRT of 35 is 75 dB HL. Wavelength = speed of sound over frequency (500 Hz in air — 0.68 m); the speed of sound rises with elasticity and falls with density; the resonant frequency of a mass-spring system falls as mass rises (4 times the mass — half the frequency). With a high-pass filter of 12 dB per octave, a tone 2 octaves below cut-off is attenuated by 24 dB. Sabine reverberation time: 0.161 times volume over absorption; classrooms should be at most 0.6 seconds with background noise at most 35 dBA, and children with hearing loss need an SNR of at least +15 dB.

Key points

Summary table

RuleFormula or valueExample from the bank
Pressure and intensityPressure: 20log; intensity: 10logDoubling pressure = +6; doubling intensity = +3
Adding sourcesAdd intensities; N identical: 10log(N)60 plus 60 = 63; four to five machines = about +1
Distance in a free field−6 dB per doubling of distanceFrom 0.5 to 4 m: 3 doublings = −18
HL, SPL and SLSPL = HL + RETSPL; SL above one's own threshold30 dB HL + 25.5 = 55.5 dB SPL
Wavelength and resonanceλ = c ÷ f; closed tube: c ÷ 4L500 Hz = 0.68 m; a 1.6 cm canal — about 5300 Hz
FiltersSlope in dB per octave; bandwidth = upper minus lower707–1414 Hz: an octave band around 1000 Hz
RoomsRT60 = 0.161 × V ÷ A; standing wavesClassroom: up to 0.6 seconds; in sound field use warble tones

A free exam question

A point sound source in a free field produces 70 dB SPL at 1 meter. What will the level be at 2 meters?

  1. A. 67 dB SPL
  2. B. 64 dB SPL
  3. C. 35 dB SPL
  4. D. 70 dB SPL

See the answer and explanation

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Sources

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Reviewed by the Examqo AI Professor · 2026-11-02. The text may contain errors; when in doubt, check the source.