- 2073 Shrawan · 5 marks
Good acoustic building, reverberation time
Short answer
Good acoustics requires optimum reverberation time, absorbent surfaces (carpets, curtains, acoustic tiles), proper loudness via reflectors, no echoes, uniform sound distribution, and noise isolation. Sabine’s formula T = 0.161V/ΣaS shows T ∝ 1/ΣaS; increasing total absorption by adding absorbents decreases reverbration time.
What are the measures of good acoustic building? Show that the reverberation time decreases with increase in absorbing factors in a hall.
Good acoustics means every listener hears the sound clearly, with enough loudness and without echo, distortion or noise. Reverberation is the persistence of sound in a hall after the source stops, due to repeated reflections.
Measures for good acoustics
- Reverberation time should be optimum (about 0.5-1 s for speech rooms, 1.5-2 s for music halls).
- Loudness: use a low ceiling or a sloping, curved reflecting sounding board behind the speaker so that sound reaches the back seats.
- Absorbing materials: line walls and ceiling with felt, acoustic tiles, heavy curtains, carpets, perforated boards. Audience and cushioned seats also absorb.
- No echoes: avoid large flat parallel walls and concave surfaces that focus sound; use irregular surfaces and diffusers.
- Sound distribution: keep the sound uniform in all parts using properly placed loudspeakers.
- Noise control: use double doors/windows, and keep the hall away from traffic and machinery.
- Resonance: avoid wooden panels or hollow boxes that resonate with particular frequencies.
Reverberation time and absorption (Sabine's formula)
Let V = volume of hall, S₁, S₂ ... surface areas with absorption coefficients a₁, a₂ ..., A = ΣaS (total absorption), E = sound energy density, c = speed of sound.
- Sound energy falling per second on the surfaces is (cE/4) per unit area, so the energy absorbed per second = (cE/4)·A.
- This equals the loss of total energy: V dE/dt = −(c A/4) E.
- Integrating: E = E₀ e^(−cAt/4V).
- Reverberation time T is the time for the energy to fall to 10⁻⁶ of the initial value: 10⁻⁶ = e^(−cAT/4V).
- Taking logs: ln(10⁶) = 13.8 = cAT/(4V), so T = 55.3 V/(cA).
- With c = 343 m/s:
T = 0.161 V / ΣaS (V in m³, S in m², T in s)
Since V is fixed for a given hall, T ∝ 1/A. When the absorbing factor (a or area S of absorbers) increases, ΣaS increases and the reverberation time decreases. Hence adding carpets, curtains and audience reduces reverberation.
✓ Answer checked by Claude Opus 5.5 (high effort)
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