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Number of persons from reverberation time

Short answer

1/T increases by 0.04762 s⁻¹ for 500 persons, giving absorption per person 9.524×10⁻⁵ s⁻¹. For T = 1.312 s, the increase is0.09553, so n =0.09553/9.524×10⁻⁵ ≈ 1000 persons.

The reverberation time for an empty hall is 1.5 sec. With 500 audiences present in the hall, the reverberation time falls to 1.4 sec. Find the number of persons present in the hall if the reverberation time falls down to 1.312 sec.

Sabine's formula gives T = 0.161V/(A₀ + n·a), where A₀ is the absorption of the empty hall and a is the absorption per person. Hence 1/T is linear in the number of persons.

Given: T₀ = 1.5 s (empty); T₁ = 1.4 s with 500 persons; T₂ = 1.312 s with n persons.

  1. Empty hall: 1/T₀ = A₀ / (0.161V), so 1/1.5 = 0.6667
  2. With 500 persons: 1/T₁ = (A₀ + 500a)/(0.161V) = 1/1.4 = 0.7143
  3. Subtract: 500a/(0.161V) = 0.7143 − 0.6667 = 0.04762
  4. Absorption per person (in these units): a/(0.161V) = 0.04762/500 = 9.524×10⁻⁵ s⁻¹
  5. For T₂ = 1.312 s: 1/T₂ = 0.7622, so n·a/(0.161V) = 0.7622 − 0.6667 = 0.09553
  6. Number of persons:
n = 0.09553 / 9.524×10⁻⁵ = 1003.0

Answer: about 1000 persons.

✓ Answer checked by Claude Opus 5.5 (high effort)

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