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Persons present from reverberation (1.32 s)

Short answer

For T = 1.32 s, 1/T increase is0.09091, so the number of persons n =0.09091/9.524×10⁻⁵ ≈ 955.

The time of reverberation of an empty hall is 1.5 sec; with 500 audiences present in the hall, the reverberation time falls to 1.4 sec. Find the no. of persons present in the hall if the reverberation time falls down to 1.32 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 = 1.32 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 = 1.32 s: 1/T₂ = 0.7576, so n·a/(0.161V) = 0.7576 − 0.6667 = 0.09091
  6. Number of persons:
n = 0.09091 / 9.524×10⁻⁵ = 954.5

Answer: about 955 persons.

✓ Answer checked by Claude Opus 5.5 (high effort)

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