---
title: "Number of persons from reverberation time"
subject: "Engineering Physics"
chapter: 3
key: "semester_1/engineering_physics:reverberation-persons-present-hall"
source: https://www.bctengineer.com/q/semester_1/engineering_physics/reverberation-persons-present-hall
site: BCT Engineering
---

# Number of persons from reverberation time

Engineering Physics · Chapter 3: Acoustics

- Asked in: 2070 Asar, 5 marks
- Video: [10. Problems on Sabine's Formula for Reverberation Time | 3 Problems Solved | Acoustics](https://www.youtube.com/watch?v=Ep1Ul_98Jy0) — MKS TUTORIALS by Manoj Sir
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/engineering_physics/reverberation-persons-present-hall
- Whole chapter: https://www.bctengineer.com/md/semester_1/engineering_physics/chapter-3

## Question

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.

## 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.

## Answer

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:

```text
n = 0.09553 / 9.524×10⁻⁵ = 1003.0
```

**Answer: about 1000 persons.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Which quantity varies linearly with the number of persons n?
   A. T
   B. T²
   C. 1/T
   D. ln T
   Answer: C. T = 0.161V/(A0 + na), so 1/T = (A0 + na)/(0.161V) is linear in n. (Key point: Linear relation)

2. Using 1/1.5 and 1/1.4, the increase in 1/T due to 500 persons is
   A. 0.0476
   B. 0.6667
   C. 0.7143
   D. 0.0952
   Answer: A. 0.7143 - 0.6667 = 0.0476. (Key point: Effect of 500 persons)

3. For T = 1.312 s, 1/T - 1/T0 equals
   A. 0.7622
   B. 0.0476
   C. 0.0955
   D. 0.1429
   Answer: C. 0.7622 - 0.6667 = 0.0955. (Key point: Target change)

4. The number of persons present is about
   A. 500
   B. 1000
   C. 750
   D. 1500
   Answer: B. n = 500 × 0.0955/0.0476 ≈ 1003. (Key point: Number of persons)

## Close variants set in other exams

- [Persons present from reverberation (1.32 s)](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Areverberation-persons-present-hall-alt) — Engineering Physics, chapter 3, asked 2×

## Related in Engineering Physics

- [Reverberation time of small hall](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Ahall-average-absorption-reverberation-bicc) — Engineering Physics, chapter 3
- [Good acoustic building, reverberation time](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Aacoustic-measures-reverberation-decrease) — Engineering Physics, chapter 3