---
title: "Persons present from reverberation (1.32 s)"
subject: "Engineering Physics"
chapter: 3
key: "semester_1/engineering_physics:reverberation-persons-present-hall-alt"
source: https://www.bctengineer.com/q/semester_1/engineering_physics/reverberation-persons-present-hall-alt
site: BCT Engineering
---

# Persons present from reverberation (1.32 s)

Engineering Physics · Chapter 3: Acoustics

- Asked in: 2069 Chaitra, 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-alt
- Whole chapter: https://www.bctengineer.com/md/semester_1/engineering_physics/chapter-3

## Question

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.

## 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.

## 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 = 1.32 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:

```text
n = 0.09091 / 9.524×10⁻⁵ = 954.5
```

**Answer: about 955 persons.**

## Check yourself

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

1. Why can the number of persons be found from a proportion of changes in 1/T?
   A. Because T is linear in n
   B. Because 1/T = (A0 + na)/(0.161V) is linear in n
   C. Because volume changes with n
   D. Because a is zero
   Answer: B. Each person adds the same absorption a, so 1/T rises in equal steps per person. (Key point: Linear relation)

2. The change in 1/T per person is
   A. 0.0476 per person
   B. 9.5×10⁻⁵ s⁻¹ per person
   C. 0.667 per person
   D. 1.4 per person
   Answer: B. 0.04762/500 = 9.524×10⁻⁵ s⁻¹. (Key point: Absorption per person unit)

3. For T = 1.32 s, 1/T - 1/T0 equals
   A. 0.0909
   B. 0.7576
   C. 0.0476
   D. 0.1500
   Answer: A. 0.7576 - 0.6667 = 0.0909. (Key point: Target 1/T)

4. The number of persons is about
   A. 500
   B. 1003
   C. 955
   D. 1320
   Answer: C. n = 0.09091/9.524×10⁻⁵ ≈ 955. (Key point: Number of persons)

## Close variants set in other exams

- [Number of persons from reverberation time](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Areverberation-persons-present-hall) — 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