---
title: "Reverberation time of small hall"
subject: "Engineering Physics"
chapter: 3
key: "semester_1/engineering_physics:hall-average-absorption-reverberation-bicc"
source: https://www.bctengineer.com/q/semester_1/engineering_physics/hall-average-absorption-reverberation-bicc
site: BCT Engineering
---

# Reverberation time of small hall

Engineering Physics · Chapter 3: Acoustics

- Asked in: 2071 Chaitra, 5 marks
- Video: [How to Calculate Reverberation Time in a Room - Acoustic Absorption Explained](https://www.youtube.com/watch?v=bMVDPkwzL8I) — Commercial Acoustics
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/engineering_physics/hall-average-absorption-reverberation-bicc
- Whole chapter: https://www.bctengineer.com/md/semester_1/engineering_physics/chapter-3

## Question

The volume of a small hall in BICC is 900m³. The wall area of the hall is 150 m², ceiling area 90 m² and floor area is 95 m². The sound absorption coefficient for wall is 0.04, for ceiling is 0.04 and for the floor is 0.08. Calculate the average sound absorption coefficient and reverberation time.

## Short answer

Total absorption ΣaS = 0.04×150 + 0.04×90 + 0.08×95 = 17.2 m². Average absorption coefficient ā = 17.2/335 ≈ 0.051. Reverberation time T = 0.161×900/17.2 ≈ 8.4 s.

## Answer

Use the average absorption coefficient ā = ΣaS / ΣS and Sabine's formula T = 0.161 V / ΣaS.

**Given:** V = 900 m³; walls: S₁ = 150 m², a₁ = 0.04; ceiling: S₂ = 90 m², a₂ = 0.04; floor: S₃ = 95 m², a₃ = 0.08.

1. Total absorption:

```text
ΣaS = 0.04×150 + 0.04×90 + 0.08×95
    = 6.0 + 3.6 + 7.6
    = 17.2 m² (open-window units)
```

2. Total surface area ΣS = 150 + 90 + 95 = 335 m²
3. Average absorption coefficient: ā = 17.2 / 335 = **0.0513**
4. Reverberation time:

```text
T = 0.161 × 900 / 17.2
  = 144.9 / 17.2
  = 8.42 s
```

**Answer: ā ≈ 0.051; T ≈ 8.4 s** (very long, so the hall needs more absorbing material).

## Check yourself

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

1. The average absorption coefficient is calculated as
   A. the plain mean of the three coefficients
   B. ΣaS / ΣS
   C. ΣS / ΣaS
   D. ΣaS × V
   Answer: B. ā = total absorption divided by total surface area, weighting each coefficient by its area. (Key point: Average absorption coefficient)

2. ΣaS for walls (150, 0.04), ceiling (90, 0.04) and floor (95, 0.08) is
   A. 17.2 m²
   B. 335 m²
   C. 0.16 m²
   D. 13.4 m²
   Answer: A. 6.0 + 3.6 + 7.6 = 17.2 m². (Key point: Total absorption)

3. Using T = 0.161V/ΣaS with V = 900 m³, T is about
   A. 0.84 s
   B. 84 s
   C. 8.4 s
   D. 2.1 s
   Answer: C. T = 144.9/17.2 ≈ 8.42 s. (Key point: Reverberation time)

4. What does such a long reverberation time tell us?
   A. The hall is perfect
   B. The hall needs more absorbing material
   C. The hall is too small
   D. The hall needs less audience
   Answer: B. 8.4 s is far above optimum, so more absorption is needed. (Key point: Interpretation)

## Related in Engineering Physics

- [Average absorption coefficient of a hall](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Areverberation-absorption-coefficient-hall) — Engineering Physics, chapter 3
- [Hall voltage, coefficient and mobility](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Ahall-numerical-copper-strip-thin) — Engineering Physics, chapter 8
- [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×
- [Good acoustic building, reverberation time](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Aacoustic-measures-reverberation-decrease) — Engineering Physics, chapter 3

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