---
title: "Sound energy flow per square cm"
subject: "Engineering Physics"
chapter: 3
key: "semester_1/engineering_physics:sound-energy-flow-frequency-amplitude"
source: https://www.bctengineer.com/q/semester_1/engineering_physics/sound-energy-flow-frequency-amplitude
site: BCT Engineering
---

# Sound energy flow per square cm

Engineering Physics · Chapter 3: Acoustics

- Asked in: 2072 Chaitra, 5 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/engineering_physics/sound-energy-flow-frequency-amplitude
- Whole chapter: https://www.bctengineer.com/md/semester_1/engineering_physics/chapter-3

## Question

A source of sound has a frequency of 256 Hz and amplitude of 0.50 cm, calculate the energy flow across a square cm per sec. The velocity of sound in air is 330 m/s and density of air is 1.29 kg/m³.

## Short answer

Intensity I = 2π²n²a²ρv = 2×π²×(256)²×(5×10⁻³)²×1.29×330 ≈ 1.38×10⁴ W/m², which is 1.38 W/cm². The energy flowing across 1 cm² per second is 1.38 J.

## Answer

The energy flowing per second across unit area (intensity) for a progressive wave is I = ½ρω²a²v = 2π²n²a²ρv.

**Given:** n = 256 Hz, a = 0.50 cm = 5×10⁻³ m, v = 330 m/s, ρ = 1.29 kg/m³.

1. Energy density of the wave = ½ρω²a², and energy crossing unit area per second = (energy density) × v.
2. ω = 2πn = 2π × 256 = 1608.5 rad/s
3. Substituting:

```text
I = 2π² n² a² ρ v
  = 2 × (9.8696) × (256)² × (5×10⁻³)² × 1.29 × 330
  = 1.377 × 10⁴ J s⁻¹ m⁻²
```

4. Convert to per cm²: 1 m² = 10⁴ cm², so
I = 1.377×10⁴ / 10⁴ = 1.377 J s⁻¹ cm⁻²

**Answer: energy flow ≈ 1.38 J per second across each square cm (1.38 W/cm²).**

## Check yourself

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

1. The energy flowing per second across unit area for a progressive sound wave is
   A. I = ½ρω²a²v
   B. I = ρωav
   C. I = ½ρωa
   D. I = ρω²a²/v
   Answer: A. Energy density ½ρω²a² multiplied by the speed v gives I = ½ρω²a²v = 2π²n²a²ρv. (Key point: Intensity formula)

2. Substituting the values gives I of about
   A. 1.38×10⁴ J s⁻¹ m⁻²
   B. 1.38 J s⁻¹ m⁻²
   C. 1.38×10⁸ J s⁻¹ m⁻²
   D. 138 J s⁻¹ m⁻²
   Answer: A. 2π²n²a²ρv = 2(9.87)(256²)(5×10⁻³)²(1.29)(330) ≈ 1.377×10⁴. (Key point: Result per m²)

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