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Mean free path and reflections in room

Short answer

(a) Mean free path d = 4V/S = 4×120/148 ≈ 3.24 m. (b) Reflections per second n = v/d = 330/3.24 ≈ 102 s⁻¹.

A room has dimensions 6×4×5 m. Find (a) the mean free path of the sound wave in the room (b) the number of reflections per second made by the sound wave with the walls of the room. (Given, velocity of sound in air is 330 ms⁻¹)

The mean free path of sound in a room is the average distance travelled between two successive reflections from the walls: d = 4V/S, where V is the volume and S the total surface area.

Given: l = 6 m, b = 4 m, h = 5 m, v = 330 m/s.

(a) Mean free path

  1. Volume V = 6 × 4 × 5 = 120 m³
  2. Surface area S = 2(lb + bh + lh) = 2(24 + 20 + 30) = 148 m²
  3. Mean free path d = 4V/S = 4 × 120 / 148 = 3.24 m

(b) Reflections per second

Number of reflections per second n = v/d = 330 / 3.243 = 101.75

Answer: (a) mean free path ≈ 3.24 m; (b) ≈ 102 reflections per second.

✓ Answer checked by Claude Opus 5.5 (high effort)

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