Room Acoustic Modes
Calculate room axial resonant frequencies and Schroeder frequency for acoustic treatment and speaker placement.
Formula
f_n = n × c / (2L)
How It Works
Worked Example
Room: 5.0 m × 3.5 m × 2.5 m, speed of sound 343 m/s. First axial modes: Length: f = 343 / (2 × 5.0) = 34.3 Hz Width: f = 343 / (2 × 3.5) = 49.0 Hz Height: f = 343 / (2 × 2.5) = 68.6 Hz Second axial modes (n=2) are at 68.6 Hz, 98.0 Hz, and 137.2 Hz respectively. Room volume: 5.0 × 3.5 × 2.5 = 43.75 m³ Schroeder frequency: 2000 × √(0.16 / 43.75) ≈ 121 Hz Below 121 Hz, the room behavior is dominated by individual modes. This room has no two first-order modes that coincide, so bass modal distribution is reasonable. Place bass traps at room corners and wall junctions where mode pressure builds up most.
Practical Tips
- ✓For home studio design, target dimension ratios such as 1.0 : 1.28 : 1.54 (Sepmeyer) or 1.0 : 1.6 : 2.33 (Bolt optimal). Avoid rooms where any two dimensions share a common ratio.
- ✓Bass traps are most effective at room corners (wall-wall, floor-ceiling, or tri-corners) because all axial modes have maximum pressure at the boundaries.
- ✓Use the Schroeder frequency as a guide: acoustic treatment (absorption, diffusion) matters most for frequencies below it; above it, the room tends to behave more uniformly.
Common Mistakes
- ✗Confusing frequency spacing with modal density — modes that are close together in frequency (within ~5 Hz) can interact and cause severe peaks or dips at those frequencies.
- ✗Forgetting that the Schroeder frequency formula (2000√(T60/V)) normally requires the room's reverberation time T60; the approximation 2000√(0.16/V) assumes T60 ≈ 0.16 s, which underestimates in reverberant rooms.
- ✗Treating room modes as a fixed-frequency problem — moving the listening position or speaker even half a metre can place you at a pressure maximum or null of a given mode.
Frequently Asked Questions
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