Skip to content
LNGR SPKRHIFI. TWO WAYS. HANDCRAFTED.
LNGR / Knowledge

Room modes: why bass changes with position

At low frequencies a room behaves like a resonant volume. The same subwoofer can therefore sound powerful at one seat and weak a short distance away.

Standing patterns, moving air

A bass note can be weak at the sofa and strong near a wall because direct and reflected sound combine differently by location. Room modes are resonant patterns of the enclosure formed by the room. Air still moves: the pressure maxima and minima occupy approximately fixed positions while their values oscillate over time.

Half a wavelength between walls

The first axial mode of an ideal rigid room axis has pressure maxima at its ends and a pressure node in the middle. Its frequency is c/(2L). Higher orders add spatial variation. Pressure and particle velocity are not the same quantity; a pressure node can coincide with strong particle motion.

Three-dimensional modes and real losses

Axial modes involve one room dimension, tangential modes two and oblique modes three. Rectangular rigid-room equations predict eigenfrequencies, not the complete measured response. Doors, openings, absorption and irregular geometry alter behaviour. Source and listener positions decide how strongly particular patterns are excited and observed.

A bass dip can have another cause

A nearby boundary reflection can interfere with direct sound, producing speaker-boundary interference. Poor phase or delay alignment between a subwoofer and main speakers can also create a dip. Measure sources separately and together, then change one position or setting. A dip that moves with geometry provides a useful clue.

Why boosting a null is inefficient

If two contributions cancel, increasing both often leaves the cancellation largely intact while demanding more power and excursion. First investigate placement, polarity and crossover timing. Reducing a repeatable peak may be more useful. Check several positions before optimising one narrow microphone location at the expense of the rest of the seating area.

More sources and acoustic treatment

Distributed subwoofers can excite different patterns and produce a more even result when level and timing are adjusted. Two sources placed together mainly increase capacity rather than diversifying room excitation. Deep-bass absorption needs suitable depth or a correctly designed resonant system; a thin decorative foam strip does not solve a deep room resonance.

A manageable measurement sequence

Save a baseline with positions and settings. Compare two or three practical placements while keeping everything else constant. Check integration with the mains, then apply appropriate EQ. Review individual seat responses as well as any average. The best solution should work in the room’s everyday use, not only for a sofa temporarily placed in a doorway.

Explore the relationship
Interactive experiment

Bass forms a pattern in the room.

Change room length and mode order. The curve shows the spatial pressure pattern of one ideal axial mode, not a frequency response at the sofa.

Model limits: Ideally rigid walls, rectangular room, c = 343 m/s. One axial mode shape only; no decay-time or absolute-level prediction.

A closer look · model and equation

fn = n·c/(2L); p(x) ∝ cos(nπx/L). For a rectangular room, f = c/2·√[(nx/Lx)² + (ny/Ly)² + (nz/Lz)²]. Real losses and source positions also determine strength.

Workshop takeaway

Before boosting a bass dip, move the listener or source and measure again. A spatial cancellation cannot be filled indefinitely with power.

Work it out · answers unfold

Write down your prediction first, then your calculation. Open the answer only when you are ready to compare.

10 · A time window is 5 ms long. Estimate characteristic frequency resolution using 1/T.

Show worked answer
  1. T = 0.005 s.
  2. Take reciprocal.
  3. About 200 Hz; not a universal hard lower accuracy limit.
Calculation
\Delta f\approx\frac{1}{T}=200\,Hz

11 · What window duration gives 50 Hz resolution by the same approximation?

Show worked answer
  1. Duration = 1/resolution.
  2. 1/50 = 0.02 s.
  3. 20 ms.
Calculation
T\approx\frac{1}{50}=20\,ms

12 · A reflected path is 1.715 m longer than the direct path. How much later does it arrive, c = 343 m/s?

Show worked answer
  1. Delay = extra path/speed.
  2. 1.715/343.
  3. Reflection arrives 5 ms later.
Calculation
\Delta t=\frac{1.715}{343}=5\,ms

19 · A frequency response looks better after stronger smoothing. Has the circuit improved? Explain.

Show worked answer
  1. No, only presentation changed.
  2. Narrow peaks and dips can be hidden.
  3. Compare before and after using identical settings.

Match drivers together →