2 × 2 = 4

As Barak suggested in his comment, the perceptible sound quality I was actually referring to in yesterday’s post is sound pressure (the actual pressure of a sound against the ear), rather than loudness, which is a convenient logarithmic scale of perception. Barak, thanks for pointing that out.

Sound pressure — the external phenomenon our ears can perceive — falls off as 1/distance, rather than 1/distance2. Why is that?

Well, sound waves consist of regions of air that are acting against each other like little springs. Each air particle is actually spinning in a kind of circle. One dimension of this circle is how rarified or compressed the air is. The other dimension of the circle is how fast each air particle is moving.

When you clap your hands together with four times as much energy, the pressure variation becomes twice as big, and each air particle also moves twice as fast: 2 × 2 = 4. You can think of the energy as the area of this pressure/velocity circle: As particles spin around in this circle with a given frequency, the area of the circle (proportional to pressure × velocity) tells us how much energy the particle has.

By the way, how may times per second each particle spins around in its energy circle is indeed the frequency of the sound. For example, the sound from a standard A-pitch tuning fork makes the air particles spin around in their little pressure/velocity circles exactly 440 times per second.

The energy you cause by clapping your hands indeed drops off as 1/distance2, but our ears can only perceive one dimension of this spinning energy — the pressure axis. The other dimension consists of velocity — how fast the air moves as it rhythmically expands and contracts. Our ears cannot perceive this velocity dimension (just as, if you get hit by a baseball, you don’t feel how fast it was going, only how hard it has hit you). The part we can perceive — the pressure variation — which varies with the diameter of the energy circle, not with its area — drops off as 1/distance.

One Response to “2 × 2 = 4”

  1. Hmm. If the pressure is cut in half, then it seems like the eardrums would be deflected half as far (in the not-too-loud regime). So the eardrums are moving back-and-forth half as far. And the force on them is half as much. Since power is force times distance, the power is (1/2)x(1/2)=1/4, giving 1/r^2. This holds if the eardrum deflection is phase lagged from the sound wave, but not if it is in phase. So it would hold at higher frequencies, but not lower ones.

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