The amplitude of a sound wave can be quantified in at least three ways:

- by measuring the maximum change in position of the particles that make up the medium,
- by measuring the maximum change in density of the medium, or
- by measuring the maximum change in pressure (the maximum gauge pressure).

None of these quantities are used much, however. In fact, the first two are unusually difficult to measure directly. For typical sound waves, the maximum displacement of the molecules in the air is only a hundred or a thousand times larger than the molecules themselves. Any resulting density fluctuations are equally miniscule and very short lived. (The period of sound waves is typically measured in milliseconds.)

Pressure fluctuations caused by sound waves are much easier to measure (animals have been doing it for hundreds of millions of years with ears), but the results of such measurements are rarely ever reported. Instead, amplitude measurements are almost always used as the raw data in some computation. When done by an electronic circuit — like the circuits in a level meter — the resulting value is called the intensity. When done by a neuronal circuit — like the circuits in your brain — the resulting sensation is called the loudness.

Briefly, the intensity of a sound wave is a combination of its rate and density of energy transfer. It is an objective quantity associated with a wave. Loudness is a perceptual response to the physical property of intensity. It is a subjective quality associated with a wave and is a bit complex. As a general rule the larger the amplitude, the greater the intensity, the louder the sound. Sound waves with large amplitudes are said to be "loud". Sound waves with small amplitudes are said to be "quiet" or "soft". (The word "low" is sometimes also used to mean quiet, but this should be avoided. Use the word "low" only to describe sounds that are low in frequency or pitch.) Loudness will be discussed at the end of this section.

By definition, the intensity of *any wave* is the time-averaged power it transfers per area through some region of space. The traditional way to indicate the time-averaged value of a varying quantity is to enclose it in angle brackets. These look similar to greater and less than symbols but are taller and less pointy. That gives us an equation that looks something like this …

I = | ⟨P⟩ |

A |

The unit of intensity is the watt per square meter — a unit that has no special name.

For simple mechanical waves like sound, intensity is related to the density of the medium and the speed, frequency, and amplitude of the wave. This can be shown with a long, horrible, calculation. Jump to the next highlighted equation if you don't care to see the sausage being made below.

Start with the definition of intensity. Replace power with energy (both kinetic and elastic) over time (one period, for convenience sake).

I = | ⟨P⟩ | = | ⟨E⟩/T | = | ⟨K + U⟩/_{s}T |

A | A | A |

Since kinetic and elastic energies are always positive we can split the time-averaged portion into two parts.

⟨P⟩ = | ⟨E⟩ | = | ⟨K + U⟩_{s} | = | ⟨K⟩ | + | ⟨U⟩_{s} | |

T | T | T | T |

Mechanical waves in a continuous media can be thought of as an infinite collection of infinitesimal coupled harmonic oscillators. Little masses connected to other little masses with little springs as far as the eye can see. On average, half the energy in a simple harmonic oscillator is kinetic and half is elastic. The time-averaged total energy in then either twice the average kinetic energy or twice the average potential energy.

⟨P⟩ = | 2⟨K⟩ | = | 2⟨U⟩_{s} | |

T | T |

Let's work on the kinetic energy and see where it takes us. It has two important parts — mass and speed.

*K* = ½*mv*^{2}

The particles in a longitudinal wave are displaced from their equilibrium positions by a function that oscillates in time and space.

Δx(x,t) = Δx sin _{max} | ⎡ ⎣ | 2π | ⎛ ⎝ | ft − | x | ⎞⎤ ⎠⎦ |

λ |

Take the time derivative to get velocity.

v(x,t) = | ∂ | Δx(x,t) = 2πƒΔx cos _{max} | ⎡ ⎣ | 2π | ⎛ ⎝ | ft − | x | ⎞⎤ ⎠⎦ |

∂t | λ |

Then square it.

v^{2}(x,t) = 4π^{2}ƒ^{2}Δx^{2} cos_{max}^{2} | ⎡ ⎣ | 2π | ⎛ ⎝ | ft − | x | ⎞⎤ ⎠⎦ |

λ |

On to the mass. Density times volume is mass. The volume of material we're concerned with is a box whose area is the surface through which the wave is traveling and whose length is the distance the wave travels.

*m* = ρ*V* = ρ*Ax*

In one period a wave would move forward one wavelength. In the volume spanned by a single wavelength, all the bits of matter are moving with different speeds. Calculus is needed to combine a multitude of varying values into one integrated value. We're dealing with a periodic system here, one that repeats itself over and over again. We can choose to do our calculation at any time we wish as long as we finish at the end of one cycle. For convenience sake let's choose time to be zero — the beginning of a sinusoidal wave.

λ | λ | λ | |||||||||||

K = | ⌠ ⌡ | dK(x,0) = | ⌠ ⌡ | 1 | (ρAdx) v^{2}(x,0) = | ⌠ ⌡ | 1 | (ρA)(4π^{2}ƒ^{2}Δx^{2}) cos_{max}^{2} | ⎡ ⎣ | − 2π | x | ⎤ ⎦ | dx |

2 | 2 | λ | |||||||||||

0 | 0 | 0 |

Clean up the constants.

1 | (ρA)(4π^{2}ƒ^{2}Δx^{2}) = 2π_{max}^{2}ρAƒ^{2}Δx^{2}_{max} |

2 |

Then work on the integral. It may look hard, but it isn't. Just visualize the cosine squared curve traced out over one cycle. See how it divides the rectangle bounding it into equal halves?

The height of this rectangle is one (as in the number 1 with no units) and its width is one wavelength. That gives an area of one wavelength and a half-area of half a wavelength.

λ | ||||||||

⌠ ⌡ | cos^{2} | ⎡ ⎣ | − 2π | x | ⎤ ⎦ | dx = | 1 | λ |

λ | 2 | |||||||

0 |

Put the constants together with the integral and divide by one period to get the time-averaged kinetic energy. (Remember that wavelength divided by period is speed.)

⟨K⟩ | = | ⎰ ⎱ | (2π^{2}ρAƒ^{2}Δx^{2})(_{max} | 1 | λ) | ⎱ ⎰ | 1 | = π^{2}ρAƒ^{2}vΔx^{2}_{max} |

T | 2 | T |

That concludes the hard part. Double the equation above and divide by area …

I = | ⟨P⟩ | = | 2⟨K⟩/T | = | 2(π^{2}ρAƒ^{2}vΔx^{2})_{max} |

A | A | A |

… and we're done.

*I* = 2π^{2}ρƒ^{2}*v*Δ*x*^{2}_{max}

Does this formula make sense? Check to see how each of the factors affect intensity.

factor | comments |
---|---|

I ∝ ρ | The denser the medium, the more intense the wave. That makes sense. A dense medium packs more mass into any volume than a rarefied medium and kinetic energy goes with mass. |

I ∝ ƒ^{2} | The more frequently a wave vibrates the medium, the more intense the wave is. I can see that one with my mind's eye. A lackluster wave that just doesn't get the medium moving isn't going to carry as much energy as one that shakes the medium like crazy. |

I ∝ v | The faster the wave travels, the more quickly it transmits energy. This is where you have to remember that intensity doesn't so much measure the amount of energy transferred as it measures the rate at which this energy is transferred. |

I ∝ Δx^{2}_{max} | The greater the amplitude, the more intense the wave. Just think of ocean waves for a moment. A hurricane-driven, wall-of-water packs a lot more punch than ripples in the bathtub. The metaphor isn't visually correct, since sound waves are longitudinal and ocean waves are complex, but it is intuitively correct. |

Don't forget to breathe, then discuss intensity and pressure amplitude. Amplitude is measured in meters [m] while pressure amplitude is measured in pascals [Pa] or more commonly millipascals [mPa] or most commonly micropascals [μPa]. Intensity is proportional to the square of pressure amplitude.

I = | ⟨power⟩ | = | ⟨F⟩v | = ⟨pressure⟩v |

A | A |

For simple harmonic motion …

*v _{max}* = 2πƒΔ

Dimensional analysis game …

…