What is sound energy or acoustic energy?

Sound is a wave, and as such, transports energy produced by a source. Therefore, sound energy or acoustic energy is the energy emitted by a sound source that propagates in the surrounding space. This energy can be proved to be proportional to the square of the amplitude of the oscillation.

At a certain point, a sound wave can reach a "receiver" (the membrane of the eardrum, a microphone, etc.) and in this case the wave energy is partially reflected away from it and partially transferred to it. The fraction of the energy transferred to the "receiver" depends on the physical property called impedance, which measures transmission efficiency. In particular, if the medium transporting the energy and the "receiver" have the same impedance (in this case, we say that we have a situation of impedance matching), the energy will be transferred with maximum efficiency. The transfer of energy can have resounding effects:

• a singer can break glasses with their voice;
• an airplane taking off can break the glass panes of windows;
• the sound of a gunshot can damage the eardrum;
• or more simply, through our sistema uditivo_en|auditory system, it allows us to hear.

What is the power of a sound source?

For power of a sound source or acoustic power, we mean the quantity of energy emitted by a sound source in a unit of time. In formula, the power is the ratio of the energy emitted E to the emission time t, in other words

${\displaystyle P=\frac{E}{t}}$

(1)

In the international system, power is measured in joule/s; this unit of measurement has been given the name watt with the symbol W.

The acoustic powers involved are usually very small: the voice in normal conversation emits about 10 millionths of a watt, a large orchestra playing fortissimo a few tens of watts, a jet aircraft at take-off tens of kilowatts. Even a loudspeaker converts only a small fraction (often around 1%) of the electrical power it receives into sound.

Actually, the ratio (1) gives the average power over time interval with a duration t; if the "instantaneous" power is desired, the average power must be calculated considering infinitesimal time intervals ${\displaystyle \mathrm{\Delta }t}$, that tend towards zero. In formulae (advanced readers will recognise the definition of derivative with respect to the time of the energy function ${\displaystyle E=E(t)}$),

${\displaystyle P_{ist}=\underset{\mathrm{\Delta }t\to 0}{\lim }\frac{\mathrm{\Delta }E}{\mathrm{\Delta }t}=E^{\prime }(t)}$

(2)

How much sound energy does a receiver "capture"?

The energy irradiated from a sound source propagates, depending on the geometry of the source and the type of emission, along well-defined directions. The information regarding the directions of energy propagation can be well represented wave fronts, which follow the symmetry of the source. The simplest cases are:

• spherical wave fronts

in this case, we are referring to a free or spherical sound field. This situation occurs in the case of point sources or of dimensions that are much smaller than the emitted wavelength (see sound diffraction) and of wave propagation in a rotationally invariant medium free of obstacles (in other words, with physical properties that do not depend on the direction of propagation). In this case, if we calculate the flux of emitted energy (in other words, the energy that passes every second through a unit area at distance r from the source), we see that it decreases as ${\displaystyle 1/r^2}$ since the sound must cross surfaces which increase as the square of the distance. Let us recall that the surface of a sphere with radius r is

${\displaystyle S_{sfera}=4\pi r^2}$

If the source emits the power P uniformly in all directions, the energy flux at distance r, that is, the sound intensity, is therefore

${\displaystyle I=\frac{P}{4\pi r^2}}$

(3)

and is measured in W/m².

• cylindrical wave fronts

If the sound source has a cylindrical symmetry (for example, a long line of automobiles in traffic), we expect the flux of emitted energy to decrease as the inverse of the distance having to cross surfaces with an area that increases linearly with the distance from the source. Let us recall that the lateral surface of a cylinder with height h and radius r is

${\displaystyle S_{lat}=2\pi r\cdot h}$

• plane wave fronts

If the sound source is planar (theoretically, an infinite plane), the wave front will be such that the energy flux crossing a plane parallel to the source remains constant, provided we neglect energy losses in the medium.

Energy (and, therefore, its flux) being directly proportional to the square of the wave amplitude means that also the oscillation amplitude decreases as the wave travels away from the source, independently from the presence of energy absorption phenomena.

The rate at which sound energy flow decreases as a function of the distance from the source has important consequences:

• If the receiver (for example, our eardrum) has a constant area, the amount of acoustic energy that it can capture depends strongly on the distance: in open environments, the amplitude of sound waves emitted from sources of modest surface (for example, musical instruments) decays very rapidly (even in the absence of absorption phenomena).

• in open environments, sound waves emitted from point sources decay much more rapidly than those emitted from cylindrical symmetry or planar sources. For example, if there is a jackhammer (which can be likened to a point sound source) working on a road with a long line of automobiles with their motors running in traffic, at a certain distance from the street, we will hear the background noise of the car engines but not the jackhammer. As we approach the street, we will begin to hear the noise of the jackhammer emerge from the background noise of the car engines. The reason is simple: when we reduce our distance from the street (for example, by half) the sound energy we hear generated by the jackhammer will increase by four times while that of the cars by only two times.

• in closed environments (for example, an auditorium) the drastic reduction of sound energy with the increase in distance should strongly penalise the spectators in the last rows. Consider, for example, a piano soloist as a point source, a spectator 50 metres away from the stage would receive sound energy that is 100 times less than a spectator 5 metres away. This does not happen in reality. Why not? The fact is that in closed environments, sound energy is reflected off the walls: what we hear is not only direct sound coming from the stage but also reflected sound.

• The phenomenon of reflection can be exploited even in open environments to guarantee a significant contribution of reflected energy to the perceived sound field: in Roman theatres, a flat wall placed behind the stage guaranteed the reflection of a quasi plane wave (therefore slowly decaying with distance)

The phenomenon of absorption

All of the above reasoning was developed attributing the decrease of flux of sound energy to the "spreading" of energy over wave fronts with increasingly larger surfaces. In real situations, above all in open environments or very large closed environments, it is no longer possible to ignore the loss of sound energy due to its partial conversion into heat. This conversion is due partly to viscous forces and to heat conduction between the compressed (warmer) and rarefied (cooler) regions of the wave, but above all to relaxation processes of the oxygen and nitrogen molecules, which absorb energy from the wave and give it back with a delay, as heat. The attenuation of the flow of sound energy due to atmospheric absorption is quantified through an attenuation coefficient, usually expressed in dB per kilometre (see also the page on architectural acoustics). It depends:

1. on the humidity conditions of the air (water vapour changes the effectiveness of the molecular relaxation processes: for this reason, depending on the frequency, absorption is largest in rather dry air, with relative humidity between about 10% and 30%, and decreases in very humid air);
2. on the air temperature;
3. on the sound wave frequency. In particular, waves at higher frequencies are more easily absorbed. A spectacular consequence of this fact is the modification of the timbre of sound waves emitted by thunder when heard from far away. Its sound intensity is weak due to the attenuation produced by the distance discussed in the previous paragraph, but above all, it sounds like a "rumble" due do the very low frequencies it contains, which are the only survivors of atmospheric absorption.

Check your understanding

In-depth study and links

• On the page perception of loudness, you will find a more in-depth look at this topic, in particular how sound intensity corresponds to sound energy flux (which in the International System is measured in ${\displaystyle W/m^2}$).
• If you are interested in further study on sound perception in closed environments, visit the page on architectural acoustics.

• If you are interested in discovering how our auditory system matches impedances of its components in order to maximize the transmission of sound energy to nerve fibres, visit the page on the anatomy of the auditory system

• If you are interested in discovering how the sound energy wasted by air friction can, in part, be reconverted into energy for motors and refrigerators, visit the page on thermoacoustics.