Differences between waves in solids, liquids and gases

• Sound waves are longitudinal elastic waves that propagate in an elastic medium (be it liquid, solid or gaseous) at an audible frequency (that is, conventionally between 20 and 20000 Hz).
• Therefore, to a first approximation, the equation for longitudinal waves that we derived, for solids, for the elastic bar also describes sound waves.

However, we must take into account some differences between liquids, solids and gases:

1. While in a liquid or a solid the density changes produced by the pressures involved in sound waves are very small (but not zero: if these media were really incompressible, sound could not propagate in them), in a gas the compression that accompanies a sound wave normally produces a non-negligible decrease in volume (that is, an increase in density).
2. Moreover, in general, a rapid compression of a gas can cause a substantial increase in its temperature, and this has an effect on the speed of sound (see speed of sound in media).

To write the wave equation for sound waves we must therefore apply, in addition to the usual Newton's equation of motion, an equation of state for the gas as well. It relates the applied pressure, the density and the temperature of the gas.

Equation of state, and elastic waves in gases

What we can say in general

We first write the equation of motion of a small volume of air ${\displaystyle dV=Adx}$, with density at rest ${\displaystyle {\rho }_0}$, which is compressed only in the x direction (in the other directions we can imagine that there are rigid walls). As a result of the compression, the left wall moves from position x to position ${\displaystyle x+\xi }$; the cross-section A stays constant, while the length of the volume dx becomes ${\displaystyle dx+d\xi }$.

• The pressure difference across the two faces of the small volume is initially

${\displaystyle dp=\frac{\partial p(x)}{\partial x}dx}$

• so the net force acting on the small volume in the x direction is

${\displaystyle dF=-Adp=-Adx\frac{\partial p(x)}{\partial x}}$

(the sign is negative because a positive dp corresponds to a force directed towards negative x).

• By Newton's equation we must equate this force to the mass of the small volume times its acceleration, that is,

${\displaystyle ma={\rho }_{0}Adx\frac{{\partial }^2\xi }{\partial t^2}}$

• Dividing both equations by the volume ${\displaystyle dV=Adx}$ we find

${\displaystyle -\frac{\partial p}{\partial x}={\rho }_0\frac{{\partial }^2\xi }{\partial t^2}}$

(1)

• The left-hand side can be expanded using

${\displaystyle \frac{\partial p}{\partial x}=\frac{dp}{d\rho }\frac{\partial \rho }{\partial x}}$.

Now we use the fact that, during the compression, the mass of the elementary volume we are considering does not change. This translates into

${\displaystyle {\rho }_{0}dx=\rho (dx+d\xi )}$,

that is

${\displaystyle \rho ={\rho }_0{(1+\frac{\partial \xi }{\partial x})}^{-1}}$,

(2)

which gives us

${\displaystyle \frac{\partial \rho }{\partial x}=-{\rho }_0\frac{{\partial }^2\xi }{\partial x^2}}$

• Our equation of motion finally becomes

${\displaystyle \frac{dp}{d\rho }\frac{{\partial }^2\xi }{\partial x^2}=\frac{{\partial }^2\xi }{\partial t^2}}$

(3)

• Now, however, we have an equation in which both the pressure and the density of the gas appear. It therefore cannot be solved until we find a relation between these two variables. Note, however, that if ${\displaystyle \frac{dp}{d\rho }}$ were constant, the equation would reduce to the simple linear wave equation. This is almost always the case when the deformation ${\displaystyle \partial \xi /\partial x}$ is small. Let us now see in detail how a compressed ideal gas behaves.

How does a compressed gas behave?

• If we slowly compress the air contained in a rigid cylinder by means of a piston, we notice that it opposes the compression like a spring. That is, the air reacts with an elastic force

${\displaystyle \mathrm{\Delta }P=-K\frac{\mathrm{\Delta }V}{V}}$,

where ${\displaystyle \mathrm{\Delta }V/V}$ is the relative change in volume, ${\displaystyle \mathrm{\Delta }P}$ the change in pressure inside the cylinder, and K the bulk modulus of the gas, which for a slow compression is equal to the pressure P of the gas.

• If instead the compression takes place rapidly, we find the relation

${\displaystyle \mathrm{\Delta }P=-1.4K\frac{\mathrm{\Delta }V}{V}}$

that is, the air behaves like a stiffer spring.

To understand this difference we must remember that in the two cases the compression of the gas takes place in different thermodynamic ways, namely:

• if the compression is slow, the work done on the gas has time to turn into heat and to pass from the gas to the walls of the container. The heat flow keeps the temperature of the gas constant. We therefore speak of isothermal compression.
• If the compression is rapid, the work done on the gas does not have enough time to be transferred to the container as heat, and remains in the gas as kinetic energy of its molecules. We speak of adiabatic compression.

In which regime does sound propagate?

• The question is: should a succession of compressions occurring many times per second, corresponding to the frequencies of audible sound waves, be considered fast or slow compared with the propagation of heat?
• More formally: is the propagation of sound waves isothermal or adiabatic?

To answer, we can compare the propagation speed of sound with the speed of heat conduction in air. The former is about 340 ms^-1, while the latter is between about 0.5 ms^-1 (at 1000 Hz) and 1.5 ms^-1 (at 10000 Hz).

Heat, in fact, does not propagate as a wave but by diffusion: a temperature change oscillating with angular frequency ω penetrates into the air with a speed of about ${\displaystyle \sqrt{2D\omega }}$, where D ≈ 2·10⁻⁵ m²/s is the thermal diffusivity of air. In one period heat therefore travels about half a millimetre at 1000 Hz, while the distance between a compression (warmer) and the neighbouring rarefaction (cooler) is half a wavelength, that is about 17 cm at 1000 Hz: heat does not have time to pass from the compressed to the rarefied regions.

We can conclude that the equation of state we must use for sound waves is the adiabatic equation:

${\displaystyle P{V}^{\gamma }=\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{s}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{t}}$,

The factor ${\displaystyle \gamma }$ takes the value 1.4 for air (as for all diatomic gases).

Using the adiabatic equation

We rewrite this equation in the form most convenient for us, still referring to our infinitesimal volume

${\displaystyle p=k{\rho }^{\gamma }}$,

where k is a constant; this gives us the expression we want to make explicit in the equation of motion (1), which we repeat for clarity:

${\displaystyle -\frac{\partial p}{\partial x}={\rho }_0\frac{{\partial }^2\xi }{\partial t^2}}$

Now using relation (2), which links ${\displaystyle \rho }$ and ${\displaystyle {\rho }_0}$, we have

${\displaystyle -\frac{\partial p}{\partial x}=k\gamma {\rho }_0^{\gamma }{(1+\frac{\partial \xi }{\partial x})}^{-(\gamma +1)}\frac{{\partial }^2\xi }{\partial x^2}}$.

Wave equation in the adiabatic regime

• Let us return to our equation of motion, and now use the equation of adiabatic transformations in the form just derived.

• We obtain a rather complex equation

${\displaystyle \frac{\gamma p_0}{{(1+\frac{\partial \xi }{\partial x})}^{\gamma +1}}\frac{{\partial }^2\xi }{\partial x^2}={\rho }_0\frac{{\partial }^2\xi }{\partial t^2}}$.

• This equation reduces to the usual wave equation when

${\displaystyle \frac{\partial \xi }{\partial x}\ll 1}$

that is, in the case of small perturbations, in which case the equation is linearised into

${\displaystyle \frac{{\partial }^2\xi }{\partial x^2}=\frac{{\rho }_0}{\gamma p_0}\frac{{\partial }^2\xi }{\partial t^2}}$

where we see that the speed of sound is related to the pressure and density of the gas at rest by the relation

${\displaystyle v=\sqrt{\frac{\gamma p_0}{{\rho }_0}}}$

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