The wave equation
The physics of the phenomenon of sound waves thus involves three features: Let us consider II first. For a gas, a liquid, or a solid, the pressure is some function of the density. Before the sound wave arrives, we have equilibrium, with a pressure $P_0$ and a corresponding density $\rho_0$. A pressure $P$ in the medium is connected to the density by some characteristic relation $P = f(\rho)$ and, in particular, the equilibrium pressure $P_0$ is given by $P_0 = f(\rho_0)$. The changes of pressure in sound from the equilibrium value are extremely small. A convenient unit for measuring pressure is the bar, where $1$ bar${}= 10^5$ N/m². The pressure of $1$ standard atmosphere is very nearly $1$ bar: $1$ atm${} = 1.0133$ bars. In sound we use a logarithmic scale of intensities since the sensitivity of the ear is roughly logarithmic. This scale is the decibel scale, in which the acoustic pressure level for the pressure amplitude $P$ is defined as  \begin{equation} \label{Eq:I:47:1} I\text{ (acoustic pressure level)} = 20\log_{10}(P/P_{\text{ref}})\text{ in dB}, \end{equation} 
 \begin{equation} \label{Eq:I:47:1} I \begin{pmatrix} \text{acoustic}\\[-.75ex] \text{pressure}\\[-.75ex] \text{level} \end{pmatrix} = 20\log_{10}(P/P_{\text{ref}})\text{ in dB}, \end{equation}  where the reference pressure $P_{\text{ref}} = 2\times10^{-10}$ bar. A pressure amplitude of $P =$ $10^3P_{\text{ref}} =$ $2\times10^{-7}$ bar1 corresponds to a moderately intense sound of $60$ decibels. We see that the pressure changes in sound are extremely small compared with the equilibrium, or mean, pressure of $1$ atm. The displacements and the density changes are correspondingly extremely small. In explosions we do not have such small changes; the excess pressures produced can be greater than $1$ atm. These large pressure changes lead to new effects which we shall consider later. In sound we do not often consider acoustic intensity levels over $100$ dB; $120$ dB is a level which is painful to the ear. Therefore, for sound, if we write \begin{equation} \label{Eq:I:47:2} P = P_0 + P_e,\quad \rho = \rho_0 + \rho_e, \end{equation} we shall always have the pressure change $P_e$ very small compared with $P_0$ and the density change $\rho_e$ very small compared with $\rho_0$. Then \begin{equation} \label{Eq:I:47:3} P_0 + P_e = f(\rho_0 + \rho_e) = f(\rho_0) +\rho_ef'(\rho_0), \end{equation} where $P_0 = f(\rho_0)$ and $f'(\rho_0)$ stands for the derivative of $f(\rho)$ evaluated at $\rho = \rho_0$. We can take the second step in this equality only because $\rho_e$ is very small. We find in this way that the excess pressure $P_e$ is proportional to the excess density $\rho_e$, and we may call the proportionality factor $\kappa$:  \begin{equation} \label{Eq:I:47:4} P_e = \kappa\rho_e, \,\text{where }\kappa = f'(\rho_0) = (dP/d\rho)_0.\quad\text{(II)} \end{equation} 
 \begin{equation} \label{Eq:I:47:4} P_e = \kappa\rho_e, \,\text{where }\kappa = f'(\rho_0) = (dP/d\rho)_0.\;\text{(II)} \end{equation}  The relation we needed for II is this very simple one. Let us now consider I. We shall suppose that the position of a portion of air undisturbed by the sound wave is $x$ and the displacement at the time $t$ due to the sound is $\chi(x,t)$, so that its new position is $x + \chi(x,t)$, as in Fig. 47–3. Now the undisturbed position of a nearby portion of air is $x + \Delta x$, and its new position is $x + \Delta x + \chi(x + \Delta x,t)$. We can now find the density changes in the following way. Since we are limiting ourselves to plane waves, we can take a unit area perpendicular to the $x$-direction, which is the direction of propagation of the sound wave. The amount of air, per unit area, in $\Delta x$ is then $\rho_0\,\Delta x$, where $\rho_0$ is the undisturbed, or equilibrium, air density. This air, when displaced by the sound wave, now lies between $x + \chi(x,t)$ and $x + \Delta x + \chi(x + \Delta x,t)$, so that we have the same matter in this interval that was in $\Delta x$ when undisturbed. If $\rho$ is the new density, then  \begin{equation} \label{Eq:I:47:5} \rho_0\,\Delta x = \rho[x + \Delta x + \chi(x + \Delta x,t) - x - \chi(x,t)]. \end{equation} 
 \begin{gather} \label{Eq:I:47:5} \rho_0\,\Delta x =\\[.5ex] \rho[x + \Delta x + \chi(x + \Delta x,t) - x - \chi(x,t)].\notag \end{gather}  Since $\Delta x$ is small, we can write $\chi(x + \Delta x,t) - \chi(x,t) = (\ddpl{\chi}{x})\,\Delta x$. This derivative is a partial derivative, since $\chi$ depends on the time as well as on $x$. Our equation then is \begin{equation} \label{Eq:I:47:6} \rho_0\,\Delta x = \rho\biggl(\ddp{\chi}{x}\,\Delta x + \Delta x\biggr) \end{equation} or \begin{equation} \label{Eq:I:47:7} \rho_0 = (\rho_0 + \rho_e)\ddp{\chi}{x} + \rho_0 + \rho_e. \end{equation} Now in sound waves all changes are small so that $\rho_e$ is small, $\chi$ is small, and $\ddpl{\chi}{x}$ is also small. Therefore in the relation that we have just found, \begin{equation} \label{Eq:I:47:8} \rho_e = -\rho_0\,\ddp{\chi}{x} -\rho_e\,\ddp{\chi}{x}, \end{equation} we can neglect $\rho_e\,\ddpl{\chi}{x}$ compared with $\rho_0\,\ddpl{\chi}{x}$. Thus we get the relation we needed for I: \begin{equation} \label{Eq:I:47:9} \rho_e = -\rho_0\,\ddp{\chi}{x}.\quad\text{(I)} \end{equation}  This equation is what we would expect physically. If the displacements vary with $x$, then there will be density changes. The sign is also right: if the displacement $\chi$ increases with $x$, so that the air is stretched out, the density must go down. We now need the third equation, which is the equation of the motion produced by the pressure. If we know the relation between the force and the pressure, we can then get the equation of motion. If we take a thin slab of air of length $\Delta x$ and of unit area perpendicular to $x$, then the mass of air in this slab is $\rho_0\,\Delta x$ and it has the acceleration $\partial^2\chi/\partial t^2$, so the mass times the acceleration for this slab of matter is $\rho_0\,\Delta x(\partial^2\chi/\partial t^2)$. (It makes no difference for small $\Delta x$ whether the acceleration $\partial^2\chi/\partial t^2$ is evaluated at an edge of the slab or at some intermediate position.) If now we find the force on this matter for a unit area perpendicular to $x$, it will then be equal to $\rho_0\,\Delta x(\partial^2\chi/\partial t^2)$. We have the force in the $+x$-direction, at $x$, of amount $P(x,t)$ per unit area, and we have the force in the opposite direction, at $x + \Delta x$, of amount $P(x + \Delta x,t)$ per unit area (Fig. 47–4):  \begin{equation} \label{Eq:I:47:10} P(x,t)\!-\!P(x + \Delta x,t) = -\ddp{P}{x}\,\Delta x = -\ddp{P_e}{x}\,\Delta x, \end{equation} 
 \begin{align} P(x,t)\!-\!P(x + \Delta x,t) &= -\ddp{P}{x}\,\Delta x\notag\\[1.5ex] \label{Eq:I:47:10} &= -\ddp{P_e}{x}\,\Delta x, \end{align}    since $\Delta x$ is small and since the only part of $P$ which changes is the excess pressure $P_e$. We now have III: \begin{equation} \label{Eq:I:47:11} \rho_0\,\frac{\partial^2\chi}{\partial t^2} = -\ddp{P_e}{x},\quad\text{(III)} \end{equation} and so we have enough equations to interconnect things and reduce down to one variable, say to $\chi$. We can eliminate $P_e$ from III by using II, so that we get \begin{equation} \label{Eq:I:47:12} \rho_0\,\frac{\partial^2\chi}{\partial t^2} = -\kappa\,\ddp{\rho_e}{x}, \end{equation} and then we can use I to eliminate $\rho_e$. In this way we find that $\rho_0$ cancels out and that we are left with \begin{equation} \label{Eq:I:47:13} \frac{\partial^2\chi}{\partial t^2} = \kappa\,\frac{\partial^2\chi}{\partial x^2}. \end{equation} We shall call $c_s^2 = \kappa$, so that we can write \begin{equation} \label{Eq:I:47:14} \frac{\partial^2\chi}{\partial x^2} = \frac{1}{c_s^2}\,\frac{\partial^2\chi}{\partial t^2}. \end{equation} This is the wave equation which describes the behavior of sound in matter.