## Pressure Variations

The sound wave travels in the form of successive compressions and rarefactions. The sound is better described in terms of its pressure variations than displacement variations. You'll see that the pressure variations of a sinusoidal sound wave is again sinusoidal.

To express the sinusoidal pressure variations of a sound wave, we consider a cylindrical element of gas of length $\Delta x$ and cross-sectional area $A'$ as shown in Figure 1. The left end of the cylinder is at $x$ and the right end is at $x+\Delta x$ in our coordinate system. As the sound wave given by the wave function $y = A \cos (kx - \omega t)$ passes through the cylinder, the left end undergoes a displacement of $y_1$ and the right end undergoes a displacement of $y_2$ at time $t$.

The volume of the gas decreases when $y_2 < y_1$ (compression) but increases when $y_2 > y_1$ (rarefaction). In our case consider that the volume decreases or the gas is compressed ($y_2 < y_1$) and the pressure increases. The change in volume is $\Delta V = (y_2 - y_1)A' = \Delta yA'$ and the initial volume is $V = \Delta x A'$. Therefore,

\[\frac{{\Delta V}}{V} = \frac{{\Delta yA'}}{{\Delta xA'}} = \frac{{\Delta y}}{{\Delta x}} \tag{1} \label{1}\]

You should not be confused with the cross-sectional area $A'$ and the amplitude $A$. Because we have considered a particular time $t$, it is constant. As $\Delta x$ approaches zero and you can find:

\[\frac{{dV}}{V} = \frac{{\partial y}}{{\partial x}} = - Ak\sin (kx - \omega t) \tag{2} \label{2}\]

The pressure variation $P$ is the excess pressure above equilibrium pressure which is positive in our case. You can find $P$ if $B$ is the bulk modulus of the gas using above equation as

\[\begin{align*} B&=-\frac{P}{\Delta V/V}=\frac{P}{Ak\sin (kx-\omega t)} \\ \text{or,}\quad P&=BAk\sin (kx-\omega t) \tag{3} \label{3}\\ \end{align*}\]

The above equation Eq. \eqref{3} shows that the pressure variation of a sinusoidal sound wave is sinusoidal. The pressure variation is the function of both position and time, so you can also write it as $P(x, t) = BAk \sin (kx - \omega t)$. The maximum value of $\sin$ function is $1$ and hence the maximum value of $P$ is $P_\text{max} = BAk$. The maximum pressure $P_\text{max}$ is called *pressure amplitude*.

## Sound Intensity

The intensity of a sound wave is the average rate of energy transfer per unit area perpendicular to the direction of wave propagation. You know the wave function of a sinusoidal sound wave is $y = A \cos (kx-\omega t)$ and the velocity of particles $v_y$ in the medium of a sound wave at a particular time $t$ is ${{v}_{y}}=\partial y/\partial t=A\sin (kx-\omega t)$.

The power of the wave is the product of force and velocity; the velocity does not mean the wave speed but it means the velocity of particles in the medium due to wave disturbance. The force is due to the pressure $P$ on area $A'$ perpendicular to the direction of wave propagation. So the power is $PA'v_y$ which is given by

\[\begin{align*} p &= PA'{v_y} = \{ BAA'k\sin (kx - \omega t)\} \{ A \omega \sin (kx - \omega t)\} \\ &= B{A^2}A'k \kern 1pt \omega \kern 1pt {\sin ^2}(kx - \omega t) \tag{4} \label{4} \end{align*}\]

The value of $\sin^2$ function oscillates between $0$ and $1$, so the average value is $1/2$. Therefore, the average power using Eq. \eqref{4} is

\[{p_{av}}{\kern 1pt} = \frac{1}{2}B{A^2}A'k\omega \tag{5} \label{5}\]

Now the intensity $I$ of the sinusoidal sound wave is

\[I = \frac{{{p_{av}}}}{{A'}} = \frac{1}{2}B{A^2}k\omega \tag{6} \label{6}\]

You know $B = \rho v^2$ from the article of wave speeds (you may need to read this article) and $v = \omega /k$ from this article about the simple harmonic wave function (you may also need to read this article) and you can rewrite the above equation as

\[I = \frac{1}{2}\rho v{A^2}{\omega ^2} \tag{7} \label{7}\]

The above equation shows that two sound waves of different frequencies should have different amplitudes for the same wave intensity; it means the two sound waves of the same intensity but different frequencies have different amplitudes. The pressure amplitude is $P_\text{max} = BkA = \rho v \omega A$, so the above equation in terms of pressure amplitude can be written as

\[I = \frac{{P_{\max }^2}}{{2\rho v}} \tag{8} \label{8}\]

The above equation shows that the two sound waves having the same sound intensity but different frequencies have the same pressure amplitude.