# Engineering Acoustics/Acoustic wave equation

To derive the wave equation, three relations are combined: the equation
of state, ideal gas law; continuity equation, conservation of mass; and
Newton\'s law, conservation of momentum. From the speed of sound, a
relation between the acoustic pressure and the bulk modulus can be
derived,

$c_o^2=\frac{\partial P}{\partial \rho} = \frac{p'}{\varrho'}= \frac{p'}{\rho-\rho_o}.$

Using the definition of the condensation, relative change of density in
a fluid, denoted by $s$,

$\frac{p'}{\rho_o}= c_o^2 \Big( \frac{\rho-\rho_o}{\rho_o} \Big) = c_o^2s,$

and introducing the bulk modulus which makes implicit use of the ideal
gas law,

$p'= \rho_o c_o^2 s = \gamma P_o s = B_o s.$

The general form of the continuity equation for a control volume from
fluid dynamics is simplified to its one-dimensional form in Cartesian
coordinates.

$\frac{\partial \rho}{\partial t} + \vec{\nabla} (\rho \vec{U}) =0$

$\frac{\partial \rho}{\partial t} + \rho_o \frac{\partial U}{\partial x} =0$

$\frac{\partial U}{\partial x} = \frac{-1}{\rho_o} \frac{\partial \rho}{\partial t} = \frac{-1}{\rho_o c_o^2} \frac{\partial P}{\partial t}$
!Pressure on fluid element inside a
tube\|left\|440x440px

Using Newton\'s law on the fluid element, the net force acting on the
fluid boundaries causes an acceleration on the fluid proportional to its
mass,

$A (P(x)-P(x+dx))= (A dx \rho_o) \frac{dU}{dt}.$

Noticing that

$\frac{P(x)-P(x+dx)}{dx} = \frac{\partial P}{\partial x}$

as $dx$ approaches zero, evaluating the derivative, and neglecting small
terms,

$\frac{\partial P}{\partial x} dx= (\rho_o dx) \frac{dU}{dt} = (\rho_o dx) \Big( \frac{\partial U}{\partial t} + \frac{\partial U}{\partial x} \frac{dx}{dt}\Big)$

$\frac{\partial U}{\partial t}= \frac{-1}{\rho_o}\frac{\partial P}{\partial x}.$

To obtain the wave equation, the partial derivative with respect to time
is taken for the continuity equation and  the partial derivative with
respect to space for the conservation of momentum equation. By equating
the two results and eliminating the density term on both sides,

$\frac{\partial^2 U}{\partial t \partial x}= \frac{-1}{\rho_o c_o^2} \frac{\partial^2 P}{\partial t^2}= \frac{-1}{\rho_o}\frac{\partial^2 P}{\partial x^2}$

the acoustic wave equation is recovered

$\frac{\partial^2 P}{\partial x^2}= \frac{1}{c_o^2} \frac{\partial^2 P}{\partial t^2}$

$\frac{\partial^2 P}{\partial x^2}- \frac{1}{c_o^2} \frac{\partial^2 P}{\partial t^2}=0.$

The equation above is the acoustic wave equation in its one-dimensional
form. It can be generalized to 3-D Cartesian
coordinates$\frac{\partial^2 P}{\partial x^2}+\frac{\partial^2 P}{\partial y^2}+\frac{\partial^2 P}{\partial z^2}- \frac{1}{c_o^2} \frac{\partial^2 P}{\partial t^2}=0.$

Using the Laplace operator, it can be generalized to other coordinate
systems

$\nabla^2 P- \frac{1}{c_o^2} \frac{\partial^2 P}{\partial t^2}=0$

### One-dimensional wave solution in Cartesian coordinates

The one-dimensional acoustic wave equation is described by the second
order partial differential equation,

$\frac{\partial^2 P}{\partial x^2}= \frac{1}{c_o^2} \frac{\partial^2 P}{\partial t^2}.$

It can be solved using separation of variables. Suppose the pressure is
the product of one function only dependent on space and another function
only dependent on time,

$P(x,t)=X(x)T(t).$

Substituting back into the wave equation,

$X''T=\frac{1}{c_o^2}XT''$

$\frac{X''}{X}=\frac{1}{c_o^2}\frac{T''}{T}=-\lambda^2.$

This substitution leads to two homogeneous second order ordinary
differential equations, one in time and one in space.

$T'' + c_o^2 \lambda^2 T=0$

$X'' + \lambda^2 X=0$

The time function is expected to be dependent on the angular frequency
of the wave

$T=Ce^{j \omega t}.$

Substituting and solving for the constant which is define as the wave
number, $K$,

$-\omega^2 + c_o^2 \lambda^2=0$

$K^2=\lambda^2=\frac{\omega^2}{c_o^2}.$

The wave number relates the angular velocity of the wave to its
propagation speed in the medium. It can be expressed in different forms,

$K=\frac{\omega}{c_o}=\frac{2 \pi f}{c_o}=\frac{2 \pi}{\lambda},$

where $f$ is the frequency in hertz and $\lambda$ is the wavelength.The
second differential equation can be solved using the wave number. The
spacial function is given a general form,

$X=Ce^{jrx}.$

Substituting and solving for $r$,

$-r^2+K^2=0$

$r=\pm K.$

The solution of the 1-D acoustic wave equation is obtained,

$P(x,t)=(C_1e^{j K x} + C_2e^{-j K x})e^{j \omega t}.$

The real and imaginary parts of the solution are also solutions to the
1-D wave equation.

$P(x,t)=C_1cos(\omega t + Kx) + C_2cos(\omega t - Kx)$

$P(x,t)=C_1sin(\omega t + Kx) + C_2sin(\omega t - Kx)$

Using phasor notation, the solution is written in more compact form,

$\mathbf{P(x,t)}=\mathbf{P}e^{j(\omega t \pm Kx)}.$

The actual solution is recovered by taking the real part of the above
complex form. The value of the constants above is determined by applying
initial and boundary conditions. In general, any function of the
following form is a solution for periodic waves,

$P(x,t)=f_1(\omega t + Kx)+f_2(\omega t - Kx),$

and similarly, for progressive waves,

$P(x,t)=f(ct + x)+g(ct - x)$

$P(x,t)=f(\xi)+g(\eta),$

where $f$ and $g$ are arbitrary functions, that represent two waves
traveling in opposing directions. These are known as the d\'Alembert
solutions. The form of these two functions can be found by applying
initial and boundary conditions.

Original document[^1]


[^1]: 1
