Themes > Science > Physics > Acoustics > Sound waves > Wave equation > Wave equation

In the last lecture we concluded our discussion of the heat equation.


TODAY
We move on to the wave equation
  • We will use sound propagation in a gas such as air as an example.
  • Sound waves are elastic waves that propagate in a fluid or solid.
  • In fluids long wavelength sound waves consist of an alternating pattern of rarefaction and compression.
  • In a solid transverse waves can also propagate.
  • In ordinary sound the changes in pressure tend to be very small.

The intensity of a sound wave is often measured in decibels

\begin{displaymath}I_{dB}=20 \log_{10}({{P_e}\over{P_{ref}}}) \end{displaymath}

The reference pressure is

\begin{displaymath}P_{ref}=2\times 10^{-10}bar \end{displaymath}

and Pe is the root mean square pressure amplitude $(1/\sqrt{2}$ of the peak excess pressure). Even at the pain level of 120 dB the peak excess pressure will be


\begin{displaymath}2\sqrt{2}\times10^{-4} bar\end{displaymath}

which is small compared to the ambient pressure $\simeq 1bar.$

The mechanism for a sound wave is that gas motion generates a change in the density, which causes a change in pressure. There will then be an unbalanced force which accelerates the gas and causes the cycle to repeat.

We write for the pressure and density

\begin{displaymath}\begin{array}{rl} P&=P_a+P_e\\ \rho &=\rho _a+\rho _e \end{array}\end{displaymath}

where the subscript a stands for average while e stands for excess.

The relationship between changes in density and pressure depends on the properties of the medium in which the sound waves propagate. We will assume that the processes are fast enough that they can be considered to be adiabatic, i.e. without any heat transport. Since the amplitudes Pe and $\rho_e$ are small compared to the ambient conditions we assume that they are proportional to each other

\begin{displaymath}P_e=\kappa \rho _e \end{displaymath}

with

\begin{displaymath}\kappa =\left.{{\partial P}\over{\partial \rho }}\right\vert _{adiabatic} \end{displaymath}



Let us consider a column of cross-section A.

\begin{figure} \epsfysize=150pt \epsffile{airvo.eps} \end{figure}


When the air is at rest in equilibrium this column extends from x to x+dx. We assume that a sound wave is traveling in the x-direction and that at some instant the left end of the column is displaced an amount $\xi (x,t)$ which is small compared to the wavelength of the sound wave.

Conservation of mass gives
\begin{displaymath}A\rho _adx=A(\rho _a+\rho _e)(d\xi +dx) \end{displaymath}
or
\begin{displaymath}\rho _edx=-(\rho _a+\rho _e)d\xi \simeq -\rho _ad\xi \end{displaymath}

since $\rho _e<<\rho _a,$ giving

\begin{displaymath}\rho _e\simeq -\rho _a{{\partial \xi }\over{\partial x}} \end{displaymath}

The column is subject to a net force

\begin{displaymath}A[P(x+\xi ,t)-P(x+\xi +dx+d\xi ,t)]\approx -A{{\partial P_e(x,t)}\over{\partial x}}dx \end{displaymath}

causing an acceleration


\begin{displaymath}Adx \rho _a{{\partial ^2\xi }\over{\partial t^2}} =-A\kapp... ...}}=A\kappa dx{{\partial ^2\xi }\over{\partial x^2}}\rho _a \end{displaymath}

We get the wave equation

\fbox{\parbox{5cm}{ \begin{displaymath} {{\partial ^2\xi }\over{\partial t^2}}=\kappa {{\partial ^2\xi }\over{\partial x^2}} \end{displaymath} }}

For an ideal gas assuming an adiabatic process

\begin{displaymath}PV^\gamma =constant \end{displaymath}

with $\gamma =C_P/C_V$ or

\begin{displaymath}P=Const.\rho ^\gamma \end{displaymath}

Differentiating we find

 

\begin{displaymath}\left.{{\partial P}\over{\partial \rho }}\right\vert _{adiaba... ...er{\rho }}={{\gamma k_BT}\over{m}}={{\gamma RT}\over{ \mu }} \end{displaymath}

where m is the mass of a molecule and $\mu $ the molecular weight. We finally get

\begin{displaymath}c=\sqrt{{{\gamma k_BT}\over{m}}}=\sqrt{{{\gamma RT}\over{ \mu }}} \end{displaymath}

It is interesting to compare the speed of sound with typical molecular speeds. From the thermodynamics of ideal gases we have for the rms speed


\begin{displaymath}v_{rms}=\sqrt{{{3k_BT}\over{m}}}=\sqrt{{{3}\over{\gamma }}}c \end{displaymath}

Since $\gamma \simeq 1.4$ for air we see that the rms speed and the sound speed are quite comparable.

By substituting into the wave equation we see that it admits solutions in the form of traveling waves $\psi(x\pm ct),$ where $c=\sqrt{\kappa }$ It can be shown (see Powers sect 3.1)that the general solution to the wave equations can be written

\fbox{\parbox{10cm}{ \begin{displaymath}\xi (x,t)=f_1(x-ct)+f_2(x+ct) \end{displaymath} }}

  • The same equation that we derived for sound waves also holds for a vibrating string (see Powers section 3.1)
  • The speed of the wave is then
    \begin{displaymath}c=\sqrt{T/\rho}\end{displaymath}

    where T is the tension and $\rho$ is the mass/unit length of the string.

  • Often one is not interested in the details of the solution but only in knowing the frequencies which can be excited. One is then led to an eigenvalue problem which is similar to what we did for the heat equation.


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