Skip to main content
Chemistry LibreTexts

3.2: A Classical Wave Equation

  • Page ID
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\)

    The easiest way to find a differential equation that will provide wavefunctions as solutions is to start with a wavefunction and work backwards. We will consider a sine wave, take its first and second derivatives, and then examine the results. The amplitude of a sine wave can depend upon position, \(x\), in space,

    \[ A (x) = A_0 \sin \left ( \frac {2 \pi x}{\lambda} \right ) \label{1} \]

    or upon time, \(t\),

    \[A(t) = A_0\sin(2\pi \nu t) \label{2} \]

    or upon both space and time,

    \[ A (x, t) = A_0 \sin \left ( \frac {2 \pi x}{\lambda} - 2\pi \nu t \right ) \label {3}\]

    We can simplify the notation by using the definitions of a wave vector, \(k = \frac {2\pi}{\lambda}\), and the angular frequency, \(\omega = 2\pi \nu\) to get

    \[A(x,t) = A_0\sin(kx − \omega t) \label {4}\]

    When we take partial derivatives of A(x,t) with respect to both \(x\) and \(t\), we find that the second derivatives are remarkably simple and similar.

    \[ \frac {\partial ^2 A (x, t)}{\partial x^2} = -k^2 A_0 \sin (kx -\omega t ) = -k^2 A (x, t) \label {5}\]

    \[ \frac {\partial ^2 A (x, t)}{\partial t^2} = -\omega ^2 A_0 \sin (kx -\omega t ) = -\omega ^2 A (x, t) \label {6}\]

    By looking for relationships between the second derivatives, we find that both involve \(A(x,t)\); consequently an equality is revealed.

    \[ k^{-2} \frac {\partial ^2 A (x, t)}{ \partial x^2} = - A (x, t) = \omega^{-2} \frac {\partial ^2 A (x, t)}{\partial t^2} \label {7}\]

    Recall that \(\nu\) and \(λ\) are related; their product gives the velocity of the wave, \(\nu \lambda = v\). Be careful to distinguish between the similar but different symbols for frequency \(\nu\) and the velocity v. If in ω = 2πν we replace ν with v/λ, then

    \[ \omega = \frac {2 \pi \nu}{\lambda} = \nu k \label {8}\]

    and Equation \(\ref{7}\) can be rewritten to give what is known as the classical wave equation in one dimension. This equation is very important. It is a differential equation whose solution describes all waves in one dimension that move with a constant velocity (e.g. the vibrations of strings in musical instruments) and it can be generalized to three dimensions. The classical wave equation in one-dimension is

    \[\frac {\partial ^2 A (x, t)}{\partial x^2} = \nu ^{-2} \frac {\partial ^2 A (x, t)}{\partial t^2} \label {9}\]

    Example \(\PageIndex{1}\)

    Complete the steps leading from Equation \(\ref{3}\) to Equations \(\ref{5}\) and \(\ref{6}\) and then to Equation \(\ref{9}\).

    This page titled 3.2: A Classical Wave Equation is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by David M. Hanson, Erica Harvey, Robert Sweeney, Theresa Julia Zielinski via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.