Wave Equation/Examples/Harmonic Wave

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Examples of Use of the Wave Equation

Let $\phi$ be a harmonic wave which is propagated along the $x$-axis in the positive direction with constant velocity $c$ and without change of shape.

From Equation of Harmonic Wave, the disturbance of $\phi$ at point $x$ and time $t$ can be expressed using the equation:

$(1): \quad \map \phi {x, t} = a \map \cos {2 \pi \paren {\dfrac x \lambda - \dfrac t \tau} }$

where:

$x$ denotes the distance from the origin along the $x$-axis
$t$ denotes the time
$\lambda$ is the wavelength of $\phi$
$\tau$ is the period of $\phi$.

This equation satisfies the wave equation.


Proof

The wave equation is expressible as:

$\dfrac 1 {c^2} \dfrac {\partial^2 \phi} {\partial t^2} = \dfrac {\partial^2 \phi} {\partial x^2} + \dfrac {\partial^2 \phi} {\partial y^2} + \dfrac {\partial^2 \phi} {\partial z^2}$


We have by partial differentiation:

\(\ds \map {\dfrac \partial {\partial t} } {\map \phi {x, t} }\) \(=\) \(\ds \map {\dfrac \partial {\partial t} } {a \map \cos {2 \pi \paren {\dfrac x \lambda - \dfrac t \tau} } }\)
\(\ds \) \(=\) \(\ds \dfrac {2 \pi a} \tau \map \sin {2 \pi \paren {\dfrac x \lambda - \dfrac t \tau} }\) Derivative of Cosine Function
\(\ds \leadsto \ \ \) \(\ds \map {\dfrac {\partial^2} {\partial t^2} } {\map \phi {x, t} }\) \(=\) \(\ds \map {\dfrac {\partial^2} {\partial t^2} } {a \map \cos {2 \pi \paren {\dfrac x \lambda - \dfrac t \tau} } }\)
\(\ds \) \(=\) \(\ds \map {\dfrac \partial {\partial t} } {\dfrac {2 \pi a} \tau \map \sin {2 \pi \paren {\dfrac x \lambda - \dfrac t \tau} } }\)
\(\ds \) \(=\) \(\ds -\dfrac {4 \pi^2 a} {\tau^2} \map \cos {2 \pi \paren {\dfrac x \lambda - \dfrac t \tau} }\) Derivative of Sine Function
\(\ds \) \(=\) \(\ds -\dfrac {4 c^2 \pi^2 a} {\lambda^2} \map \cos {2 \pi \paren {\dfrac x \lambda - \dfrac t \tau} }\) Period of Harmonic Wave

and:

\(\ds \map {\dfrac \partial {\partial x} } {\map \phi {x, t} }\) \(=\) \(\ds \map {\dfrac \partial {\partial x} } {a \map \cos {2 \pi \paren {\dfrac x \lambda - \dfrac t \tau} } }\)
\(\ds \) \(=\) \(\ds -\dfrac {2 \pi a} \lambda \map \sin {2 \pi \paren {\dfrac x \lambda - \dfrac t \tau} }\) Derivative of Cosine Function
\(\ds \leadsto \ \ \) \(\ds \map {\dfrac {\partial^2} {\partial x^2} } {\map \phi {x, t} }\) \(=\) \(\ds \map {\dfrac {\partial^2} {\partial x^2} } {a \map \cos {2 \pi \paren {\dfrac x \lambda - \dfrac t \tau} } }\)
\(\ds \) \(=\) \(\ds \map {\dfrac \partial {\partial x} } {-\dfrac {2 \pi a} \lambda \map \sin {2 \pi \paren {\dfrac x \lambda - \dfrac t \tau} } }\)
\(\ds \) \(=\) \(\ds -\dfrac {4 \pi^2 a} {\lambda^2} \map \cos {2 \pi \paren {\dfrac x \lambda - \dfrac t \tau} }\)


As $y$ and $z$ do not appear in $(1)$, the partial derivative of $(1)$ with respect to $y$ and $z$ is identically zero.


Hence we have:

$\dfrac {\partial^2 \phi} {\partial t^2} = c^2 \paren {\dfrac {\partial^2 \phi} {\partial x^2} + \dfrac {\partial^2 \phi} {\partial y^2} + \dfrac {\partial^2 \phi} {\partial z^2} }$

and the result follows.


Sources

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