Example: Harmonic Function Satisfying the Wave Equation

In this post I’m going to work through the math showing that a simple harmonic wave profile/function satisfies the one-dimensional wave equation. The simple harmonic function is given in the following equation. The variable $v$ is the wave speed in the $x$ direction, while $\psi$ is the shape of the profile of the wave.

(1) $\begin{equation*} \psi (x,t) = A\sin[k (x-vt)] = A\sin(kx-kvt) \end{equation*}$

The 1D wave equation that we wish to satisfy is given below.

(2) $\begin{equation*} \frac{\partial ^2 \psi}{\partial x^2} = \frac{1}{v^2} \frac{\partial ^2 \psi}{\partial t^2} \end{equation*}$

To see whether the wave function satisfies the wave equation, we only need to take a couple of partial derivatives. Let’s start with the left side of the equation. The first step is to take the partial derivative of $\psi$ with respect to $x$ while holding $t$ constant.

(3) $\begin{equation*} \frac{\partial \psi}{\partial x} = A \cos (kx-kvt) k = Ak\cos(kx-kvt) \end{equation*}$

Now let’s take another partial derivative with respect to $x$ while holding $t$ constant again.

(4) $\begin{equation*} \frac{\partial^2 \psi}{\partial x^2} = -Ak\sin(kx-kvt)k = -Ak^2\sin(kx-kvt) <span id="d316603af0">To be on purchase viagra online <a href="http://greyandgrey.com/personal-injury-1/">http://greyandgrey.com/personal-injury-1/</a> the safer side, look for generic medicines. Hefner, all of these great artists were confined to the underground <a href="http://greyandgrey.com/wp-content/uploads/2018/07/NYSBA2-June-2000.pdf">Full Article</a> discount levitra of American society. <a href="http://greyandgrey.com/wp-content/uploads/2018/07/Spector.pdf">sildenafil 100mg canada</a> And yes, every foreigner doesn't like how Chinese normally function in their society. It's time that ICC rewards them with the Test <a href="http://greyandgrey.com/wp-content/uploads/2020/07/The-Experience-of-Immigrants-and-Low-Wage-Workers-in-The-New-York-State-Workers-Compensation-.pdf">cialis online order</a> Status. </span>\end{equation*}$

That’s it for the left hand side of the equation. Now we can move to the right hand side, and take a derivative of $\psi$ with respect to $t$ while holding $x$ constant.

(5) $\begin{equation*} \frac{\partial \psi}{\partial t} = A\cos(kx-kvt)(-kv) = -Akv\cos(kx-kvt \end{equation*}$

Now we can take another partial derivative with respect to $t$ while holding $x$ constant.

(6) $\begin{equation*} \frac{\partial^2 \psi}{\partial t^2} = Akv\sin(kx-kvt)(-kv) = -Ak^2v^2\sin(kx-kvt) \end{equation*}$

We now have all the terms we need to plug into the wave equation, as seen below.

(7) $\begin{equation*} -Ak^2\sin(kx-kvt) = \frac{1}{v^2}[-Ak^2v^2\sin(kx-kvt)] \end{equation*}$

The last step is to cancel the $v^2$ term.

(8) $\begin{equation*} -Ak^2\sin(kx-kvt) = -Ak^2\sin(kx-kvt) \end{equation*}$

We can see that the left hand side is equal to the right hand side, indicating that our wave function does indeed satisfy the wave equation.

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