Simple Harmonic Wave Function and Wave Equation

We consider a simple harmonic wave either longitudinal or transverse which travels in positive x-direction of our coordinate system. The graph of displacement $y$ of particles with position on the x-axis at an instant of time $t$ is shown in Figure 1.

Let $v$ be the speed of the wave. Consider that a particle at $x$ is at point $p$ in time $t$. The time the wave takes to reach the position of the particle is $x/v$. So the displacement of the particle in time $t$ is the same as the displacement of the particle at $x = 0$ in the earlier time $t - x/v$. We can write the wave function as $y=f(x)=A\cos \omega t$ where $w$ is the angular frequency of the simple harmonic wave and $A$ is the amplitude. Now you can replace $t$ in $y=f(x)=A\cos \omega t$ by $t - x/v$ and,

\[y=A\cos \omega \left( t-\frac{x}{v} \right)=A\cos \omega \left( \frac{x}{v}-t \right) \tag{1} \label{1}\]

We know that $\omega = 2 \pi f$, $f = 1/T$ and $T = \lambda /v$, and you can simplify the above equation in the form

\[y=A\cos 2\pi \left( \frac{x}{\lambda } -\frac{t}{T} \right) \tag{2} \label{2}\]

We define a quantity called wave number denoted by $k$ which is $k = 2\pi / \lambda$ and you know that $2\pi/T$ is the angular frequency $\omega$ and therefore we can again write Eq. \eqref{2} in another form as

\[y=A\cos (kx - \omega t) \tag{3} \label{3}\]

If the wave travels in negative x-direction, the wave takes time $x/v$ to reach the point $x=0$, and hence the displacement of the point at $x$ in time $t$ is the same as the displacement of the point at $x=0$ in time $t + x/v$. Therefore, we can replace $t$ in $y=f(x)=A\cos \omega t$ by $t + x/v$ for the wave travelling in negative x-direction. Thus, you can finally get the expression for the wave travelling in negative x-direction as shown below:

\[y=A\cos (kx+\omega t) \tag{4} \label{4}\]

Therefore, we can write the expression of the wave function for both negative and positive x-direction as

\[y=A\cos (kx \pm \omega t) \tag{5} \label{5}\]

You can pick "$-$" sign for positive direction and "$+$" sign for negative direction. You just saw various forms of wave function of the simple harmonic wave and all are in the form of cosine function. The cosine function can be easily turned into sine function and therefore we also call the simple harmonic wave as sinusoidal wave or the sine wave.

You saw two similar quantities $k = 2\pi/\lambda$ and $\omega = 2 \pi /T$, both have similar expressions. You know the wave speed is $v = \lambda f = \lambda/T$, and $\omega / k$ is also equal to $\lambda /T$, so you can find,

\[v=\frac{\omega }{k} \tag{6} \label{6}\]

Wave Equation

In determining the wave equation we use the method of partial derivative. First you know the wave equation for the wave travelling in positive x-direction from Eq. \eqref{3} which is

\[y=A\cos (kx-\omega t)\]

Differentiating the above equation with respect to $t$ keeping $x$ constant we get the velocity of the particle $v_\text{y}$ at $x$.

\[\frac{\partial y}{\partial t}={{v}_{y}}=A \omega \sin (kx-\omega t) \tag{7} \label{7}\]

Note that the magnitude of velocity $v_\text{y}$ is not the wave speed. Here the velocity of the particle $v_\text{y}$ is the velocity when the particle moves up and down or back and forth in simple harmonic motion as the wave disturbance reaches the particle. You can find the acceleration $a_y$ of the particle at position $x$ by again differentiating Eq. \eqref{7} with respect to $t$ keeping $x$ constant:

\[{{a}_{y}}=\frac{\partial {{v}_{y}}}{\partial t}=\frac{{{\partial }^{2}}y}{\partial {{t}^{2}}}=-A{{\omega }^{2}}\cos (kx-\omega t) \tag{8} \label{8}\]

Now differentiating equation $y=A\cos (kx-\omega t)$ with respect to position $x$ keeping $t$ constant, you get the slope of the curve at position $x$:

\[\frac{\partial y}{\partial x}=-Ak \sin (kx-\omega t) \tag{9} \label{9}\]

Again differentiating Eq. \eqref{9} with respect to $x$ keeping $t$ constant, you'll get the curvature of the curve at position $x$:

\[\frac{{{\partial }^{2}}y}{\partial {{x}^{2}}}=-A{{k}^{2}}\cos (kx-\omega t) \tag{10} \label{10}\]

Now using Eqs. \eqref{8} and \eqref{10}, and you know $v = \omega/k$ from Eq. \eqref{6}, you can get a form:

\[\begin{align*} \frac{{{\partial }^{2}}y/\partial {{t}^{2}}}{{{\partial }^{2}}y/d{{x}^{2}}}&=\frac{{{\omega }^{2}}}{{{k}^{2}}} \\ \text{or,}\quad \frac{{{\partial }^{2}}y}{\partial {{x}^{2}}}&=\frac{1}{{{v}^{2}}}\frac{{{\partial }^{2}}y}{\partial {{t}^{2}}} \tag{11} \label{11} \end{align*}\]

The above equation Eq. \eqref{11} is called linear wave equation which gives total description of wave motion. This equation is obtained for a special case of wave called simple harmonic wave but it is equally true for other periodic or non-periodic waves. This is one of the most important equations of physics. You will get the same wave equation for the wave travelling in negative x-direction.