# 16.2 Mathematics of waves  (Page 4/11)

 Page 4 / 11

## Velocity and acceleration of the medium

As seen in [link] , the wave speed is constant and represents the speed of the wave as it propagates through the medium, not the speed of the particles that make up the medium. The particles of the medium oscillate around an equilibrium position as the wave propagates through the medium. In the case of the transverse wave propagating in the x -direction, the particles oscillate up and down in the y -direction, perpendicular to the motion of the wave. The velocity of the particles of the medium is not constant, which means there is an acceleration. The velocity of the medium, which is perpendicular to the wave velocity in a transverse wave, can be found by taking the partial derivative of the position equation with respect to time. The partial derivative is found by taking the derivative of the function, treating all variables as constants, except for the variable in question. In the case of the partial derivative with respect to time t , the position x is treated as a constant. Although this may sound strange if you haven’t seen it before, the object of this exercise is to find the transverse velocity at a point, so in this sense, the x -position is not changing. We have

$\begin{array}{ccc}\hfill y\left(x,t\right)& =\hfill & A\phantom{\rule{0.2em}{0ex}}\text{sin}\left(kx-\omega t+\varphi \right)\hfill \\ \hfill {v}_{y}\left(x,t\right)& =\hfill & \frac{\partial y\left(x,t\right)}{\partial t}=\frac{\partial }{\partial t}\left(A\phantom{\rule{0.2em}{0ex}}\text{sin}\left(kx-\omega t+\varphi \right)\right)\hfill \\ & =\hfill & \text{−}A\omega \phantom{\rule{0.2em}{0ex}}\text{cos}\left(kx-\omega t+\varphi \right)\hfill \\ & =\hfill & \text{−}{v}_{y\phantom{\rule{0.2em}{0ex}}\text{max}}\phantom{\rule{0.2em}{0ex}}\text{cos}\left(kx-\omega t+\varphi \right).\hfill \end{array}$

The magnitude of the maximum velocity of the medium is $|{v}_{{y}_{\text{max}}}|=A\omega$ . This may look familiar from the Oscillations and a mass on a spring.

We can find the acceleration of the medium by taking the partial derivative of the velocity equation with respect to time,

$\begin{array}{cc}\hfill {a}_{y}\left(x,t\right)& =\frac{\partial {v}_{y}}{\partial t}=\frac{\partial }{\partial t}\left(\text{−}A\omega \phantom{\rule{0.2em}{0ex}}\text{cos}\left(kx-\omega t+\varphi \right)\right)\hfill \\ & =\text{−}A{\omega }^{2}\phantom{\rule{0.2em}{0ex}}\text{sin}\left(kx-\omega t+\varphi \right)\hfill \\ & =\text{−}{a}_{y\phantom{\rule{0.2em}{0ex}}\text{max}}\text{sin}\left(kx-\omega t+\varphi \right).\hfill \end{array}$

The magnitude of the maximum acceleration is $|{a}_{{y}_{\text{max}}}|=A{\omega }^{2}.$ The particles of the medium, or the mass elements, oscillate in simple harmonic motion for a mechanical wave.

## The linear wave equation

We have just determined the velocity of the medium at a position x by taking the partial derivative, with respect to time, of the position y . For a transverse wave, this velocity is perpendicular to the direction of propagation of the wave. We found the acceleration by taking the partial derivative, with respect to time, of the velocity, which is the second time derivative of the position:

${a}_{y}\left(x,t\right)=\frac{{\partial }^{2}y\left(x.t\right)}{\partial {t}^{2}}=\frac{{\partial }^{2}}{\partial {t}^{2}}\left(A\phantom{\rule{0.2em}{0ex}}\text{sin}\left(kx-\omega t+\varphi \right)\right)=\text{−}A{\omega }^{2}\phantom{\rule{0.2em}{0ex}}\text{sin}\left(kx-\omega t+\varphi \right).$

Now consider the partial derivatives with respect to the other variable, the position x , holding the time constant. The first derivative is the slope of the wave at a point x at a time t ,

$\text{slope}=\frac{\partial y\left(x,t\right)}{\partial x}=\frac{\partial }{\partial x}\left(A\phantom{\rule{0.2em}{0ex}}\text{sin}\left(kx-\omega t+\varphi \right)\right)=Ak\phantom{\rule{0.2em}{0ex}}\text{cos}\left(kx-\omega t+\varphi \right).$

The second partial derivative expresses how the slope of the wave changes with respect to position—in other words, the curvature of the wave, where

$\text{curvature}=\frac{{\partial }^{2}y\left(x,t\right)}{\partial {x}^{2}}=\frac{{\partial }^{2}}{{\partial }^{2}x}\left(A\phantom{\rule{0.2em}{0ex}}\text{sin}\left(kx-\omega t+\varphi \right)\right)=\text{−}A{k}^{2}\phantom{\rule{0.2em}{0ex}}\text{sin}\left(kx-\omega t+\varphi \right).$

The ratio of the acceleration and the curvature leads to a very important relationship in physics known as the linear wave equation    . Taking the ratio and using the equation $v=\omega \text{/}k$ yields the linear wave equation (also known simply as the wave equation or the equation of a vibrating string),

$\begin{array}{cc}\hfill \frac{\frac{{\partial }^{2}y\left(x,t\right)}{\partial {t}^{2}}}{\frac{{\partial }^{2}y\left(x,t\right)}{\partial {x}^{2}}}& =\frac{\text{−}A{\omega }^{2}\phantom{\rule{0.2em}{0ex}}\text{sin}\left(kx-\omega t+\varphi \right)}{\text{−}A{k}^{2}\phantom{\rule{0.2em}{0ex}}\text{sin}\left(kx-\omega t+\varphi \right)}\hfill \\ & =\frac{{\omega }^{2}}{{k}^{2}}={v}^{2},\hfill \end{array}$

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