Question

Question: Show that the standing wave $f\left(z,t\right)=Asin\left(kz\right)cos\left(kvt\right)$satisfies the wave equation, and express it as the sum of a wave traveling to the left and a wave traveling to the right (Eq. 9.6).

Short answer

It is proved that the standing wave $f\left(z,t\right)=A\sin \left(kz\right)\cos \left(kvt\right)$satisfies the wave equation, and the given equation is expressed as the sum of a wave traveling to the left and a wave traveling to the right.

Step-by-step solution

Expression for the sum of a wave traveling to the left and a wave traveling to the right:

Write the expression for the sum of a wave traveling to the left and a wave traveling to the right.

$f\left(z,t\right)=g\left(z-vt\right)+h\left(z+vt\right)$

Determine the differentiation of the given equation with respect to z and t :

Differentiate the given equation with respect to $z$.

$\frac{\partial f}{\partial z}=Ak\cos \left(kz\right)\cos \left(kvt\right)$

Again differentiate the above equation with respect to z.

$$\begin{gathered} \frac{{\partial }^{2}f}{\partial z^2}=-A{k}^2\sin \left(kz\right)\cos \left(kvt\right) \\ =-k^{2}f\left(z,t\right) \end{gathered}$$ …… (1)

Differentiate the given equation with respect to $t$.

$\frac{\partial f}{\partial t}=-Akv\sin \left(kz\right)\sin \left(kvt\right)$

Again differentiate the above equation with respect to $t$.

$$\begin{gathered} \frac{{\partial }^{2}f}{\partial t^2}=-A{k}^2{v}^2\sin \left(kz\right)\cos \left(kvt\right) \\ =-k^2{v}^{2}f\left(z,t\right) \end{gathered}$$ …… (2)

From equations (1) and (2).

$\frac{{\partial }^{2}f}{\partial z^2}=\frac{1}{v^2}\frac{{\partial }^{2}f}{\partial t^2}$

Therefore, it is proved that the standing wave $f\left(z,t\right)=A\sin \left(kz\right)\cos \left(kvt\right)$satisfies the wave equation.

Express the given equation as the sum of a wave traveling to the left and a wave traveling to the right:

Simplify the equation as follows.

$$\begin{gathered} f\left(z,t\right)=\frac{A}{2}\left[2\sin \left(kz\right)\cos \left(kvt\right)\right] \\ =\frac{A}{2}\left[\sin \left(kz+kvt\right)+\sin \left(kz-kvt\right)\right] \\ =\frac{A}{2}\sin k\left(z-vt\right)+\frac{A}{2}\sin k\left(z+vt\right) \\ =g\left(z-vt\right)+h\left(z+vt\right) \end{gathered}$$

Therefore, it is expressed as the sum of a wave traveling to the left and a wave traveling to the right.