Chapter 15, Problem 36P

To determine

Find the correct options which satisfy the wave equation:

(a) $\text{y}\left(\text{x,t}\right)\text{=k}{\left(\text{x+vt}\right)}^{\text{3}}$

(b) $\text{y}\left(\text{x,t}\right){\text{=Ae}}^{\text{ik}\left(\text{x-vt}\right)}$

(c) $\text{y}\left(\text{x,t}\right)\text{=ln}\left[\text{k}\left(\text{x-vt}\right)\right]$

Answer to Problem 36P

Option (a), (b) and (c) are correct as it satisfies the general wave equation that is $\frac{{\partial }^{\text{2}}\text{y}/\partial {\text{x}}^{\text{2}}}{{\partial }^{\text{2}}\text{y}/\partial {\text{t}}^{\text{2}}}=\frac{\text{1}}{{\text{v}}^{\text{2}}}$ .

Explanation of Solution

Given:

Equation:

1. $\text{y}\left(\text{x,t}\right)\text{=k}{\left(\text{x+vt}\right)}^{\text{3}}$ .
2. $\text{y}\left(\text{x,t}\right){\text{=Ae}}^{\text{ik}\left(\text{x-vt}\right)}$
3. $\text{y}\left(\text{x,t}\right)\text{=ln}\left[\text{k}\left(\text{x-vt}\right)\right]$

Calculation:

To show the given function satisfies the wave equation.

The wave equation:

  $\frac{{\partial }^{\text{2}}\text{y}}{\partial {\text{x}}^{\text{2}}}\text{=}\left(\frac{\text{1}}{{\text{v}}^{\text{2}}}\right)\frac{{\partial }^{\text{2}}\text{y}}{\partial {\text{t}}^{\text{2}}}$

So, initially need to find their first and second derivatives with respect to $\text{x}$ and

$\text{t}$ and then substitute these derivatives in the wave equation.

The first two spatial derivatives of $\text{y}\left(\text{x,t}\right)\text{=k}{\left(\text{x+vt}\right)}^{\text{3}}$ ,

  $\frac{\partial \text{y}}{\partial \text{x}}\text{=3k}{\left(\text{x+vt}\right)}^{\text{2}}$

And

  $\frac{{\partial }^{\text{2}}\text{y}}{\partial {\text{x}}^{\text{2}}}\text{=6k}\left(\text{x+vt}\right)$ …….(1)

Now, first two temporal derivatives of $\text{y}\left(\text{x,t}\right)\text{=k}{\left(\text{x+vt}\right)}^{\text{3}}$ ,

  $\frac{\partial \text{y}}{\partial \text{t}}\text{=3kv}{\left(\text{x+vt}\right)}^{\text{2}}$

And

  $\frac{{\partial }^{\text{2}}\text{y}}{\partial {\text{t}}^{\text{2}}}{\text{=6kv}}^2\left(\text{x+vt}\right)$ ……(2)

The ratio of the equation (1) to equation (2) is,

  $\begin{array}{l}\frac{\frac{{\partial }^{\text{2}}\text{y}}{\partial {\text{x}}^{\text{2}}}}{\frac{{\partial }^{\text{2}}\text{y}}{\partial {\text{t}}^{\text{2}}}}=\frac{\text{6k}\left(\text{x+vt}\right)}{{\text{6kv}}^{\text{2}}\left(\text{x+vt}\right)}\\ \frac{\frac{{\partial }^{\text{2}}\text{y}}{\partial {\text{x}}^{\text{2}}}}{\frac{{\partial }^{\text{2}}\text{y}}{\partial {\text{t}}^{\text{2}}}}=\frac{\text{1}}{{\text{v}}^{\text{2}}}\end{array}$

Thus, option (a) is correct answer as it is confirming that $\text{y}\left(\text{x,t}\right)\text{=k}{\left(\text{x+vt}\right)}^{\text{3}}$ satisfies the general wave equation.

Option (b) is correct answer as it is confirming that $\text{y}\left(\text{x,t}\right){\text{=Ae}}^{\text{ik}\left(\text{x-vt}\right)}$ satisfies the general wave equation.

Reason:

Find the first two spatial derivatives of $\text{y}\left(\text{x,t}\right){\text{=Ae}}^{\text{ik}\left(\text{x-vt}\right)}$ ,

  $\frac{\partial \text{y}}{\partial \text{x}}{\text{=ikAe}}^{\text{ik}\left(\text{x-vt}\right)}$ , $\frac{{\partial }^{\text{2}}\text{y}}{\partial {\text{x}}^{\text{2}}}{\text{=i}}^{\text{2}}{\text{k}}^{\text{2}}{\text{Ae}}^{\text{ik}\left(\text{x-vt}\right)}$

Or

  $\frac{{\partial }^{\text{2}}\text{y}}{\partial {\text{x}}^{\text{2}}}{\text{=-k}}^{\text{2}}{\text{Ae}}^{\text{ik}\left(\text{x-vt}\right)}$ ……(1).

Now, first two temporal derivatives of $\text{y}\left(\text{x,t}\right){\text{=Ae}}^{\text{ik}\left(\text{x-vt}\right)}$ ,

  $\frac{\partial \text{y}}{\partial \text{t}}{\text{=-ikvAe}}^{\text{ik}\left(\text{x-vt}\right)}$ , $\frac{{\partial }^{\text{2}}\text{y}}{\partial {\text{t}}^2}{\text{=i}}^{\text{2}}{\text{k}}^{\text{2}}{\text{Ae}}^{\text{ik}\left(\text{x-vt}\right)}$

Or     $\frac{{\partial }^{\text{2}}\text{y}}{\partial {\text{t}}^2}{\text{=-k}}^{\text{2}}{\text{v}}^{\text{2}}{\text{Ae}}^{\text{ik}\left(\text{x-vt}\right)}$ ……(2).

The ratio of the equation (1) to equation (2) is,

  $\begin{array}{l}\frac{\frac{{\partial }^{\text{2}}\text{y}}{\partial {\text{x}}^{\text{2}}}}{\frac{{\partial }^{\text{2}}\text{y}}{\partial {\text{t}}^2}}=\frac{{\text{-k}}^{\text{2}}{\text{Ae}}^{\text{ik}\left(\text{x-vt}\right)}}{{\text{-k}}^{\text{2}}{\text{v}}^{\text{2}}{\text{Ae}}^{\text{ik}\left(\text{x-vt}\right)}}\\ \frac{\frac{{\partial }^{\text{2}}\text{y}}{\partial {\text{x}}^{\text{2}}}}{\frac{{\partial }^{\text{2}}\text{y}}{\partial {\text{t}}^{\text{2}}}}=\frac{\text{1}}{{\text{v}}^{\text{2}}}\end{array}$

Thus, confirming that $\text{y}\left(\text{x,t}\right){\text{=Ae}}^{\text{ik}\left(\text{x-vt}\right)}$ satisfies the general wave equation.

Option (c) is correct answer as it is confirming that $\text{y}\left(\text{x,t}\right)\text{=ln}\left[\text{k}\left(\text{x-vt}\right)\right]$ satisfies the general wave equation.

Reason:

first two spatial derivatives of $\text{y}\left(\text{x,t}\right)\text{=ln}\left[\text{k}\left(\text{x-vt}\right)\right]$ ,

  $\frac{\partial \text{y}}{\partial \text{x}}\text{=}\frac{\text{k}}{\text{x-vt}}$ , $\frac{{\partial }^{\text{2}}\text{y}}{\partial {\text{x}}^{\text{2}}}\text{=}\frac{{\text{k}}^{\text{2}}}{{\left(\text{x-vt}\right)}^{\text{2}}}$ …….(1)

Now, first two temporal derivatives of $\text{y}\left(\text{x,t}\right)\text{=ln}\left[\text{k}\left(\text{x-vt}\right)\right]$ ,

  $\frac{\partial \text{y}}{\partial \text{t}}\text{=}\frac{\text{vk}}{\text{x-vt}}$ , $\frac{{\partial }^{\text{2}}\text{y}}{\partial {\text{t}}^{\text{2}}}\text{=}\frac{{\text{v}}^{\text{2}}{\text{k}}^{\text{2}}}{{\left(\text{x-vt}\right)}^{\text{2}}}$ …….(2)

The ratio of the equation (1) to equation (2) is,

  $\begin{array}{l}\frac{\frac{{\partial }^{\text{2}}\text{y}}{\partial {\text{x}}^{\text{2}}}}{\frac{{\partial }^{\text{2}}\text{y}}{\partial {\text{t}}^2}}=\frac{\frac{{\text{k}}^{\text{2}}}{{\left(\text{x-vt}\right)}^{\text{2}}}}{\frac{{\text{v}}^{\text{2}}{\text{k}}^{\text{2}}}{{\left(\text{x-vt}\right)}^{\text{2}}}}\\ \frac{\frac{{\partial }^{\text{2}}\text{y}}{\partial {\text{x}}^{\text{2}}}}{\frac{{\partial }^{\text{2}}\text{y}}{\partial {\text{t}}^{\text{2}}}}=\frac{\text{1}}{{\text{v}}^{\text{2}}}\end{array}$

Thus, confirming that $\text{y}\left(\text{x,t}\right)\text{=ln}\left[\text{k}\left(\text{x-vt}\right)\right]$ satisfies the general wave equation.

Conclusion:

Thus, all three options satisfy the given general wave equation that is $\frac{{\partial }^{\text{2}}\text{y}/\partial {\text{x}}^{\text{2}}}{{\partial }^{\text{2}}\text{y}/\partial {\text{t}}^{\text{2}}}=\frac{\text{1}}{{\text{v}}^{\text{2}}}$ .