91Ó°ÊÓ

Any function can be considered as a travelling wave if it satisfies the wave equation,

$\frac{{∂}^{2}f}{∂t^2}=v^2\frac{{∂}^{2}f}{∂x^2}$

Therefore, we have to verify whether the function $D\left(x,t\right)=\left(3mm\right)e^{i\left(2.0x+8.0x+5.0\right)}$, satisfies the wave equation.

The equation can be also written as,$D\left(x,t\right)=\left(3×{10}^{-3}m\right)e^{i\left(2.0x+8.0x+5.0\right)}$

Therefore, taking the partial derivative we get,

$$\begin{gathered} \frac{∂D}{∂t}=\left(3×{10}^{-3}m/s\right)\left(8i\right)e^{i\left(2.0x+8.0x+5.0\right)} \\ ⇒\frac{{∂}^{2}D}{∂t^2}=\left(3×{10}^{-3}m/s^2\right){\left(8i\right)}^2{e}^{i\left(2.0x+8.0x+5.0\right)} \end{gathered}$$

Also, taking partial derivative w.r.t we get,

$$\begin{gathered} \frac{∂D}{∂x}=\left(3×{10}^{-3}\right)\left(2i\right)e^{i\left(2.0x+8.0x+5.0\right)} \\ ⇒\frac{{∂}^{2}D}{∂x^2}=\left(3×{10}^{-3}{m}^{-1}\right){\left(2i\right)}^2{e}^{i\left(2.0x+8.0x+5.0\right)} \end{gathered}$$

Therefore, substituting the values obtained in the wave equation, we get,

$$\begin{gathered} \frac{{∂}^{2}D}{∂t^2}=v^2\frac{{∂}^{2}D}{∂x^2} \\ ⇒\left(3×{10}^{-3}m/s^2\right){\left(8i\right)}^2{e}^{i\left(2.0x+8.0x+5.0\right)} \\ =v^2\left(3×{10}^{-3}{m}^{-1}\right){\left(2i\right)}^2{e}^{i\left(2.0x+8.0x+5.0\right)} \\ ⇒64m/s^2=v^2\left(4m^{-1}\right) \\ ⇒v^2=16m^2/s^2 \\ ⇒v=4m/s \end{gathered}$$

Thus, the given function is a possible travelling wave and it has a wave speed of$4m/s.$