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$D\left(x,t\right)=\left(0.10-0.10x^2+xt-2.5t^2\right)m$,satisfies the wave equation.

The given function can be written as,

$$\begin{gathered} D\left(x,t\right)=\left(0.10-0.10x^2+xt-2.5t^2\right)m \\ D\left(x,t\right)=0.10\left[1-{\left(x-5t\right)}^2\right]m \end{gathered}$$

Therefore, taking the partial derivative we get,

$$\begin{gathered} \frac{∂D}{∂t}=0.10×\left(-2\right)\left(x-5t\right)\left(-5\right)m/s \\ ⇒\frac{∂D}{∂t}=\left(x-5t\right)m/s \\ ⇒\frac{{∂}^{2}D}{∂t^2}=-5m/s^2 \end{gathered}$$

Also, taking partial derivative w.r.t $\text{'}x\text{'},$we get,

$$\begin{gathered} \frac{∂D}{∂x}=0.10×\left(-2\right)\left(x-5t\right)\left(1\right) \\ ⇒\frac{∂D}{∂x}=-0.2×\left(x-5t\right)m^{-1} \\ ⇒\frac{{∂}^{2}D}{∂x^2}=-0.2m^{-1} \end{gathered}$$

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

$$\begin{gathered} \frac{{∂}^{2}D}{∂t^2}=v^2\frac{{∂}^{2}D}{∂x^2} \\ ⇒-5m/s^2=v^2\left(-0.2m^{-1}\right) \\ ⇒v^2=25m^2/s^2 \\ ⇒v=5m/s \end{gathered}$$

Thus, the given function is a possible travelling wave and it has a wave speed of$5m/s.$