Question

Is the displacement $D\left(x,t\right)=\left(0.10-0.10x^2+xt-2.5t^2\right)m$,

where x is in m and t is in s, a possible traveling wave? If so, what

is the wave speed?

Short answer

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

Step-by-step solution

Given information

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

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

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.

Verification of the function

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,

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

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

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

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

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

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