Question

Show that the displacement $D(x,t)=c{x}^2+d{t}^2$, where c and d are constants, is a solution to the wave equation. Then find an expression in terms of c and d for the wave speed.

Short answer

The given displacement is a solution to the wave equation.

The expression for the wave speed is$v=\sqrt{\frac{d}{c}}$.

Step-by-step solution

Given :

The displacement is given by$D(x,t)=c{x}^2+d{t}^2$where c, d are constants.

Calculating the partial derivatives to show that the given displacement satisfy the wave equation:

The given displacement must satisfy the wave equation Take the partial derivatives of D(x, t) with respect to x,

$$\begin{gathered} \frac{\partial D}{\partial x}=\frac{\partial \left(c{x}^2+d{t}^2\right)}{\partial x} \\ =2cx \\ \frac{{\partial }^{2}D}{\partial x^2}=2c \end{gathered}$$

Take the partial derivatives of D(x, t) with respect to t,

$$\begin{gathered} \frac{\partial D}{\partial t}=\frac{\partial \left(c{x}^2+d{t}^2\right)}{\partial t} \\ =2dt \\ \frac{{\partial }^{2}D}{\partial t^2}=2d \end{gathered}$$

Substituting the above values in the wave equation:

$$\begin{gathered} \frac{{\partial }^{2}D}{\partial t^2}=v^2\frac{{\partial }^{2}D}{\partial x^2} \\ 2d=v^2.2c \\ {v}^2=\frac{d}{c} \end{gathered}$$

Therefore, it satisfies the wave equation.

Calculating the speed of the wave:

From the above equation, the wave speed is obtained as

$$\begin{gathered} v^2=\frac{d}{c} \\ v=\sqrt{\frac{d}{c}} \end{gathered}$$

Therefore, the wave speed is$v=\sqrt{\frac{d}{c}}$.