Show that the displacement D(x,t) = cx² + dt², 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.

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Welcome back, everyone. We are making observations about this equation for a disturbance in a given medium. And we are told that A and B are two positive and constant coefficients. And we need to find the relationship between A and B that satisfies the wave equation. Now, what is our wave equation? Well, the wave equation is as follows, we have that the second partial derivative of Y with respect to X minus one over V squared times the second partial derivative of Y with respect to T is equal to zero. But in order to solve this, we need to find both of those partial derivatives. So the partial derivative, second partial derivative of Y with respect to X and the second partial derivative of Y with respect to T. Now, the way I'm gonna do this, let me go and change color real quick is I'm just going to take one derivative at a time here. So first, let's just take the first derivative of Y with respect to X. Remember we are referencing this equation up top which is the position as a function of X and T is given by A X plus BT cubed So the first derivative of that equation with respect to X is just going to be three A times A X plus BT squared. And the second partial derivative with respect to X is going to be six A squared times A X plus BT Great. So now let's go ahead and move on with the partial derivatives with respect to T. The first derivative is going to be three B times A X plus BT squared. And the second derivative with respect to T is equal to six B squared times A X plus BT. Now that we have both of our second partial derivatives, we can plug that into our wave equation. Now, before I do that, I'm actually going to add the second term of our wave equation to both sides. Therefore, getting rid of the equal zero and just setting each term equal to one another. And this will make things a little bit easier. All right. So plugging into our equation here, what we have is six A squared times A X plus BT is equal to uh my apologies, one over V squared times six B squared times A X plus BT. Now, what I'm gonna do is I'm gonna divide both sides by six times A X plus BT. What this gives us is A squared is equal to B squared over V squared. Finally, I just want to multiply both sides by one over B squared. And what we get is we get that A squared over B squared is equal to one over V squared. And finally, to simplify this equation further, taking the square root of both sides of our equation, we finally find the relationship between A and B or A divided by B is equal to one divided by V which corresponds to our final answer. Choice of C. Thank you all so much for watching. I hope this video helped. We will see you all in the next one.

Show that the displacement D(x,t) = cx² + dt², 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.

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Key Concepts

Wave Equation

Displacement Function

Wave Speed