# Determine the direction and speed of the wave from a given wave equation

1. Sep 13, 2012

### Ryker

1. The problem statement, all variables and given/known data
Given an equation for a wave $\psi(x,t) = A e^{-a(bx+ct)^{2}}$ determine the direction of its propagation if you know $\psi(x,t) = f(x \pm vt)$ and use this to find its speed.

2. Relevant equations

3. The attempt at a solution
I figured I would just rearrange the expression in the exponent, so as to yield $x + \frac{c}{b}t$, and then just read off $v = \pm\frac{c}{b}$. However, if we don't know whether $\psi(x,t) = f(x + vt)$ or $\psi(x,t) = f(x - vt)$, can we really determine the direction of its propagation?

Also, I found somewhere the answer to this question would uniquely be $v = -\frac{c}{b}$ by letting $bx + ct = C$, and then after solving for x, $x = \frac{C}{b} - \frac{c}{b}$, taking the derivative with respect to time, yielding the above unique solution with the minus sign. Is this the proper way of doing things instead of just rearranging the expression like I did?

Thanks!

2. Sep 16, 2012

### klawlor419

Either way is fair game!

3. Sep 16, 2012

### vela

Staff Emeritus
The only difference between f(x+vt) and f(x-vt) is the direction that the wave propagates. f(x-vt) represents a wave moving in the +x direction, and f(x+vt) represents a wave moving in the -x direction, where v>0 is the speed of the wave. Whether v=±c/b depends on the signs of b and c.

4. Sep 16, 2012

### Ryker

Thanks for the replies! After thinking about it some more, that's what I figured, as well, as I just couldn't justify why there would necessarily be a minus sign.