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

[Ryker]

Homework Statement

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.

Homework Equations

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!

 

[klawlor419]

Either way is fair game!

 

[vela]

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?

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.

 

[Ryker]

Either way is fair game!

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.

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.