# Why does f(x-vt) represent a wave along +x?

1. Jan 31, 2012

### binrdow

I just can't understand how this makes sense. Why does f(x-vt) represent a transverse wave along +x? Where v is the velocity, t is time, and x is position along the x axis. It seems to be exactly the opposite to what I would think makes sense, with f(x+vt) representing a wave along +x and f(x-vt) representing a wave along -x. But this isn't the case! I've been trying to wrap my head around it, not matter how I look at it.
Can anyone explain this concept in a way that makes sense?

Thanks!

2. Jan 31, 2012

### HallsofIvy

Staff Emeritus
Let's for simplicity set our coordinates so that x= 0 when t= 0. If the object is moving at speed v, then at any future time, t, x= vt. So x-vt= 0. That is, f(0) "moves" along the x-axis at speed v.

3. Jan 31, 2012

### Ken G

Think of the function f(x-vt) as being a pattern of some kind. The pattern will propagate in the +x direction at speed v, because at x=0 and t=0, the pattern is f(0) (think of this as the center of the feature that is propagating), but at any other time t and x=vt, the pattern will still be f(0). So the center of the feature is always found at some x=vt, so is moving at speed v in the +x direction, because that's just what x=vt means.

4. Jan 31, 2012

### binrdow

Oh! Okay, so you just have to look at the equation for the position x first as x=vt. Thanks much!

5. Feb 1, 2012

### Studiot

Just remember that this statement does not apply to every f(x-vt) only suitable ones, although there are a great many such.

6. Feb 1, 2012

### TheBlackNinja

Which would be those restrictions? continuity?

7. Feb 1, 2012

### Studiot

The simplest example would be

f(x-vt) = k

8. Feb 1, 2012

### sophiecentaur

I had this problem too and eventually came to this conclusion (a nice arm waving one).
f(x-vt) tells you what the function will be at x in terms of what it was, at the origin, t seconds ago (i.e. t is the time it took to propagate to x from the origin) - hence the negative t sign.
Confused more? - Sorry if you are but it helped me, once.

9. Feb 1, 2012

### Ken G

I believe the issue was the idea that the wave is "transverse", which is not a requirement of the f(x-vt) form. Anything that propagates at v has the form f(x-vt), including longitudinal and water waves.

10. Feb 1, 2012

### binrdow

Actually, this helps allot! Thanks! :)

11. Feb 1, 2012

### Pengwuino

If you have mathematica or a scientific calculator, you can plot the functions f(x) = cos(x - vt), let v equal some random number, and plot it as you vary t between different graphs. You'll actually SEE the graph moving to the right as you increase t.

I think it's a fantastically convincing argument for seeing how this works.