# Deriving the displacement equation for a sinusoidal wave

Hi everyone,

I'm trying to understand the derivation of

$$D(x,t) = A \sin{(kx - \omega t + \phi_o)}$$

which is the displacement equation for a sinusoidal wave.

The way my textbook (Physics for scientists and engineers by Knight) does it:

Look at the graph of displacement versus position at time t = 0. The function that describes this graph is

$$D(x,t=0) = A \sin{(2 \pi (\frac{x}{\lambda}) + \phi_o)}$$

Now what I don't get is the next step. We replace x with the quantity (x-vt), where v is the speed of the wave. The resulting equation is

$$D(x,t) = A \sin{(2 \pi (\frac{x-vt}{\lambda}) + \phi_o)}$$

Where does this (x-vt) term come from? I don't understand what this equation means, because I thought we were considering t = 0, and now we are throwing in a t variable.

edit: I found this: http://en.wikipedia.org/wiki/D'Alembert's_formula

I think it is related, but I don't understand the article at all...

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After a time t, the wave will have moved a distance vt in the positive x direction.

After a time t, the wave will have moved a distance vt in the positive x direction.

Thank you, but I still don't understand.

I'm really trying to get some intuition behind why we are able to do this. I can see that vt is equal to the distance that one crest will move in time t if it is moving at wave speed v.

I can also see that if we put t = 0 in the new equation, we get the original equation.

But,

1) Why do we put a minus sign for positive x direction and a plus sign for negative x direction? I remember from math that, for example, sin(2pi - 1) is the graph of sin(2pi) shifted to the right by 1 unit, but I still don't understand what this means for the graphs and the actual waves. But this leads me to my next question...

2) What does this even mean? How can we just put a time variable into a equation that assumed from the beginning that t = 0?

Thanks.

1) The graph of f(x-a) is the graph of f(x) shifted a to the right (positive x direction)
if you have f(x) = C for x=D then you have f(x-a) = C for x = D+a

2) This isn't really a derivation. They just gave a function with the property that D(x,t) = D(x-vt, 0)