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My book states:

The wave disturbance travels from $x = 0$ to some point $x$ to the right of the origin in an amount of time given by $t=\frac{x}{v}$, where $v$ is the wave speed. So the motion of point $x$ at time $t$ is the same as the motion of point $x = 0$ at the earlier time $\left(t - \frac{x}{v}\right).$

I do not really get why it says to use negative $\frac{x}{v}$ when the sinusoidal wave is moving to the $+x$ direction.

Why is it that?

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2 Answers 2

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It is basically a notation to represent a sinusoidal wave (which is travelling in $+ x$ direction ) in the form : $\sin(wt-kx)$. Had the wavefunction been $\sin(wt + kx) $ the wave function would correspond to a wave travelling in a $-x$ direction. So it's only a notation as far; nothing so conceptual about it.

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  • $\begingroup$ It is something more than a notation. Think deep , you will find why +kx means -x direction and converse. There is a CONCEPT. $\endgroup$ Sep 24, 2015 at 16:16
  • $\begingroup$ @Aniket: There may be a CONCEPT but that is not deep. It is just shifting of the graph; that's it. $\endgroup$
    – user36790
    Sep 24, 2015 at 16:26
  • $\begingroup$ @user36790 You consider "shifting of graph" something not conceptual? Perhaps you took "deep" too seriously. BTW Akshay did not mention it, he had referred it to be ONLY a notation. That's what I objected to. $\endgroup$ Sep 24, 2015 at 16:37
  • $\begingroup$ @Aniket: Don't take me wrong! I never told there is no concept underneath; what I've told is it is not too deep; that's it. $\endgroup$
    – user36790
    Sep 24, 2015 at 17:22
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The wave travels to the right, so a disturbance that passes x=0 at t=0 arrives at point x (>0) at a later t>0. I can "predict" the motion of the wave at point $x$ by looking back in time - what was the wave doing at an earlier time at the origin? Since the disturbance travels at velocity $v$, we know how long ago it passed an earlier point. And that is the meaning of the $-\frac{x}{v}$ term, and the reason you need the negative sign.

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