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I'm trying to learn wave mechanics and came across the equation that states wave as a function of $x$ and $t$ and specifically $(kx - \omega t)$. In my faulty imagination, the $kx$ is always equal to $\omega t$, as both seem to be a measure of angular "distance" from a reference point of zero (start).

So my question is, why can't the quantity $(kx - \omega t)$ be always equal to zero?

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    $\begingroup$ Because $x$ and $t$ are two independant variables? $\endgroup$ – samjoe Mar 12 '17 at 14:17
  • $\begingroup$ But for a uniform linear velocity, which is natural for a wave, x and t are no longer independent, right? $\endgroup$ – Viswa Mar 12 '17 at 14:19
  • $\begingroup$ they are still independent. There is nothing preventing you from selecting $x$ and $t$ at your heart's desire. Once you have fixed a pair of (x,t), you have located a point on the wave, and this point will "travel" towards the right (assuming $k>0$ so as to keep $kx-\omega t$ constant, but you can select any pair $(x,t)$ with $x$ and $t$ chosen completely independently. $\endgroup$ – ZeroTheHero Mar 12 '17 at 14:20
  • $\begingroup$ I still don't understand. I thought value of x is automatically determined when we fix the value of t, because the wave is moving along x with a fixed velocity? Am I missing something? $\endgroup$ – Viswa Mar 12 '17 at 14:27
  • $\begingroup$ $x(t)= v_0t + x_0$ where $x_0$ is the initial position, which you are free to choose independently of $t$. $\endgroup$ – ZeroTheHero Mar 12 '17 at 14:31
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Consider this animation: Sine Wave

Along the x axis is our x and the sine wave is moving with time. If you were to stop the animation and change x you're looking at x with a constant t and we can see this covers all values of the sine wave. If we were instead to only look at one value of x and plot out all the positions the wave has through time, this would also cover all values of the sine wave.

The x and t are independent of each other but to get a value from the sine wave we have to know where and when we're looking.

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  • $\begingroup$ Thanks for the nice animation. I realized this just before you have posted this. But this is is the answer I was looking for! $\endgroup$ – Viswa Mar 12 '17 at 15:02

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