# For a sine (or cosine) wave, how can the $kx$ be different from $\omega t$?

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?

• Because $x$ and $t$ are two independant variables? – samjoe Mar 12 '17 at 14:17
• But for a uniform linear velocity, which is natural for a wave, x and t are no longer independent, right? – Viswa Mar 12 '17 at 14:19
• 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. – ZeroTheHero Mar 12 '17 at 14:20
• 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? – Viswa Mar 12 '17 at 14:27
• $x(t)= v_0t + x_0$ where $x_0$ is the initial position, which you are free to choose independently of $t$. – ZeroTheHero Mar 12 '17 at 14:31