# Why is the Plane progressive wave equation $y= a\sin (kx-wt)$ for positive direction of x-axis?

Let's take the argument of the function i.e, $kx-\omega t$. The argument of the function should remain constant,(equivalently the phase must remain constant)for a particular section of the wave. $$kx-\omega t=\lambda$$ where $\lambda$ is a constant. Differentiating both sides we get, $$k\frac{dx}{dt}=\omega$$ which is positive, thus this represents a wave travelling in positive direction. Similarly for the wave travelling in negative direction.

This can be illustrated by plotting also, the blue wave represents wave at $t=0$ and the orange wave represents wave at a later time, As you can see, the wave has moved in left direction for the first figure for the equation $y=a\sin(kx+\omega t)$ and in the right for the second i.e for $y=a\sin(kx-\omega t)$ case.Consider the blue wave below at some $x$, now if you want to make the wave move, then you need the same $y$ at some other $x$ and $t$, thus for the first case,it happens when $x$ becomes less positive, or moves to the left, implying that the wave has traveled in negative direction.

Suppose an x-axis which points to the right.
A wave produced by a source travels to the right.

The displacement due to the wave at position $$x$$ at a time $$t$$ is $$y = f(kx \pm \omega t)$$ where $$f(kx \pm \omega t)$$ is a function which satisfies the wave equation.

Since the wave and hence the "information" is travelling to the right at some later time $$t + \Delta t$$ a particle at $$x + \Delta x$$ will have the same displacement $$y$$.
In other words the peak has travelled a distance $$\Delta x$$ in a time $$\Delta t$$ with both $$\Delta x$$ and $$\Delta t$$ both positive.

So $$y = f(kx \pm \omega t) = f(k(x + \Delta x) \pm \omega (t + \Delta t))$$

which implies $$kx \pm \omega t = k(x + \Delta x) \pm \omega (t + \Delta t) \Rightarrow 0 = k\Delta x \pm \omega \Delta t$$

Since in the case of a right travelling wave $$k, \Delta x, \omega$$ and $$\Delta t$$ are all positive the only way to satisfy this equation is to have $$0 = k\Delta x - \omega \Delta t$$ which implies that the original function was $$y = f(kx - \omega t)$$.

Out of this analysis you also get that $$\frac {\Delta x }{\Delta t} = \frac \omega k$$ which is the speed of the wave.

When the information is travelling to the left then either $$-\Delta t$$ and $$+\Delta x$$ or $$+\Delta t$$ and $$-\Delta x$$.

To satisfy $$0 = k(-\Delta x) \pm \omega \Delta t$$ or $$0 = k\Delta x \pm \omega (-\Delta t)$$ it must be that $$y = f(kx + \omega t)$$.

So if you want to watch the progress of a right travelling wave, ie follow a peak as time $$t$$ and position of peak $$x$$ increases you need to keep $$kx - \omega t$$ constant.