Why is the Plane progressive wave equation $y= a\sin (kx-wt)$ for positive direction of x-axis? Likewise, why is  $y= a\sin(kx+\omega t)$ for negative direction?  What is the basis/derivation for this?
 A: 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.
\begin{equation} 
kx-\omega t=\lambda
\end{equation}
where $\lambda$ is a constant.
Differentiating both sides we get, 
\begin{equation}
k\frac{dx}{dt}=\omega
\end{equation}
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.


A: 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.
It might help your visualisation if you think of that displacement corresponding to a peak.
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. 
A: The points at which the sine is maximum will satisfy $kx+\omega 
t = \frac{1}{2}\pi + 2\pi n$ with $n$ an arbitrary integer. So at a given time $t_1$ these peaks are at $x_n (t_1) = \frac{1}{k}(\frac{1}{2}\pi + 2\pi n - \omega t_1)$.
Now look at these peaks a short time $t_2 = t_1 + \Delta t$ later. You will easily find
\begin{equation}
x_n (t_2) = \frac{1}{k}(\frac{1}{2}\pi + 2\pi n - \omega t_2 )
= \frac{1}{k}(\frac{1}{2}\pi + 2\pi n - (\omega t_1 + \omega \Delta t)) = x_n (t_1) - \frac{\omega}{k} \Delta t
\end{equation}
so the peaks move somewhat to the left (smaller $x$). This provided that $k$ and $\omega$ have the same sign. If they have the opposite sign they will move to the right.
