Equation of reflected wave (fixed end/free end) I have an equation of a wave as
$y = 2 \sin\left( \dfrac{\pi}{6}x - \dfrac{\pi}{4}t \right)$. I want to find the equation of the wave which is formed when it gets reflected from (i) a fixed end or (ii) a free end
So, first of all, the wave will now travel in negative direction, so I have to put a negative sign, either in the $\omega t$ term or the $kx$ term. Then for a fixed end I have to introduce a phase difference of $\pi$. So I'll put the negative sign in front of the equation.
Where do I put the negative sign to change the direction of the wave's velocity? In front of $kx$ or $\omega t$? Also, does this sign depend on the distance after which I have fixed the end or left the end free?
Following is a simple observation:


Red => $y = \sin( x - t )$  [moving right]
Blue => $y = \sin( x + t )$  [moving left]
Green => $y = -\sin( x + t )$   [moving left]
Red + Green => antinode at $x=0$  [standing 1]
Red + Blue => antinode at $x=4.5$  [standing 2]
So, I have free end and fixed end at different x for these two standing waves.
Then, how can I decide which one is correct?
 A: You have to put it in front $\omega t$ since 
$$\sin(kx-\omega t)$$ will give you a wave travelling to the right (with respect to the time $t$). Then $$\sin(kx+\omega t)$$ is a wave travelling to the left. You can see this by fixing the location $x$ and vary the time $t$.
The sign does not depend on the distance where you fix the end. The wave will travel to the end and then will be reflected.
A: $y(x,t)=\sin(\omega\,t -k\, x)$ and $y(x,t)=\sin(-\omega\,t +k\, x) $ are examples of the wave equation for a wave travelling in the positive x-direction.  
$y(x,t)=\sin(\omega\,t +k\, x)$ and $y(x,t)=\sin(-\omega\,t -k\, x)$ are examples of the wave equation for a wave travelling in the negative x-direction.  
You can change the phase by adding or subtracting a phase angle $\phi$.
For example $y(x,t)=\sin(\omega\,t -k\, x+\phi)$ or $y(x,t)=\sin(\omega\,t -k\, x-\phi)$
A: at a fixed (hard) boundary, the displacement  remains zero and the reflected wave changes its polarity (undergoes a $180^\circ{}$ phase change) and direction opposite (replace $\omega t$ by    $-\omega t$) 
so,in this case reflected wave will be $y=-2\  sin \left(\dfrac{\pi}{6}x+\dfrac{\pi}{4} t\right)$  because $sin(x+\pi)=-sinx$
at a free (soft) boundary end, reflected wave will not be inverted and rest will be same as above 
so, in this case reflected wave will be $y=2 \ sin \left(\dfrac{\pi}{6}x+\dfrac{\pi}{4} t\right)$
above, i've assumed boundary as $x=0$
if boundary was $x=x_{0}$  replace everywhere $x$ by $x-x_{0}$
A: If you don't want to be confused by all these signs, just write down the general solution of your problem: $$ y(x,t)=A\sin(kx-\omega t ) + B\sin(kx+\omega t)$$ In your case $A=2$. Stating that a reflection occurs means that $B$ is not null. If reflection arises at $x=l$, just solve for B in the proper equation $y(x=l,t)=0$ or $\frac{\partial y}{\partial x} (x=l,t) =0$ depending on your boundary conditions.
