Explanation of the Wave Equation $y(x,t)=f(t-x/v)$ Why  is the vertical displacement $y$ of a particle of a string disturbed by a wave given by
$y(x,t)=f(t-x/v)$? I can't find a proper explanation of this equation. Which time $t$ is referred to in this equation, and why is the equation $f(t-x/v)$ for positive $x$ direction and $f(t+x/v)$ for negative $x$ direction?
 A: Consider a horizontal string whose right end is fixed to wall. Now suppose you jerk (just once) the left end at time $t = 0$. Assume that at left end $x = 0$. Now, the jerk will move towards right end with velocity $v$. Let the positive direction of X-axis points towards right so that wave moves in positive direction. Now suppose after some time $t'$, the jerk reaches point $x'$. Now at $x'$, at time $t'$ shape of the string will be exactly same as that was at $x=0$ and $t=0$. So we can say that, $y(x', t') = y(0, 0)$.
Note that $t' = \frac{x'}{v}$. So we can see that shape of string at point $x=x'$ and $t=t'$ is same as that of point $x=0$ and $t=0$. Similarly, shape of string at any point $x=x'$ at any time $t$ is same as that of $x=0$ at time $t-\frac{x'}{v}$.
Therefore you can say that $y(x', t) = y\left(0, t-\frac{x'}{v}\right)$. Generalizing for any $x$,
$$y(x, t) = y\left(0, t-\frac{x}{v}\right)$$
Now $y\left(0, t-\frac{x}{v}\right)$ is function of single variable as first variable $x$ is always zero. So we can replace $y\left(0, t-\frac{x}{v}\right) = f\left(t-\frac{x}{v}\right)$ to get,
$$y(x, t) = f\left(t-\frac{x}{v}\right).$$
Now if you reverse the velocity, same logic will work to give 
$$y(x, t) = f\left(t+\frac{x}{v}\right).$$
A: Firstly, I'll settle the confusion about positive and negative displacements in the $x$ direction. As an example, consider the function $\cos(x)$ as well as its shift, $\cos(x-1)$:

The one 'in front' is $\cos(x-1)$, so you can see that a shift $x \to x -1$ corresponds to a shift in the $x$ direction to the right. Analogous, $\cos(x+1)$ moves to the left.

The displacement at a point $x$ at time $t$ denoted $y(x,t)$ satisfies the wave equation:
$$\frac{\partial^2 y}{\partial t^2} - v^2 \frac{\partial^2 y}{\partial x^2} = 0.$$
If we perform a change of variables, $U = x + vt$ and $V = x - vt$ you can check for yourself that the equation in the new coordinates is,
$$\frac{\partial^2 y(U, V)}{\partial U \partial V} = 0.$$
Notice a general solution is, $y(U,V) = F(U) + G(V) = F(x+vt) + G(x-vt)$ and from our interpretation of what a shift does, we see that $F$ is a wave travelling to the left with speed $v$ and $G$ is moving to the right with speed $v$ as well.
You can see as we go from $t$ to $t + \Delta t$, the argument of $G(x-vt)$ is shifted by $-v\Delta t$ which means it is moving with a constant velocity $v$ and shifted to the right by an amount $v\Delta t$.
A: Basically the traveling wave is a function of $x -vt$ for a wave travelling to the right and of $x + vt$ for a wave traveling to the left. If you look at a travelling wave as opposed to being a function of $x$ and $t$ independently. It has a simple harmonic motion component in time with angular frequency $\omega = \frac{2\pi}{T}$ where $T$ is the period of one oscillation, and a translational component with constant propagation speed v which we can assume to be to the right without a loss of generality. This means that if we fix a coordinate system $x'y'$ to the traveling wave and treat it as a standing wave moving uniformly to the right with velocity $v$, the equation of motion is of the form by $y' = \sin x'$. Of course $x'$ is measured in meters and so is $y'$ and we must provide an angle argument to the $\sin$ function by non-dimensionalizing the $x'$ in the equation and also multiplying the $\sin x'$ by $A$ the amplitude of the string vibration to arrive at the correct value $y'$. Furthermore $y = y'$ and $x= x' + vt$ or $x' = x - vt$.
Substituting these:
$$y(x,t) = A \sin (x - vt)
= A \sin \left[ -v\left(t - \frac{x}{v}\right)\right]
= -A \sin \left[v\left( t-\frac{x}{v}\right)\right]
= f\left(t-\frac{x}{v}\right)$$
since $v$ is a constant. I hope this clarifies the problem at hand.
