# Static wave's equation

In my book equation for two advanced wave which created this static wave are $$y_1=asin{\frac{2\pi}{\lambda}(vt+x)}$$ $$y_2=asin{\frac{2\pi}{\lambda}(vt-x)}$$

and equation for static wave is derived by adding this two equation: $\ Y=y_1+y_2$

Q1: Though this two wave comes from opposite direction why it is added rather then subtracted? Shouldn't it be $\ Y=y_1-y_2$

Q2: Why sign of $\ x$ is opposite for two equation though it is said that it's the distance of the particular point from the reference and whose displacement for the wave is $\ y$

Well to be hones both of your questions are related. Let me start by rewriting your equations of $y_1$ and $y_2$ (this will make the discussion easier), your version of $y_1$ and $y_2$ can be rewritten as: $$y_1=a\sin\left(\frac{2\pi}{\lambda}(x+vt)\right),$$$$y_2=-a\sin\left(\frac{2\pi}{\lambda}(x-vt)\right),$$where i put a minus outside of $y_2$ by using $\sin(-x)=\sin(x)$.

Let's now first look at question 2.

question 2: The answer to this question can be found by looking at the wavefronts of $y_1$ and $y_2$, which in its turn can be done by looking at a constant value for the arguments of $y_1$ and $y_2$ (since a constant argument yields a constant value of $y_1$ and $y_2$ and hence a wavefront). Let's call this constant value of the argument $x_0$, then the arguments of $y_1$ and $y_2$ become: $$x_0=x+vt \text{ for the argument of y_1},$$$$x_0=x-vt \text{ for the argument of y_2}.$$ Both arguments can be rewritten as:$$x=x_0-vt \text{ for the argument of y_1},$$$$x=x_0+vt \text{ for the argument of y_2},$$ Where we see that the introduced constant $x_0$ denotes the position of the wavefront at $t=0$. What this tells us is that $y_1$ represents a wave travelling in the negative $x$-direction (when the time increases the value of $x$ decreases) and $y_2$ represents a wave travelling in the positive $x$-direction (when time increases the value of $x$ increases). In general. You can do this analysis for every kind of wave, and you will find that waves with an argument of the form $(x+vt)$ are waves travelling in the negative x-direction (als called ''left travalling waves'') and waves with arguments of the form $(x-vt)$ are wave travelling in the positive x-direction (also called ''right travelling waves''). On hyperphysics a few more drawings and discussions are available (should you be interested).



question 1: The answer to question 1 can be given by the fact that te amplitudes of waves can be summed. This is because of the fact that each wave tells you what displacement $y$ is causes at a given point $x$ on a time $t$, when you have two waves which interact, the displacements should be summed. Now when you look at reflection (of sound our light or whatever wave you are looking at), there are 3 things that can happen:

1. (first case on the figure): when you reflect the reflected wave picks up a phase $\phi=\pi/2$ (this happens when you reflect on a dense medium). In that case your reflected wave picks up a minus sign since $\sin(x+\pi/2)=-\sin(x)$, this is probably the case you are looking at).
2. (second case on the figure): when you reflect the reflected and indecent wave are in anti-phase, so they cancel eachother out.
3. (third case on the figure): when you reflect, the reflected wave doesn't pick up a sign (and they are in phase), the amplitude of the wave doubles.

It doesn't matter whether you add or subtract $y_1$ and $y_2$, because all this does is introduce a phase shift in the resulting standing wave. To see this you need the identities:

$$\sin(a + b) = \sin(a)\cos(b) + \cos(a)\sin(b)$$

$$\sin(a - b) = \sin(a)\cos(b) - \cos(a)\sin(b)$$

Let's take your two equations and try adding them. Using the identity we rewrite $y_1$ and $y_2$ as:

$$y_1 = a\sin\left(\frac{2\pi}{\lambda}vt\right) \cos\left(\frac{2\pi}{\lambda}x\right) + a\cos\left(\frac{2\pi}{\lambda}vt\right) \sin\left(\frac{2\pi}{\lambda}x\right)$$

$$y_2 = a\sin\left(\frac{2\pi}{\lambda}vt\right) \cos\left(\frac{2\pi}{\lambda}x\right) - a\cos\left(\frac{2\pi}{\lambda}vt\right) \sin\left(\frac{2\pi}{\lambda}x\right)$$

So we get:

$$y_1 + y_2 = 2a\sin\left(\frac{2\pi}{\lambda}vt\right) \cos\left(\frac{2\pi}{\lambda}x\right)$$

$$y_1 - y_2 = 2a\cos\left(\frac{2\pi}{\lambda}vt\right) \sin\left(\frac{2\pi}{\lambda}x\right)$$

These don't immediately look the same, but note that:

\begin{align} \sin(x) &= \cos(x - \tfrac{\pi}{2}) \\ \cos(x) &= \sin(x + \tfrac{\pi}{2}) \end{align}

and if we use these to rewrite $y_2$ we get:

$$y_1 + y_2 = 2a\sin\left(\frac{2\pi}{\lambda}vt\right) \cos\left(\frac{2\pi}{\lambda}x\right)$$

$$y_1 - y_2 = 2a\sin\left(\frac{2\pi}{\lambda}vt+ \frac{\pi}{2}\right) \cos\left(\frac{2\pi}{\lambda}x- \frac{\pi}{2}\right)$$

So the two functions are the same but just shifted a bit along the $t$ and $x$ axes.