Wave Function for a Sinusoidal Wave (Why minus sign?) I was trying to understand how the wave function for a sinusoidal wave was derived, but did not understand one specific sign, the minus sign in the following formula:
$$y(x,t) = A \sin(k x – \omega t + \theta_0)$$
Can anybody explain me what does the negative sign between $k x$ and $\omega t$ means in physical and mathematical context? I would really appreciate detailed explanation.
 A: If your phase is zero $\theta_0=0$ then your wave has zero amplitude when $k x = \omega t$ or $x = c t$ where $c = \frac{\omega}{k}$ is the wave speed.
So it represents a wave moving in the $+x$ direction with speed $c$.
A: First, let's let $\theta_0=0$ for simplicity.  
Now, consider $y(x,0)=Asin(kx)$.  That means that when $t=0$ then $y=0$ when $x=0$.  That is, $y(0,0)=0$.  
Now, let's follow that point $y=0$ as time increases.  Consider $t=1$, then what is $x$ such that $y(x,1)=0$?  Plugging in $t=1$ we see that $0=Asin(kx-w)$.  Therefore $kx-w=0$, i.e. $x=w/k$.  Since both $w$ and $k$ are greater than zero, we see that $x$ increases as $t$ increases.  
That is, the point we first considered $y(0,0)=0$ moved to the right a distance $x=w/k$ after $t$ increased by $1$.  
Thus,  we find that the negative sign infront of $w$ means the wave is moving to the right (increasing $x$ as $t$ increases).  If the negative sign were instead positive, we could do a similar analysis to find that the wave moves to the left (decreasing $x$ as $t$ increases).
A: If the positive $x$-axis is pointing to the right, the minus sign between $kx$ and $\omega t$ in the equation means that the wave is travelling to the right. If the sign is plus instead of minus, the wave is travelling to the left. 
To verify, draw $\sin(kx - \omega t + \theta_0)$ for any given $k$, $\omega$, and $\theta_0$, and for two different $t$. For simplicity, choose $k=1$, $\omega=1$, and $\theta_0 = 0$. You may find that when $t$ is increased by a small amount, the sine curve is shifted to the right. 
