waves
Is there a method to identify which wave is leading and which wave is lagging from their equations?
For example, if the two waves are $A\sin(\omega t+\pi/6)$ and $A\sin(\omega t+2\pi)=A\sin(\omega t)$ (by trigonometric relations), is the second wave leading in phase or lagging in phase compared to the first? Or we can say both is true?
What is the convention to be followed when we say "wave A leads wave B…" ?
I'm pretty sure that we can't do any better with trig identities; it looks like any such expression would have a term vaguely like $\sqrt{1 + r \cos \theta}$, and then the square root ruins the nice intuition.
In lieu of that I'll offer a derivation of the group velocity that uses no fancy math at all! Hopefully that's what you wanted, even if you didn't directly ask it.
First off, let's figure out what group velocity is. Apparently, group velocity is the velocity of the envelope of a wave. But what is the envelope of a wave? I can create a wave with any initial position and velocity, so it can be an arbitrarily weird shape. There might be no discernible envelope at all.
So let's back up and find some examples. When we talk about the envelope of a wave, we mean some curve you draw around an oscillation, like this.
In order to do this, there must be a well defined oscillation to draw the envelope around. That means that our wave must be made up of individual frequencies that are close to one central frequency. To make things convenient, let's write that schematically as $$\text{wave} = \sum_{k'} \sin(k'x - \omega(k') t) \text{ for a bunch of } k' \approx k$$ However, if we just have one frequency, the wave is just an infinite sinusoid $\sin(kx)$. This doesn't have an envelope, strictly speaking, because it just goes on forever at the same amplitude. We must have waves of other frequencies, which will constructively and destructively interfere with each other, to actually get an envelope.
So we've concluded that the envelope is defined by where a bunch of sinusoids making up our wave constructively or destructively interfere. Their phases are, as a function of space and time, $$\phi(k') = k'x - \omega(k') t$$ Now let's assume for simplicity that the top of the envelope is at $x = 0$ at time $t = 0$. That means that the waves must constructively interfere there, so all the $\phi(k')$ are about the same.
As time goes on, the envelope will move, but the peak will still be where the phases of the component waves are the same. That means $$\text{peak of envelope satisfies } \frac{d\phi(k')}{dk'} = 0$$ Performing the differentiation, we have $$x - \frac{d\omega(k')}{dk'} t = 0$$ Since we said $k' \approx k$, let's drop the primes and rearrange for $$\frac{x}{t} = \frac{d \omega(k)}{dk}$$ But $x/t$ is exactly the speed of the peak of the envelope, so this is the group velocity.
Fairly easy, just add the two waves using superposition.
$\phi^\prime = 2 \phi = 2 \frac{A_0}{r} sin(kr - \omega t)$ for the resulting wave (the prime denotes superposition)
Using $I = h(A_0/r)^2$ we see that $I^\prime = I \times (2A_0/A_0)^2 = 4I$.
Therefore the intensity has become four times larger.
Best Answer
If you start from $y_1= A\sin(\omega t)$ and compare it with $y_2= A\sin(\omega t+ \phi)$ you find that time $t=0$ motion $1$ has a displacement of $y_1=0$ and motion $2$ has a displacement of $y_1=\sin \phi$.
You will see that whatever motion $2$ does motion $1$ does a little later in time so motion $2$ leads motion $1$ by phase angle $\phi$.
So the more positive $\phi$ represents the motion which is leading.
However it is also true the motion $1$ lags motion $2$ by phase angle $\phi$.
The way you have written the two motions your second motion leads your first motion by $2\pi-\frac \pi 6$.
However if your two motions had been given as $A\sin(\omega t + \frac \pi 6)$ and $A\sin(\omega t)$ and no other information was given you would say that the first motion leads second motion by phase angle $\frac \pi 6$.