How do you find the velocity function of a mechanical wave?

With the form $y(x,t)=A\sin(kx-\omega t+\phi_0)$, there are two variables, How do I find the velocity? I don't know I can apply derivative with two variables.

-

You can (and, in fact, have to) apply the derivative to find the velocity, but it takes a bit of careful reasoning.

First, think about this: what exactly is the speed of a wave? It's the speed at which a particular point on the wave's structure moves. Points on the wave's structure are identified by their phase, which is the argument of the $\sin$ function. For instance, a peak is identified by phase $\phi = \frac{n\pi}{2}$, where $n$ is an odd integer. So you're looking for the speed of a point of constant phase.

Once you know that, you can just implicitly differentiate the expression for phase,

$$\phi = kx - \omega t + \phi_0$$

keeping in mind that $\phi$ is constant:

$$\frac{\mathrm{d}}{\mathrm{d}t}\phi = \frac{\mathrm{d}}{\mathrm{d}t}[kx - \omega t + \phi_0]$$

giving

$$0 = k\frac{\mathrm{d}x}{\mathrm{d}t} - \omega$$

or

$$\frac{\mathrm{d}x}{\mathrm{d}t} = \frac{\omega}{k}$$

which is the expression for the speed of a sinusoidal wave.

-
I wouldn't say you have to take a derivative. If you have $kx - \omega t = c$ for some constant $c$, then $x = \frac{c + \omega t}{k}$ by algebra, so $x$ increases linearly in time with the velocity $v = \omega/k$. – Mark Eichenlaub Feb 12 '11 at 4:00
@Mark: ok, true, but that's just a way of taking the derivative in disguise. – David Z Feb 12 '11 at 5:01
No argument here. It can be strange to look in some non-calculus physics textbooks where the author goes to absurd lengths to avoid writing a derivative. – Mark Eichenlaub Feb 12 '11 at 6:26

David and Mark explained how one can estimate the velocity $v_x$ of the form propagation along the propagation direction.

There is another velocity, say, the vertical velocity $v_y$ at a given place which is quite different and is determined with the wave amplitude, frequency, and time: $v_y = A\omega cos(\omega t - kx - \phi_0)$. It is variable.

-

What do you think is meant by the velocity of a wave?

Well, it's the velocity of any point on the wave, so pick one where y(x1,t1) = C say. Over an additional time t2, the point will have moved an additional distance x2 and since we're looking at the same point, this means y(x1+x2, t1+t2) = C also.

You're told that y(x,t)=Asin(kx-wt+O), so Asin(kx1-wt1+O) = C, so kx1-wt1+O = D. At an additional time t2 and distance x2, k(x1+x2) - w(t1+t2) + O = D. Subtracting these two expressions from one another gives,

k(x1+x2)-K(x1) - w(t1+t2) + w(t1) = 0,

((x1+ x2)-(x1))/((t1+t2)-(t1)) = w/k

v = w/k

-