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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.

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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.

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    $\begingroup$ 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 $v = \omega/k$. $\endgroup$ – Mark Eichenlaub Feb 12 '11 at 4:00
  • $\begingroup$ @Mark: ok, true, but that's just a way of taking the derivative in disguise. $\endgroup$ – David Z Feb 12 '11 at 5:01
  • $\begingroup$ 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. $\endgroup$ – Mark Eichenlaub Feb 12 '11 at 6:26
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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.

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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 $y(x_1,t_1) = C$ say. Over an additional time $t_2$, the point will have moved an additional distance $x_2$ and since we're looking at the same point, this means $y(x_1+x_2, t_1+t_2) = C$ also.

You're told that $y(x,t)=A\sin(k_x-w_t+O)$, so $A\sin(k_{x_1}-w_{t_1}+O) = C$, so $k_{x_1}-w_{t_1}+O = D$. At an additional time $t_2$ and distance $x_2, k(x_1+x_2) - w(t_1+t_2) + O = D$. Subtracting these two expressions from one another gives,

$$\begin{align} k(x_1+x_2)-k(x_1) - w(t_1+t_2) + w(t_1) &= 0,\\[5pt] \frac{(x_1+ x_2)-(x_1)}{(t_1+t_2)-(t_1)} &= \frac wk\\[5pt] v &= \frac wk \end{align}$$

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