Find the maximum velocity of the particle of the wave given by $Y=A\sin(\omega t-kx)$.
My book says that the maximum velocity of the particle of a wave is $A\omega$. But I have a doubt that why don't we take the resultant of the wave velocity and $A\omega$ for the maximum velocity of the particle.
The particles of the wave are undergoing pure transverse motion (motion in $y$ axis): The velocity of particles is completely in the $y$ direction. Therefore, the velocity of a particle located at position $x$ at time $t$ is given by
$$v=\frac{\partial y}{\partial t} \Biggr|_{(x,t)}=A \omega \cos(\omega t - kx)$$
Suppose I had a row of children and I asked them to jump one after the other, you would observe that each child jumps/moves only in the vertical direction but the disturbance they create travels in the horizontal direction. This analogy is not exactly the same as the sine wave but hopefully illustrates my point.
