Particle velocity in a standing wave Consider a standing wave of the equation $y=A\cos(\omega t)\sin(kx)$ on an l-long string vibrating in one segment(fundamental mode). I'm supposed to find the maximum particle velocities at specific points, so I differentiated the equation and got $$v_{particle}=2A\omega \cos(\omega t)\cos(kx)$$
As $ k= \frac{\pi}{l}$, $$v_{particle}=2A\omega \cos(\omega t)\cos(\frac{\pi x}{l})$$.
However, this seems to imply that the particle at $x=\frac{l}{2}$ will always have velocity 0. Likewise, the ends of the string have a velocity that may be non-zero.
Where did I mess up in obtaining the particle velocity equation?
Also' the solution to the problem just uses velocity =$a\omega $.  Why is that supposed to be applicable here?
 A: Your mistake is saying that $v_{particle}=2A\omega \cos(\omega t)\cos(kx)$, which is incorrect.
Start off with the standing wave equation, $$y(x,t)=2A\sin(kx)\cos(\omega t)$$
To get the $y$-velocity of a particle on the wave, take the partial derivative wrt time.
$$\dfrac {\partial y} {\partial t}=\dfrac{\partial}{\partial t}\left(2A\sin(kx)\cos(\omega t) \right)$$
The $2A \sin(kx)$ term does not depend on $t$, and you get,
$$\dfrac {\partial y} {\partial t}=2A\sin(kx)\dfrac{\partial}{\partial t}\left(\cos(\omega t) \right)$$
Which finally becomes $$\dfrac {\partial y} {\partial t}=-2A\omega\sin(kx)\cos(\omega t) \\ \implies v_y=-2A\omega\sin(kx)\cos(\omega t)$$
Then, knowing that the wave number $k=\dfrac{2\pi}{\lambda}$ and that $\lambda=2L$, as you correctly stated, you obtain $k=\dfrac{\pi}{L}$. This gives
$$v_y=-2A\omega\sin\left(\dfrac{\pi}{L}x\right)\cos(\omega t)$$
As you can see, for $x=L/2$, the y-velocity won't always be zero. Similarly, the end points will always have a velocity of zero.
Your mistake was computational.
A: To find velocity the wave equation should be partially differentiated with respect to time and which will  give time variance under sine function and position variation also under sine function. So if you find it's velocity at l/2 it will give  non zero value unless the function incorporating time variation becomes zero.
