Transverse speed of a wave Equation of a wave is : $$y=Asin(kx-wt)$$ What is the maximum transverse speed?
To answer this we take the derivative of the above equation. $$dy/dt =v= Acos(kx-wt).(-w) = -wAcos(kx-wt).$$ Till now everything is fine. But when we say , the maximum transverse velocity is the amplitude of this equation therefore wA, what is the logic behind this?
 A: 
$dy/dt =v= Acos(kx-wt).(-w) = -wAcos(kx-wt)$

Here $v$ is the speed of the oscillating particle which are moving in the y-direction (transverse) whilst the wave is moving in the x-direction.
All the particles are oscillating with shm in the y-direction with an amplitude $A$ and a frequency $f = \frac {2\pi}{\omega}$.
The maximum speed of the particles is $\omega A$ which happens as they pass their equilibrium position when  
$y=0 =A\sin(kx-\omega t) \Rightarrow kx- \omega t =0 \Rightarrow \cos (kx -\omega t ) = \pm 1$.
The speed of the wave in the x-direction is $\frac {\omega}{k}$. 
The gif file shows the wave moving in the x-direction whilst the particle oscillates in the y-direction.

A: The cosine function goes from -1 to 1. Therefore, its maximum value is 1. Since the amplitude and the angular frequency​ don't change, it is this cosine which makes velocity vary. If we maximize the cosine, we maximize the velocity. Since the maximum value of the cosine is 1, we have that the maximum value of the velocity is the amplitude times the angular frequency times 1, which is the amplitude times the angular frequency itself. 
