# Maximum Upwards Velocity of a wave In a transverse wave the particle motion is perpendicular to the wave direction. I understand that the particles oscillate from a point of equilibrium, and in a transverse wave, up and down. On a displacement distance graph, if you draw two waves, one relatively soon after the other, you can see in which direction the particles are travelling. However, I am not quite wrapping my head around the idea that when the particle crosses the equilibrium point, it is at its maximum velocity, especially since there is no mention of time in the graph. I can see that at that point is the greatest gradient of the graph, but I have not found any resource that tells me that this is the reason why it is at its greatest velocity at that point. Furthermore, when a point is pictured as the next one after after a crest that is at the equilibrium point, why is this the maximum upwards velocity rather than the maximum downards velocity?

I am referring to points P and Q

At $$Q$$ the particle is at rest so it has no kinetic energy and has maximum potential energy.
At $$P$$ and $$R$$ - the particle's (static) equilibrium position - the particle has a minimum potential energy and hence the kinetic energy of the particle, and hence its speed, is a maximum.
If the up and down arrows at positions $$P$$ and $$Q$$ on your diagram to show the velocity directions of the particles had not been there then there is no way of ascertaining as to which was the direction of travel of the wave.
As shown with the velocity arrow pointing downwards at position $$P$$ at little later in time the particle would be positioned below the axis and the only way that could have happened is if the wave profile had moved to the right - ie it is a right-travelling wave.