I'm interested in the maximum flow $(Q\max)$ requirements of a hydraulic system. The system moves an actuator in a sinusoidal pattern from position 0 to $\pm A$ where $A$ is the volume of oil in the actuator. As I understand it, to find the flow rate required at a time t, the derivative of $\sin(t), \cos(t)$ can be used. I would like to know the maximum possible flow rate needed by the system for a given amplitude.
Expanding on the first assumption, the general form of the sine wave form is $y = A\sin(wx)$ (ignoring phase shift and vertical offset) where $w$ is the angular frequency. From the following answer as the maximum of a cosine is 1, the maximum derivative of $\sin(x)$ is simpified to $Aw$. Noting that $w = \frac{2\pi}{\text{period}}$ and that I am interested a period of $10 s$, the maximum rate of flow is then simply the volume of oil multiplied by $\frac{2\pi}{10}$
Is this correct? I have tried to verify and all other guidance points to the maximum of the cosine being equal to the maximum of the sine. So a volume of $1L$ would require a maximum flow rate of $1\frac{L}{s}$. Or is the $Q\max = Aw$ approach correct? If we set the volume to 1 litre then the maximum flow rate is $2\pi$. This could make sense to me given that the period is also $2\pi$. And the flow rate has to be a function of the period somehow - an amplitude of 1 litre can have different flow rates depending on how quickly you need to shift it. I'm just looking for confirmation that it is as simple as $A\left(\frac{2\pi}{P}\right)$