Suppose we have a mass $m$, connected to a spring with spring stiffness $k$, and that there exists a dampening force $F$ acting against the motion of the mass. The mass is only able to oscillate in one dimension, up or down. Suppose we define our coordinate system such that positive is upwards in let's say our variable $y$, and that our origin lies exactly in equilibrium.
We have four cases, first, the mass is above our origin with positive velocity. Newtons eq. yields us that $m\ddot{y} = -F-ky \Leftrightarrow F = -(m\ddot{y}+ky) $, similarly, if the mass is above our origin with negative velocity, we get: $m\ddot{y} = F-ky \Leftrightarrow m\ddot{y} + ky = F$. If the mass is below our origin with positive velocity, then, by the argument above, $F = (-m\ddot{y}+ky)$, and for the last case $F = -(-m\ddot{y}+ky)$
So, all in all, we should be able to describe $F = -\operatorname{sgn} (\dot{y} y)(\operatorname{sgn}(y) m\ddot{y}+ky)$
Is this correct? If not, why? What makes me doubt this is whether I should replace $y$ and $\ddot{y}$ with its absolute values inside the second parenthesis.
Thanks.
There are two forces, a Hookean restoring force toward the origin $- ky$, and a Coulomb friction force opposite to the direction of motion: $ - F \text{sgn}(\dot{y})$. Here, $F$ is a friction coefficient. Therefore, Newton's second law takes the form $$ m \ddot{y} = - k y - F \text{sgn}(\dot{y}).$$ This is a challenging non-linear differential equation for the displacement $y$ which is closely related to a "stick-slip system".