# How is the simple damper equation derived?

I know the spring is modeled as $$F_{\text{elastic}} = k\cdot x$$ when the displacements are small since this is empirically based, but what happens with $$F_{\text{damping}}=c\cdot\dot{x}$$? It is the same? Or it is derived from theory?

• Note that $F_\mathrm{elastic}=k\cdot x$ can be derived. Jan 18, 2019 at 16:31

A damping force opposes motion, i.e. it points in the direction $$-\Delta\mathbf{r}$$.
This is not the same as a restoring force, which always points in the direction $$-\mathbf{r}$$, the latter being position.

The easiest way to mathematically convey a force in the direction $$-\Delta\mathbf{r}$$ is to note that the velocity is defined as $$\mathbf{v} = -\frac{\Delta\mathbf{r}}{\Delta t}$$.

So "damping" is usually just written off as some constant times the velocity, $$\mathbf{F}_{\text{damping}} = -c \mathbf{v}$$.

Other times, damping (or drag) is written as $$\propto v^2$$. In which case I would recommend looking here.

• But opposition to motion points in the direction of $-\mathbf{\dot{r}}$, not $-\mathbf{\Delta r}$. If I displace the mass and then forcefully push it back to its equilibrium position, for example, then the damping force points in the direction of $-\dot{\mathbf{r}}$ but $\mathbf{+\Delta r}$. Jan 18, 2019 at 16:37
• @Chemomechanics Look at the definition of $\vec{\Delta r}$ in the answer. It's in the same direction as $\vec{\dot{r}}$. $\vec{\Delta r}$ is not measured from the origin, but from an immediately previous location. Jan 18, 2019 at 17:07

The system which are considering is the mass $$m$$ and it has three external forces acting on it.

I know the spring is modeled as $$F_{\text{elastic}} = k\cdot x$$.

The force on the mass due to the spring is $$-kx$$.
When the extension $$x$$ is positive (assumed downwards from the diagram) the spring is pulling the mass up.

The damping force opposes the motion so when the mass is moving down with $$\dot x$$ positive the force opposing the motion must be upwards.
So the retarding force is $$-c\dot x$$.
This is the right form for this equation because when the mass is moving up $$\dot x$$ is negative then \$-c\dot x is positive ie downwards and so still opposing the motion.

To set up the equation of motion using Newton’s second law you should gather up all the forces on the left hand side

$$F_0\sin \omega t -kx -c\dot x = m\ddot x$$

and put this net force equal to the mass times its acceleration.