# How to calculate the viscous damping coefficient given drag force

Force due to drag at low velocities, is equal to some constant times negative velocity

$$F_{d}=-c_{1}V$$

The viscous damping coefficient equals decay constant divided by 2 times mass

$$\gamma = \frac{c_{2}}{2m}$$

So, is $$c_{1}$$ the same as $$c_{2}$$?

How is the drag force related to the viscous damping coefficient, what equation is there to relate them? I think the relationship is linear but I'm not certain.

For context, this is for a mass-spring system inside a beaker of water being damped by the friction of the water.

What you have is not actually fully submersed drag equation, or else it would be correct to handle it as damping.

The drag force on a submersed streamlined object is:$$F_D=\frac{1}{2} \rho\cdot u^2C_DA$$

• $$F_{D} =$$ the drag force

• $$\rho=$$ mass density of the fluid

• $$u=$$ the flow velocity

• $$C_D=$$ the drag coefficient (function of the Reynold's number)

• $$A=$$ is the crosssectional area

• This is not true at low Reynolds numbers, where the drag is proportional to Re and therefore to $v$. Of course you can include that in your formula by making $C_D$ proportional to $1/u$ as $u$ tends to 0, but that isn't very enlightening. Aug 2, 2019 at 0:12
• OP has skipped a couple of steps ahead of this and already decided that they're in a regime where $C_{\textrm{D}} \sim 1/\textit{Re}$, so that drag force is proportional to velocity. Jan 18 at 9:39

If it really is the case that $$F_{\textrm{d}} = -c_1V$$, with constant $$c_1$$, then there's an acceleration $$\mathsf{d}V/\mathsf{d}t =-\left(c_1/m\right)V$$. That differential equation is solved by an exponential decay with decay constant $$c_1/m$$. So it looks like, by the definition of $$c_2$$ you provide, $$c_1 = c_2/2$$.