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


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.


2 Answers 2


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

  • $\begingroup$ 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. $\endgroup$
    – alephzero
    Aug 2, 2019 at 0:12
  • $\begingroup$ 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. $\endgroup$ 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, $c_1 = c_2/2$.


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