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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.

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3 Answers 3

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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

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    $\begingroup$ 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
    Commented Aug 2, 2019 at 0:12
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    $\begingroup$ 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. $\endgroup$
    – user28774
    Commented Jan 18, 2022 at 9:39
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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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The definition or point you have made is wrong.

The viscous damping coefficient equals decay constant divided by two times mass
$$\gamma = \frac{c_{2}}{2m}$$

The decay constant or damping coefficient $(\gamma)$ equals the ratio of the viscous damping constant to two times the mass. Note that some authors may use terms differently. $$\color{red}{\boxed{\text{Decay constant }=\frac{\text{Damping constant }}{2\times\text{mass }};\ \gamma=\frac{c_2}{2m}}}$$

If the magnitude of the velocity is small, meaning the mass oscillates slowly, the damping force is proportional to the velocity and acts against the direction of motion. $$\large \color{red}{\boxed{F_d=-c_1V=-c_1\dot x}}$$ The constant $c_1$ is called the "damping constant". Both $c_1$ and $c_2$ are essentially the same.

To understand why $c_2=c_1$, consider a damped free vibration. The equation of motion is derived using Newton's second law of motion: $m\ddot x=F_d+F_s=-c\dot x-kx$.

$$\large m\ddot x+c\dot x+kx=0\implies\ddot x+2\zeta\omega_n\dot x+\omega_n^2x=0$$ where

  • damping ratio, $\zeta=\frac{c}{c_c}=\frac{\text{damping coefficient}}{\text{critical damping coefficient}}=\frac{c}{2\sqrt{km}}=\frac{c}{2m\omega_n}$
  • natural frequency, $\omega_n=\sqrt{\frac km}$

The underdamped solution can be expressed in the form: $$\large x(t)=A_0e^{-\zeta\omega_nt}\cos\left(\omega t-\varphi\right)=A_0e^{-\gamma t}\cos\left(\omega t-\varphi\right)$$ As you can see, the decay constant $\gamma=\zeta\omega_n=\frac {c}{2m}$. It's also called the damping coefficient by some authors. And $c$ in $F_d=-cv=-c\dot x$ is called the damping constant. One may call $c$ as the damping coefficient as well. So, it's essential to know what notation is used for a term. It depends on the context and authors.

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