# Do the terms “damping constant” and “damping coefficient” have standard uses?

I've heard the terms "damping constant" and "damping coefficient" used to describe both the $$c$$ from the viscous damping force equation $$F = -c\dot{x}$$ and the $$\gamma$$ from the definition $$\gamma = \frac{c}{2m}$$. Is there standard usage for each term, or is it context dependent? If it is context dependent, are there terms which uniquely refer to each of $$c$$ and $$\gamma$$?

• I think damping constant is defined as '$b$' in $F=-bv$ – HS Singh May 18 at 18:08
• Thanks for your comment, I accidentally forgot to include the dot above the $x$. Regarding $b$, I've not seen that used before: the only usage of $b$ in relation to oscillation that I think I've seen is shown [here][en.wikipedia.org/wiki/…. $F = -cv$ is used by [HyperPhysics][hyperphysics.phy-astr.gsu.edu/hbase/oscda.html] – VortixDev May 18 at 18:30
• The damping coefficient is $\gamma = \frac{c}{2 \sqrt{ m k}}$. Check your units and see that you are missing a term. – ja72 May 18 at 18:31
• @ja72 HyperPhysics (hyperphysics.phy-astr.gsu.edu/hbase/oscda.html) uses the definition as I have. I assume you were thinking of the damping ratio, which Wikipedia defines as you have (en.wikipedia.org/wiki/Damping_ratio)? – VortixDev May 18 at 18:37

## 2 Answers

The answer is yes. Talk to any engineer and if say the terms "damping constant" and "damping ratio" they know exactly what you mean without any further explanations.

• Damping coefficient $$c$$ signifies the contribution of velocity to force, as in $$F = \ldots + c \dot{x} + \ldots$$
• Damping ratio $$\zeta$$ is a number the signifies the region of damping. When $$\zeta<1$$ the problem is underdamped: when $$\gamma>1$$ it is overdamped: and when $$\zeta=1$$ it is critically damping. This means the form of the solution (as in the equation used to solve for motion) is different depending on the region.

The complementary terms of the above but with respect to stiffness are

• Stiffness coefficient $$k$$, which is the contribution of deflection to force, as in $$F = \ldots + k x + \ldots$$
• Natural frequency $$\omega_n$$ which combines stiffness and mass to tell you how fast the system responds to inputs.

In practive the above terms are using the transform the standard equation of motion $$m \ddot{x} = -k x - c \dot{x}$$ into one with known solutions. By substituting \begin{aligned} k & = m \omega_n^2 & c & = 2 \zeta m \omega_n \end{aligned} you get

$$\ddot{x} + 2 \zeta \omega_n \dot{x} + \omega^2_n x =0$$

with the well-known solutions

$$x(t) = \begin{cases} {\rm e}^{-\zeta \omega_n t} \left( A \cos\left( \omega_n \sqrt{1-\zeta^2} \right) + B \sin\left( \omega_n \sqrt{1-\zeta^2} \right) \right) & \zeta < 1 \\ {\rm e}^{-\zeta \omega_n t} \left( A \cosh\left( \omega_n \sqrt{\zeta^2-1} \right) + B \sinh\left( \omega_n \sqrt{\zeta^2-1} \right) \right) & \zeta> 1 \\ (A+t B) {\rm e}^{-\omega_n t} & \zeta = 1 \end{cases}$$

• Thanks for the answer. I notice that HyperPhysics (hyperphysics.phy-astr.gsu.edu/hbase/oscda.html) defines $\gamma$ as distinct from the damping ratio (which Wikipedia uses $\zeta$ to represent), and labels it "damping coefficient". UT Austin's page (farside.ph.utexas.edu/teaching/315/Waves/node10.html) uses the same definition of $\gamma$. Is $\gamma$ more commonly associated with the damping ratio, rather than the definitions used there? – VortixDev May 18 at 19:35
• Hyperphysics groups together $\gamma = \zeta\omega_n$. I change my notation in order to avoid confusion. The damping ratio is now $\zeta$. That $c^2 - 4 m k$ term they examine is looking at $\zeta$, since $c^2 = 4 \zeta^2 m k$. – ja72 May 18 at 19:57
• @Farcher - thank you. I edited the answer. – ja72 May 19 at 14:12

Lets see where the damping ratio $$\gamma$$ come from ?

Start with:

$$m\,\ddot{x}+c\,\dot{x}+k\,x=0\tag 1$$

where $$c$$ is the damping constant with the unit $$[N\,s/m]$$ and

$$k$$ is the spring constant with the unit $$[N/s]$$

to find a solution for equation (1) we make the ansatz: $$x=A\,e^{\lambda\,t}$$ and get:

$$m\,\lambda^2+c\,\lambda+k=0\tag 2$$

the solution for $$\lambda$$ is now:

$$\lambda_{1,2}=\frac{-c\pm\sqrt{c^2-4\,m\,k}}{2\,m}=-\frac{c}{2\,m}\pm\sqrt{\frac{c^2-4\,m\,k}{4\,m^2}}=-\frac{c}{2\,m}\pm \sqrt{\frac{k}{m}\left(\frac{c^2}{4m^2}\frac{m}{k}-1\right)}\tag 3$$

with $$\frac{k}{m}:=\omega_n^2$$ we get for equation (3) $$\lambda_{1,2}=-\frac{c}{2\,m}\pm\omega_n\sqrt{\frac{c^2}{4m^2\,\omega_n^2}-1}=-\frac{c}{2\,m}\pm\omega_n\sqrt{\gamma^2-1}\tag 4$$

so $$\gamma^2=\frac{c^2}{4m^2\,\omega_n^2}$$ and $$c=2\,\gamma\,\omega_n\,m$$

and equation (4) $$\mapsto$$ $$\lambda_{1,2}=-\gamma\,\omega_n\pm\,\omega_n\sqrt{\gamma^2-1}\tag 5$$ the solution for $$\lambda_{1,2}$$ depend on 2 parameters $$\omega_n$$ unit $$[1/s]$$ and $$\gamma$$ unit $$[-]$$

we get real solution for for $$\lambda_{1,2}$$ from equation (5) if $$\gamma^2 \ge 1$$ and a complex solution if $$\gamma^2 \lt 1$$, so I think it is just convenience to define $$\gamma^2=\frac{c^2}{4m^2\,\omega_n^2}$$.