# Dimensonal analysis of damping constant?

What are the units of the damping constant from the following equation by dimensional analysis?

$$\zeta = \frac{c}{2\sqrt{mk}}$$

I'm assuming the units have to be s^-1, as the damping constant is present in the exponential equation which plots damping of y=Ae^kt (which plots amplitude vs time). Is that a correct assumption?

If somebody could do a quick dimensional analysis to confirm it would be great.

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Maybe you could tell, what $c$, $m$, $k$ and $\zeta$ are? (once you know what these other constants are you might be in the position to answer the question yourself) – Fabian May 12 '11 at 8:43

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Is this homework? – Georg May 12 '11 at 8:44
Well mass is kg, k is spring constant (N/m) and ζ is obviously damping constant. I have no idea what the units of c area however. – user3511 May 12 '11 at 9:19

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@user3511 If you know what dimensional analysis is, you can easily perform it yourself. If you do not know what it is, use the internet to learn. If you try to use the internet to learn, and run into trouble because there's something about it you don't understand, ask a question here. As it is, your question makes no sense because you just threw up an equation with undefined quantities, and when someone asked you what the variables meant, you replied that you didn't know. Why are you asking about an equation with quantities you can't even identify? -1 for lack of effort. – Mark Eichenlaub May 12 '11 at 10:26
My apologies, i needed help in confirming the steps and the units of each. Next time I will post them in the question. – user3511 May 12 '11 at 11:43

## 2 Answers

I'm assuming the units have to be s^-1, as the damping constant is present in the exponential equation which plots damping of y=Ae^kt (which plots amplitude vs time). Is that a correct assumption?

No, the damping ratio $\zeta$ is dimensionless:

$$[c] = \frac{[F]}{\left[\frac{dx}{dt}\right]} = \frac{\mathrm{N}}{\mathrm{m}\cdot\mathrm{s}^{-1}} = \frac{\mathrm{kg}\cdot\mathrm{m}\cdot\mathrm{s}^{-2}}{\mathrm{m}\cdot\mathrm{s}^{-1}} = \mathrm{kg}\cdot\mathrm{s}^{-1}$$

$$[\zeta] = \frac{[c]}{\sqrt{[m][k]}} = \frac{\mathrm{kg}\cdot\mathrm{s}^{-1}}{\sqrt{\mathrm{kg}\cdot\mathrm{N}\cdot\mathrm{m}^{-1}}} = \frac{\mathrm{kg}\cdot\mathrm{s}^{-1}}{\sqrt{\mathrm{kg}^2\cdot\mathrm{s}^{-2}}} = \frac{\mathrm{kg}\cdot\mathrm{s}^{-1}}{\mathrm{kg}\cdot\mathrm{s}^{-1}} = 1$$

The solution of the damped harmonic oscillator differential equation (when underdamped) is

$$x(t) = A e^{-\zeta \omega_0 t} \ \sin \left( \sqrt{1-\zeta^2} \ \omega_0 t + \varphi \right)$$

so the exponent is dimensionless (as it must be):

$$[\zeta \omega_0 t] = 1\cdot\mathrm{s}^{-1}\cdot\mathrm{s} = 1$$

Dimensionless and dimensionful parameters

The differential equation for a damped harmonic oscillator is

$$m\frac{d^2x}{dt^2} + c\frac{dx}{dt} + kx = 0$$

We can reduce the number of parameters to 2 just by dividing by $m$

$$\frac{d^2x}{dt^2} + \frac{c}{m}\frac{dx}{dt} + \frac{k}{m}x = 0$$

Then we can transform the two remaining parameters to get a dimensionless one, controlling the shape of the solution, and a dimensionful one, setting the timescale. One way of doing that is to define

$$\omega_0 = \sqrt{\frac{k}{m}}$$

$$\zeta = \frac{\frac{c}{m}}{\omega_0} = \frac{c\sqrt{m}}{m\sqrt{k}} = \frac{c}{\sqrt{k\,m}}$$

so that the differential equation takes the form:

$$\frac{d^2x}{dt^2} + \zeta\omega_0\frac{dx}{dt} + \omega_0^2x = 0$$

The reason to choose $\omega_0$ as the dimensionful parameter is physical: when the system is underdamped, $\omega_0$ is the angular frequency of oscillation.

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I was posting the same thing, but you beat me to it! +1. For a perfect answer, you could add something about the characteristic equation. – Gerben May 12 '11 at 23:01
Thanks, thats great! Can you explain further why the damping constant will be dimensionless in terms of the differential equation? (I don't quite understand the last line). – user3511 May 13 '11 at 8:59
@user3511 I added some more details. – mmc May 13 '11 at 16:43
– user3511 May 15 '11 at 5:57
@user3511 The $\gamma$ shown there is just $\zeta\,\omega_0$. So, while $\zeta$ is dimensionless, $[\gamma] = \mathrm{s}^{-1}$. The same solution can be expressed in many different ways, emphasizing different things :-) – mmc May 15 '11 at 15:04
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The above is unit less. How? well damping is always force/speed thus $c=[\rm{N}\,\rm{s}\,\rm{m^{-1}}]$, and stiffness is force/distance $k=[\rm{N}\,\rm{m^{-1}}]$, and of course a newton is $[\rm{N}]=[\rm{kg}\,\rm{m}\,\rm{s^{-2}}]$

Combine them all to make

$$\dfrac{\mathrm{N\,/(m/s)}}{\sqrt{{\rm kg\, N/m}}}=\sqrt{\dfrac{{\rm N}\,{\rm s}^{2}}{{\rm kg\, m}}}=\sqrt{\dfrac{{\rm kg\, (m/s^{2})\, s^{2}}}{{\rm kg\, m}}}=1$$

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