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