Viscously damped system

[Dustinsfl]

A viscously damped system has a stiffness of \(5000\) N/m, critical damping constant of \(0.2\) N-s/m, and a logarithmic decrement of \(2.0\). If the system is given an initial velocity of \(1\) m/s, determine the maximum displacement.

From the question, we have that \(k = 5000\), \(\delta = 2.0\), \(c_c = 0.2\), and \(\dot{x}(0) = 1\). I suppose we are also assuming then that \(x(0) = 0\) then for no initial displacement.

Then
\[
\zeta = \frac{\delta}{\sqrt{(2\pi)^2 + \delta^2}}\approx 0.303314
\]
and
\[
\zeta = \frac{c}{c_c}\Rightarrow c = c_c\zeta\approx 0.0606629
\]

Our general equation of motion is
\begin{align}
x(t) &= e^{-\zeta\omega_nt}\Bigg[x(0)\cos\bigg(\omega_nt\sqrt{1 - \zeta^2}\bigg) +
\frac{\dot{x}(0) + \zeta\omega_nx(0)}{\omega_n\sqrt{1 - \zeta^2}}\sin\bigg(\omega_nt\sqrt{1 - \zeta^2}\bigg)\Bigg]\\
&= e^{-\zeta\omega_nt}\frac{\dot{x}(0)}{\omega_n\sqrt{1 - \zeta^2}}\sin\bigg(\omega_nt\sqrt{1 - \zeta^2}\bigg)
\end{align}
Since \(c_c = 2\sqrt{km}\), \(m = \frac{c_c^2}{4k} = 2\times 10^{-6}\).

I feel wary of the mass being so small which leads to \(\omega_n = 50000\).

Then to find the maximum displacement, I set \(\dot{x} = 0\), and since this is an underdamped system, the max displacement will be at the first \(t\) critical for \(t > 0\).

So \(t_c = 0.000026501\) and \(x_{\max} = 0.0000133809\).

Is this correct is or something wrong or is this method incorrect?

 

[DaveE]

That all looks good to me. 2mg doesn't seems ridiculously small to me. For example small accelerometers, or mechanical (electronic) filters.