- #1
Dustinsfl
- 2,217
- 5
Homework Statement
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.
Homework Equations
The Attempt at a Solution
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?
