# Homework Help: Viscous damped system

1. Sep 27, 2014

### Dustinsfl

1. The problem statement, all variables and given/known data
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.

2. Relevant equations

3. 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?

2. Sep 28, 2014

### Staff: Mentor

Check how you derived the formula for your mass calculation. I get hundreds of kg.

The effort you put into casting this in Latex is appreciated.

3. Sep 28, 2014

### rude man

I couldn't fid anything wrong including the mass and radian frequency calculations but I did not do the final part to find xmax (setting dx(t)/dt = 0 etc.). Straightforward but laborious ...

4. Sep 28, 2014

### Dustinsfl

How did you find the mass then? I ask because rude man doesn't see an issue so I have two different views of the same problem.

5. Sep 28, 2014

### rude man

Welcome to the club! Happens a lot on these forums.

6. Sep 29, 2014

### Staff: Mentor

mx'' + Cx' + Kx = f(t)

K = mω²

Just check, but isn't C = mc?

7. Sep 29, 2014

### Dustinsfl

Zeta = c/c_c

Where c is the damping coefficient and c_c is the critical damping so c=c_c*zeta

8. Sep 29, 2014

### Staff: Mentor

I think you'll find that where the coefficient of x'' is m, the coefficient of x' is mc and written as C.

Does the textbook give you their answers?

9. Sep 29, 2014

### rude man

If the coefficient of x'' is m the the coefficient of x' is c.
mx'' is force and so is cx'. And so are kx and any forcing function F(t).

10. Sep 29, 2014

### Staff: Mentor

You're right. So I need to find the missing thousands some other way. Dustinsfl, could you check the original question and see whether $c_c$ isn't actually given in N-s/mm? The textbook answer would be useful here.

11. Sep 29, 2014

### Dustinsfl

The units for c_c are in the problem statement post one. The book has no answer to this problem.

12. Sep 29, 2014

### Staff: Mentor

Similar questions in your textbook may reveal that $c_c$ is being given typically in kN-s/m (or N-s/mm), suggesting a typo in this question.

13. Sep 30, 2014

### rude man

What "missing thousands"?

14. Sep 30, 2014

### Staff: Mentor

The factor that turns it into something realistic. :w