Finding maximum displacement in viscously damped system with logarithmic decrement

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Dustinsfl
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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?
 
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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. :cool:
 
Dustinsfl said:

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


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 ...
 
NascentOxygen said:
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. :cool:

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.
 
Dustinsfl said:
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.

Welcome to the club! Happens a lot on these forums. :smile:
 
mx'' + Cx' + Kx = f(t)

K = mω²

Just check, but isn't C = mc?
 
Zeta = c/c_c

Where c is the damping coefficient and c_c is the critical damping so c=c_c*zeta
 
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?
 
NascentOxygen said:
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?

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

What "missing thousands"?
 
The factor that turns it into something realistic. :w