Viscously damped system

In summary, the question asks to determine the maximum displacement of a viscously damped system with a stiffness of \(5000\) N/m, critical damping constant of \(0.2\) N-s/m, and a logarithmic decrement of \(2.0\), given an initial velocity of \(1\) m/s. By calculating the damping ratio and using the general equation of motion for an underdamped system, the maximum displacement is found to be \(0.0000133809\) m at a time of \(0.000026501\) s.
  • #1
Dustinsfl
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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?
 
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  • #2
That all looks good to me. 2mg doesn't seems ridiculously small to me. For example small accelerometers, or mechanical (electronic) filters.
 

FAQ: Viscously damped system

What is a viscously damped system?

A viscously damped system is a mechanical system that experiences resistance to motion due to the presence of a viscous fluid, such as oil or air. The resistance is proportional to the velocity of the system and is typically represented by a damping coefficient.

How does a viscously damped system behave?

A viscously damped system will exhibit a decaying oscillatory motion in response to an external force. The damping force reduces the amplitude of the oscillations over time, eventually bringing the system to rest.

What factors affect the damping in a viscously damped system?

The damping coefficient, the mass of the system, and the viscosity of the fluid all affect the amount of damping in a viscously damped system. A higher damping coefficient or a more viscous fluid will result in a greater amount of damping.

How is damping ratio related to a viscously damped system?

The damping ratio is a dimensionless parameter that describes the amount of damping in a system. In a viscously damped system, the damping ratio is equal to the damping coefficient divided by the critical damping coefficient, which is the minimum amount of damping needed to prevent oscillations.

What are the applications of viscously damped systems?

Viscously damped systems are commonly used in engineering and physics to control the motion of structures and machines. They are also used in seismology to dampen the effects of earthquakes on buildings and other structures. In addition, viscously damped systems are used in shock absorbers and other vibration control devices.

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