Mass - Spring - damper in Parallel

In summary, the conversation is about determining the transfer function for a mass-spring-damper mechanism, where force is the input and displacement is the output. The conversation includes discussions on the equations involved such as spring force, damping force, and force, as well as the attempt at finding a solution and using the inverse Laplace identity to determine the output in the time domain. There is also a mention of how to represent a situation where the damping ratio is greater than one in a graph.
  • #1
kjay262
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Homework Statement


The problem is to determine the transfer function where force F is input and displacement x is output in the mass-spring-damper mechanism.

Homework Equations


Spring Force = kx [k:spring constant]
Damping Force = Cx [C:damping coefficient]
Force = (Mass)(acceleration)

The Attempt at a Solution


Attempt at solution is in picture. I am interested to know if I am following the correct methodology and if I am missing anything.
 

Attachments

  • 10273955_10154109711405307_3235781433148029459_n.jpg
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  • #2
****EDIT**** not in parallel, IN SERIES
 
  • #3
The differential equations look correct. That was the hard part. I haven't look over the part about the development of the transform, but that shouldn't have been a problem.

Chet
 
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  • #4
If the input force is a unit impulse the transfer function is equal to the output displacement in the complex domain. I'm struggling to determine the x(t) that is the output in the time domain, which is basically the inverse laplace of the transfer function. I using the identity (attached below), however the answer introduces a complex value, am I missing something.
 
Last edited:
  • #5
Inverse Laplace Identity

Using the identity attached to determine the inverse laplace of the transfer function
 

Attachments

  • Screenshot 2014-05-21 13.11.30.png
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  • #6
Multiply numerator and denominator by cs+k. See if this simplifies things.

Chet
 
  • #7
What if you have a situation where [itex]\xi[/itex] is greater than one (according to the identity image attached above) this would result in a complex number, yes? How would that be represented in graph?
 
  • #8
kjay262 said:
What if you have a situation where [itex]\xi[/itex] is greater than one (according to the identity image attached above) this would result in a complex number, yes? How would that be represented in graph?
Factor the denominator using the quadratic formula, and then resolve the transform into partial fractions, and you will then be able to answer your own question.
 
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1. What is a mass-spring-damper system in parallel?

A mass-spring-damper system in parallel is a mechanical system that consists of a mass, a spring, and a damper connected in parallel. This means that the components are arranged in such a way that the same input force is applied to all three components simultaneously.

2. How does a mass-spring-damper system in parallel behave?

A mass-spring-damper system in parallel behaves according to Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement of the mass from its equilibrium position. The damper, on the other hand, dissipates energy and reduces the amplitude of the oscillations.

3. What is the equation of motion for a mass-spring-damper system in parallel?

The equation of motion for a mass-spring-damper system in parallel is given by F = m\u0394x + kx + c\u0394x, where m is the mass, k is the spring constant, c is the damping coefficient, x is the displacement of the mass from its equilibrium position, and F is the external force applied to the system.

4. What is the natural frequency of a mass-spring-damper system in parallel?

The natural frequency of a mass-spring-damper system in parallel is given by \u03C9n = \u221A(k/m). This is the frequency at which the system will oscillate without any external forces acting on it.

5. How does changing the parameters of a mass-spring-damper system in parallel affect its behavior?

Changing the parameters of a mass-spring-damper system in parallel can have a significant impact on its behavior. Increasing the mass or the spring constant will result in a lower natural frequency and slower oscillations. Increasing the damping coefficient will result in faster damping and a decrease in the amplitude of the oscillations. Similarly, changing the external force will affect the amplitude and frequency of the oscillations.

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