# Homework Help: Mass - Spring - damper in Parallel

1. May 20, 2014

### kjay262

1. The problem statement, all variables and given/known data
The problem is to determine the transfer function where force F is input and displacement x is output in the mass-spring-damper mechanism.

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

3. 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.

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• ###### 10273955_10154109711405307_3235781433148029459_n.jpg
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2. May 20, 2014

### kjay262

****EDIT**** not in parallel, IN SERIES

3. May 20, 2014

### Staff: Mentor

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

4. May 20, 2014

### kjay262

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: May 20, 2014
5. May 20, 2014

### kjay262

Inverse Laplace Identity

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

#### Attached Files:

• ###### Screenshot 2014-05-21 13.11.30.png
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6. May 20, 2014

### Staff: Mentor

Multiply numerator and denominator by cs+k. See if this simplifies things.

Chet

7. May 20, 2014

### kjay262

What if you have a situation where $\xi$ 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. May 21, 2014

### Staff: Mentor

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.