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Mass-two spring-damper system

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ja5

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Hesch

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MDS.jpg

I assume that F is the input to the system, and that you will have to find x1, x2 ( velocities? ) as a function of F.

If you are confident with Laplace transforms, I'd build a model by means of Laplace function blocks.
 
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ja5

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After I gotten this equation, I tryed to excrete x1, but I'm missing one equation to make a state space model.
I think I need to get matrix A 4x4 and I don't know how to get x2 double derivate when it doesn't exist, if you know what I mean.
 

Hesch

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but I'm missing one equation to make a state space model.
I think that state state space is effective by multiple input/output systems, but you have a system with only one input ( F ), thus a classical model is more effective here.
( And, yes, I have to admit that I have only limited experience with state space ).
When using the classical model, you can actually "see" what you are doing. In a 4x4 matrix, a certain number just disappears in the amount of numbers.
 

ja5

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So, I try with Laplace then?
 

Hesch

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Hesch

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No, this is not made by function blocks, here is what I mean:
upload_2015-6-20_15-29-43.jpeg

You draw these blocks/loops and reduce them afterwards. For example the 4 blocks in lower left corner can be reduced to one block:

k2/s+c2

The loops can be reduced by Masons rule.

At last you will have the transfer functions: x1(s)/F(s) and x2(s)/F(s). For example you put on some step function: F(s) = 1/s, and you will find x1(s) and x2(s).
 
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ja5

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Alright, I know what you mean, but how will I draw a loop to apply mason rule, I get confused when I need to put x2
 

Hesch

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how will I draw a loop to apply mason rule
Say you have a closed loop (negative feed back).
Say that the forward path has the transfer function: A(s) and that the backward path has the transfer function: B(s), then Masons rule says that the transfer function for the loop as a whole =

A(s) / ( 1 + A(s)*B(s) ) . . . . ( It's easy to prove ).

If the feedback is positive, you simply change the "+" to a "-".
 

LvW

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Say you have a closed loop (negative feed back).
Say that the forward path has the transfer function: A(s) and that the backward path has the transfer function: B(s), then Masons rule says that the transfer function for the loop as a whole =
A(s) / ( 1 + A(s)*B(s) ) . . . . ( It's easy to prove ).
If the feedback is positive, you simply change the "+" to a "-".
Perhaps not too important for solving this task - but for the sake of clarity I like to mention that the given formula for the closed-loop transfer function is known as Blacks formula; Mason`s rule is different from that.
Questions:
* What is x2? Input or output (according to your block diagram)?
* Where is c1 in the diagram?
 

Hesch

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* What is x2? Input or output (according to your block diagram)?
* Where is c1 in the diagram?
x2 is the velocity(?)/position(?) in the diagram shown in #2. This diagram is just a copy from one of the links in #1. I don't know, I have assumed it's an output, and that F is the input.

c1 ( and k1 ) is coming up when the diagram is expanded in #8 to the right: ( "Finding x2" reference ).

The diagram in #8 has just not been finished ( something left for ja5 to do ).
 
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LvW

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OK - so I wait until it will be finished. The procedure of block diagram reduction is a standard one and shouldn`t be a problem.
 

Hesch

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The procedure of block diagram reduction is a standard one and shouldn`t be a problem.
No, maybe it should not, but it is.
 

LvW

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That`s why I said that I will wait - and then I can offer my help (if needed).
 

Hesch

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Capture.jpg

This is not right. By Laplace it could be written: k2*s*x2 = k2*s*x1 + c2(x1-x2) , and that's a 1. order system. A spring and a mass will oscillate which means that the system must be a 2. order system. You may not assume a mass (A), when it is not there.

In the (half) model in #8, you must complete the model. Then you can see what you are doing. Complete it and reduce it.

In #8 a force, F, is induced. Dividing this force by m you will get the acceleration of m: ( dx1/dt ). If you integrate this acceleration ( divide by s ) you will get x1.
The back-force ( as to F ) will be induced by:

1) by c2: Fback = x1*c2.
2) by spring: Fback = (x1/s)*k2 : Dividing x1 by s, you integrate velocity to position, thus Fback = Δposition * k2.

But in (2) the velocity is missing: Correctly it must be: Fback = ( (x1-x2)/s ) * k2.

The Fback is subtracted from F ( left sum block ) so that the resulting force, pushing m, is: F - Fback. ( Loop closed ).
 
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ja5

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Can you draw me whole loop?
 

ja5

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but what is F2?
 

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