Mass-two spring-damper system

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In summary: He needs to get matrix A and find x2 double derivate when it doesn't exist, if you know what I mean. He also needs to find c1 in the diagram.
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  • #2
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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  • #3
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.
 
  • #4
ja5 said:
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.
 
  • #5
So, I try with Laplace then?
 
  • #6
ja5 said:
So, I try with Laplace then?
Yes, I recommend that.
Post a sketch of the ( classical ) model.
 
  • #8
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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  • #9
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
 
  • #10
ja5 said:
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 "-".
 
  • #11
Hesch said:
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?
 
  • #12
LvW said:
* 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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  • #13
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.
 
  • #14
LvW said:
The procedure of block diagram reduction is a standard one and shouldn`t be a problem.
No, maybe it should not, but it is.
 
  • #15
That`s why I said that I will wait - and then I can offer my help (if needed).
 
  • #17
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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  • #18
Can you draw me whole loop?
 
  • #19
ja5 said:
Can you draw me whole loop?
Something like this:
 

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  • #20
but what is F2?
 
  • #21
ja5 said:
but what is F2?
It's the force at x2.
 

1. What is a mass-two spring-damper system?

A mass-two spring-damper system is a physical system that consists of two springs and two dampers connected to a mass. It is commonly used in engineering and physics to model the behavior of objects that are subject to forces from multiple springs and dampers.

2. How does a mass-two spring-damper system work?

In a mass-two spring-damper system, the two springs and two dampers work together to resist the motion of the mass. The springs provide a restoring force that is proportional to the displacement of the mass, while the dampers provide a force that is proportional to the velocity of the mass. This results in a system that can oscillate or dampen depending on the parameters of the springs and dampers.

3. What are the equations used to describe a mass-two spring-damper system?

The equations used to describe a mass-two spring-damper system are Newton's second law of motion (F=ma) and Hooke's law. These equations can be combined to create a differential equation that describes the motion of the mass in the system.

4. What factors affect the behavior of a mass-two spring-damper system?

The behavior of a mass-two spring-damper system is affected by several factors, including the stiffness and damping coefficients of the springs and dampers, the mass of the object, and the initial conditions of the system (such as initial position and velocity).

5. What applications is a mass-two spring-damper system commonly used for?

A mass-two spring-damper system is commonly used in engineering and physics to model various physical systems, such as vehicle suspensions, building structures, and oscillating pendulums. It is also useful for understanding and analyzing the behavior of systems in mechanical and electrical engineering.

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