Modelling: Second Order System

In summary, the conversation discusses finding differential equations to describe the motion of a second order system with a spring and damper. The equations for potential energy and damper dissipation are provided, and the discussion focuses on building a free body diagram for each mass and sorting out the signs of the forces. It is concluded that the forces from the spring and damper oppose the motion of mass m1 from left to right.
  • #1
MarkD88
2
0
I have attached an image of a question I am trying to do, I want to find the differential equations that describe the second order system in the image.

I know for a spring, potential energy = 1/2.K.x (where k is the spring constant, and x is the distance the spring is stretched).

I know that damper dissipation is = 1/2.B.(dx/dt) (where B is the damping constant and dx/dt is the rate of change of position with respect to time).

My solution:

I try to build a free body diagram for each mass on its own:

I have showed a snapshot of my attempt for the D.E. describing the motion of m1.

If any of you guys could tell me if I am correct or you have a handy way of doing these types of questions please feel free to respond.

Thank you
 

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  • Second Order System.JPG
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  • Snapshot.jpg
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  • #2
Hello Mark, welcome to PF :smile: !

For some reason the template has disappeared. Pity (*). The relevant equation is ##\Sigma F = ma##. So you want to sort out the signs a little more carefully (of the forces; ##-k## is not a force, nor is ##-B## ).(*) Its use is mandatory in PF, for very good reasons; see the guidelines.
 
  • #3
Hi BvU,

Apologies about the template, I will not post without a template again.

My notation there was a little careless, I did not mean to label K and B as forces, I was trying to imply that there is a force acting on mass m1 due to both the spring and the damper.

Spring: k.x1
Damper: B.(x2dot-x1dot)

And that these two forces oppose the motion of mass m1 from left to right.
 
  • #4
And that these two forces oppose the motion of mass m1 from left to right
is good. And when you state that, you should also write something like ##m_1\ddot x_1 = -kx_1## and not e.g. ##m_1\ddot x_1 -kx_1 = 0 ## :rolleyes:
 

Related to Modelling: Second Order System

1. What is a second order system?

A second order system is a mathematical model used to describe the behavior of dynamic systems that exhibit oscillatory behavior. It is characterized by two state variables and two poles in the transfer function.

2. What is the transfer function of a second order system?

The transfer function of a second order system is given by:
G(s) = K/((s^2 + 2ζω_ns + ω_n^2),
where K is the system gain, ζ is the damping ratio, and ω_n is the natural frequency.

3. How do you determine the stability of a second order system?

The stability of a second order system can be determined by analyzing the poles of the transfer function. If the poles are in the left half of the complex plane, the system is stable. If the poles are on the imaginary axis or in the right half of the complex plane, the system is unstable.

4. What is the significance of the damping ratio in a second order system?

The damping ratio, ζ, determines the rate at which the oscillations in the system will decay. A higher damping ratio results in faster decay and a lower damping ratio results in slower decay.

5. How is a second order system modeled in practice?

In practice, a second order system can be modeled using differential equations, state-space representation, or transfer function representation. These models can then be used to simulate and analyze the behavior of the system under different conditions.

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