Modeling a mass-spring-damper system

In summary: Yes, when m2 is displaced to the right, this lengthens the spring but also acts to shift m1 (in effect to contract the spring) so we compensate for that with the minus.
  • #1
VitaX
184
0

Homework Statement



EDMwuAy.png


Homework Equations



12b7d08830147608e122c8206841515d.png


The Attempt at a Solution



This is one of our past homework assignments with the solution given to us. I'm trying to work my way through each part of this assignment but am getting stuck on a few aspects of the model.

My approach:

I first set up a FBD modeling each mass in the system and analyzing the forces acting on that mass. So for mass 1 in part a) I would say there is the spring and damping force acting to the left of the mass and another spring force acting to the right of the mass. What I'm having a bit of difficulty understanding is the the displacement of each mass and how its oriented in the diagram. Do we always take the origin at the center of mass for the object? Is it wrong to place the origin at the wall where spring 1 is connected? I'm getting a little confused on such small matters like this but its effecting my ability to move forward. For instance in part b) I'm confused about the 2nd differential equation modeling mass 2. Why are both forces negative?
 
Last edited by a moderator:
Physics news on Phys.org
  • #2
Think of moving mass 2 a bit to the right. Don't both springs force it to the left?
 

Attachments

  • EDMwuAy.png
    EDMwuAy.png
    26.4 KB · Views: 1,505
Last edited by a moderator:
  • #3
Hmmm but how I understand the forces on springs in part b) is that the force from spring 1 acting on the wall is towards the right. Thus equal and opposite on mass 1 makes that spring force to the left. That's what I thought you do for spring 3 in part b) as well. You look at the force spring 3 is exerting on the wall and that should be to the left I thought? Then I would say equal and opposite force on mass 2 would make that spring force 3 acting on mass 2 towards the right. Am I looking at this completely wrong?
 
  • #4
You have to relate the spring forces to the displacements. If m2 displaces to the positive direction (to the right) it makes spring 3 shortened. The spring pushes both the wall and m2 away from its centre. The wall experiences a force to the right, and m2 experiences a force to the left from spring 3, opposite to its displacement.

ehild
 
  • #5
VitaX said:
Hmmm but how I understand the forces on springs in part b) is that the force from spring 1 acting on the wall is towards the right. Thus equal and opposite on mass 1 makes that spring force to the left. That's what I thought you do for spring 3 in part b) as well. You look at the force spring 3 is exerting on the wall and that should be to the left I thought? Then I would say equal and opposite force on mass 2 would make that spring force 3 acting on mass 2 towards the right. Am I looking at this completely wrong?

Never mind what the forces on the walls are. Look at the motion of the two masses. For each mass, if you push the mass to the right, both springs apply force to the left. One pushes, one pulls, but they both apply force to the direction opposite to motion.
 
  • #6
Ah now that makes sense. I should have thought of it like that. By the way, when dealing with the spring k2 in these problems, is the reason why you have to take the position x2 into account when looking at mass 1 because the masses are connected via that spring? I guess what I'm saying is at first I was a little confused why its written as k2(x2-x1) for the differential equation for mass 1.
 
  • #7
VitaX said:
Ah now that makes sense. I should have thought of it like that. By the way, when dealing with the spring k2 in these problems, is the reason why you have to take the position x2 into account when looking at mass 1 because the masses are connected via that spring?
Yes, when m2 is displaced to the right, this lengthens the spring but also acts to shift m1 (in effect to contract the spring) so we compensate for that with the minus.

I guess what I'm saying is at first I was a little confused why its written as k2(x2-x1) for the differential equation for mass 1.
The spring force on m2 due to k2 acts to the left. Since m1 and m2 are connected via k2, there is a force on m1 to the right due to this spring. That is why there is a plus '+k2(x2-x1)'
 
  • #8
VitaX said:
Ah now that makes sense. I should have thought of it like that. By the way, when dealing with the spring k2 in these problems, is the reason why you have to take the position x2 into account when looking at mass 1 because the masses are connected via that spring? I guess what I'm saying is at first I was a little confused why its written as k2(x2-x1) for the differential equation for mass 1.

Assume that the positive direction is to the right and x2>x1.
The increase of length of spring 2 is x2-x1. It exerts the force k(x2-x1) inward on the masses attached to the ends: So m1 experiences k(x2-x1) positive force (to the right) and m2 experiences -k(x2-x1) force (that is, force to the left).

Note that the spring forces on mass mi are of opposite sign as the "own" displacement xi and of the dame signs as the displacements of the other masses.

ehild
 

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

A mass-spring-damper system is a physical model used to represent the motion of an object that is connected to a spring and a damper (frictional force) in a linear fashion. This system is commonly used to study oscillations and vibrations in mechanical and electrical systems.

2. How do you model a mass-spring-damper system?

To model a mass-spring-damper system, you need to determine the mass of the object, the spring constant of the spring, and the damping coefficient of the damper. You can then use Newton's second law of motion and Hooke's law to create a differential equation that describes the motion of the system. This equation can then be solved to find the displacement, velocity, and acceleration of the object at any given time.

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

The behavior of a mass-spring-damper system is affected by several factors, including the mass of the object, the stiffness of the spring, the damping coefficient of the damper, and the initial conditions (displacement, velocity, and acceleration) of the object. Additionally, external forces and the properties of the environment in which the system is located can also impact its behavior.

4. What is the difference between an underdamped, critically damped, and overdamped mass-spring-damper system?

An underdamped mass-spring-damper system oscillates with decreasing amplitude and takes longer to return to its equilibrium position. A critically damped system returns to its equilibrium position quickly without any oscillations. An overdamped system takes the longest to return to its equilibrium position and does not undergo any oscillations.

5. What are the practical applications of modeling a mass-spring-damper system?

Modeling a mass-spring-damper system has various practical applications, including designing and analyzing suspension systems in cars and buildings, studying the behavior of electrical circuits, and creating mechanical and electronic devices that involve oscillations or vibrations. It is also used in fields such as robotics, aerospace engineering, and seismology.

Similar threads

  • Introductory Physics Homework Help
Replies
31
Views
972
  • Introductory Physics Homework Help
Replies
3
Views
830
  • Introductory Physics Homework Help
Replies
2
Views
821
  • Introductory Physics Homework Help
Replies
17
Views
238
  • Introductory Physics Homework Help
Replies
3
Views
705
  • Introductory Physics Homework Help
Replies
1
Views
907
  • Introductory Physics Homework Help
Replies
17
Views
280
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
2
Replies
53
Views
4K
  • Introductory Physics Homework Help
Replies
8
Views
825
Back
Top