Help on a two masses - two springs system

In summary, the system behaves as if there is a single spring between the two masses. The stiffness coefficient of the new spring is determined by the equation for the new k.
  • #1
gushazm
6
0
Hi, I need to know where I can find information on the web so as to analytically solve the system on the attached jpg where:

m1, m2 are the values of each mass
v1, v2 are the speeds of each mass
K1, K2 are the stiffness coefficient of each spring
L1, L2 are the lenghts of each spring.

I am assuming a 1D motion.

The system is a VERY simplified model of a head-on collision between two automobiles. I am an engineer from Argentina and I need to analyze for a given time (t) the values of v1 and v2, and the values of L1 and L2.

I would greatly appreciate any kind of help you can give me. I wish you a very good year.
 

Attachments

  • Masas y resortes 4.JPG
    Masas y resortes 4.JPG
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  • #2
Further work on the model

Hi, I did some further work on the model, and modified it by adding a mass in front of each spring, so that in the moment of the collision the system would look as in the attached figure.

Now I have an important question to ask to anyone who is interested in the subject. Is it right to suppose that while v1 and v2 are greater than 0, mass 3 does not move? (this is so no matter what are the values of k1 and k2?)

The second supposition after this one is that after v1 or v2 become 0, the system behaves itself as a reduced mass one. Is this right too?

I would greatly appreciate any help. Thank you
 

Attachments

  • Gráfico sistema masas y resortes 3.JPG
    Gráfico sistema masas y resortes 3.JPG
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  • #3
Equations of forces:
m1d2x1/dt + k1x1+m2d2x2/dt + k2x2 = 0
x1(0) = L1
x2(0) = L2
dx1/dt(0) = V1
dx2/dt(0) = V2

Equations of energy:
Before the collision:
1/2m1V12 + 1/2m2V22

After the collision:
1/2m1v12 + 1/2m2v22 +
1/2K1(L1-x1)2
1/2K2(L2-x2)2
 
  • #4
Thanks

I thank you very much for your reply SGT. I will do some calculations and ask again if any questions arise.
 
  • #5
I forgot to say that v1 = dx1/dt and v2 = dx2/dt are time varying quantities, whereas V1 and V2 are the initial values of those quantities.
 
  • #6
Hi, gushazm; welcome to PF!
I assume that you're interested in finding a typical measure of the deformation of each car (or what?).
Although not precisely the same, this thread touches upon issues relevant for that concern:
https://www.physicsforums.com/showthread.php?t=60323
Feel free to pose further questions.
 
  • #7
Thanks again

Thank you Arildno. What I am really wanting to know is which one of the two vehicles stops first, an their speeds at any moment. On the one hand, if the total kinetic energy of the system if lower than the potential energy that the system of springs is able to absorb, one of the vehicles is going to "push" the other one (because it has some remaining kinetic energy) and I would like to know the extent of the change of speed. On the other hand (an event that happens at high-speed collisions) when the system of springs is unable to absorb the whole kinetic energy of the system, an elastic collision happens (or the deformation extends to the cokpit which I am not modelling right now). In this second case I also want to know the remaining kinetic energy (basically through the speeds of each mass) to calculate the delta v, and/or the energy that will deform the cockpit. Therefore, I need a system of equations that help me to calculte for any given time the position of the masses (or the length of each spring) and their speeds. Thank you again for your help. Kind Regards.
 
  • #8
Further help

SGT: As I said I am grateful for your help. Yet, I have been working on the equations you posted, but I cannot obtain an eauation for x1(t), x2(t) and v1(t), v2(t) from the mentioned sistem of equations. I would appreciate very much further help on the issue. Regards.

Arildno: I analyzed the other example in the thread you told me. But I have a doubt: is the system I am talking about behaving as a system formed by two masses and a single spring formed by the two original springs? In that case, what would be the resulting stiffness coefficient of the new spring? I found some bibliography on the matter but I am not sure that the two springs will behave as a single one (if I use the equation for the new k I found, the resulting k is lower than the two original ones. This means that if two equal vehicles traveling at the same speed sustain a head on collision, each one of them will have a different change of speed or possition than the one the would have in case of colliding against a fixed objet, and this does not seem right to me). Regards.
 
  • #9
gushazm
I will try to elaborate a little more on the equations.
About your observation to Arildno for the combination of ks, I think the resulting coefficient should really be less then each of the originals. Since I am an EE, I think always in terms of electrical components. Springs in series behave as capacitors in series. The inverse resulting k (or C ) is the sum of the inverses of the components in series.
And yes, the two vehicles colliding will have different variation of velocities than if colliding with a fixed object.
 
  • #10
Deriving the equation of energy you have:
m1v1dv1/dt + m2v2dv2/dt - k1(L1-x1)dv1/dt - k1(L1-x1)dv2/dt = 0

You can solve for dv1/dt = d2x1/dt2 and replace in the equation of forces to have a closed form.
This nonlinear equation has no analytical solution, but you can solve it numerically.
 
  • #11
Thanks

Thanks SGT!, you've been very helpful. I will try to solve the equations by iteration on the weekend.
 

1. What is a two masses - two springs system?

A two masses - two springs system is a physical system consisting of two masses connected by two springs. The system is often used to study harmonic motion and oscillations in physics.

2. What are the equations of motion for a two masses - two springs system?

The equations of motion for a two masses - two springs system can be derived from Newton's second law of motion, where the forces acting on each mass are equal to the mass times its acceleration. These equations are often represented as coupled differential equations.

3. How do the masses and springs affect the motion of the system?

The masses and springs in a two masses - two springs system affect the motion of the system by determining the stiffness and natural frequency of the oscillations. Heavier masses and stiffer springs will result in a higher natural frequency and faster oscillations.

4. Can the two masses have different values?

Yes, the two masses in a two masses - two springs system can have different values. This will affect the overall behavior of the system, as the heavier mass will have a greater influence on the motion of the system.

5. How can a two masses - two springs system be used in real-world applications?

A two masses - two springs system can be used in various real-world applications, such as modeling the motion of a car's suspension system or studying the behavior of a double pendulum. It can also be used as a simplified model for systems with multiple degrees of freedom.

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