# Help on a two masses - two springs system

1. Dec 21, 2004

### gushazm

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.

#### Attached Files:

• ###### Masas y resortes 4.JPG
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2. Jan 17, 2005

### gushazm

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

#### Attached Files:

• ###### Gráfico sistema masas y resortes 3.JPG
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3. Jan 18, 2005

### SGT

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. Jan 18, 2005

### gushazm

Thanks

I thank you very much for your reply SGT. I will do some calculations and ask again if any questions arise.

5. Jan 19, 2005

### SGT

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. Jan 19, 2005

### arildno

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:
Feel free to pose further questions.

7. Jan 19, 2005

### gushazm

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 lenght of each spring) and their speeds. Thank you again for your help. Kind Regards.

8. Jan 28, 2005

### gushazm

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 travelling 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. Jan 28, 2005

### SGT

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. Jan 28, 2005

### SGT

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. Jan 28, 2005

### gushazm

Thanks

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