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System of 2 Masses – 3 Springs

  1. Oct 31, 2012 #1
    1. The problem statement, all variables and given/known data

    Two particles, each of mass M, are hung between three identical springs. Each spring is massless and has spring constant k. Neglect gravity. The masses are connected as shown to a dashpot of negligible mass.

    The dashpot exerts a force of bv, where v is the relative velocity of its two ends. The force opposes the motion. Let x1 and x2 be the displacement of the two masses from equilibrium.


    a. Find the equation of motion for each mass.

    b. Show that the equation of motion can be solved in terms of the new dependent variables y1 = x1 + x2 and y1 = x1 - x2.

    c. Show that if the masses are initially at rest and mass 1 is given initial velocity v0, the motion of the masses after a sufficiently long time is
    x1=x2
    = (v0/2ω)* sin(ωt)

    Evaluate ω.


    2. Relevant equations

    x.. + γx.+ ω2x = 0

    x = Ae-γt/2cos(ωt + ∅)

    3. The attempt at a solution

    I think I managed to get the first two by just using different F = -kx formulae and adding -bv to it.

    The second part I just added and subtracted the equations to get:
    y..1 + ω2y1 + [itex]\frac{2bv}{m}[/itex] = 0

    and

    y..2 + 3ω2y2 = 0

    I think these are right, as far as I know. The thing is, I'm not sure how to proceed from here. I tried using:

    x = Ae-γt/2cos(ωt + ∅)

    I can get it very close to the answer, but I don't think that's how you're supposed to do this... using what I got in part 2 would be more relevant, I think.
     
    Last edited: Oct 31, 2012
  2. jcsd
  3. Oct 31, 2012 #2
    Seeing the diagram or at least its description would be helpful.
     
  4. Oct 31, 2012 #3
    Sorry, this is the image:

    35cf45ff-190f-4bc2-acc7-636af40a0b06.png
     
  5. Oct 31, 2012 #4
    Could you show your equations for a) and the derivation for b)?
     
  6. Oct 31, 2012 #5
    ma1=-kx1-k(x1-x2) - bv

    ma2=-kx2-k(x2-x1) - bv

    These were my first two equations of motion.

    I just added and subtracted them, using k = ω2m to get the equations for 2.

    Quick question: can I take relative velocity v as equal to the derivative of y2 with respect to time?
     
  7. Oct 31, 2012 #6
    But what is v in your equations? And how come it is the same in both equations?
     
  8. Oct 31, 2012 #7
    "... where v is the relative velocity of its two ends. The force opposes the motion."

    So yeah, it should be the same for both cases – both magnitude and direction.
     
  9. Oct 31, 2012 #8
    "Relative velocity" is relative to the velocity of the opposite end. It is the same (= zero) only of the ends have the same velocity. Express it in terms of x1 and x2 for both ends.
     
  10. Oct 31, 2012 #9
    Yeah, it's the relative velocity of one end of the dashpot to the other (thus the relative velocity of one mass to another).

    So for the first mass it'll be derivate of x2 minus that of x1 with respect to time... and for the second mass it'll be negative of that? But the force opposes the motion. Therefore, for the first mass that is displaced by x1, the force will be negative (to the left). Similarly, if the second mass is displaced by x2, the force will be negative as well. Are my signs wrong?

    If relative velocity is in terms of x2 - x1, then I could express it in terms of y2, right?

    On another note, in the third part, if x1 = x2 then relative velocity should be 0.
     
  11. Nov 1, 2012 #10
    I am with you till this point.

    This is less clear. The net force acting on either mass is the sum of three forces, two due to springs and one due to damping. I think should write down the equations in terms of x1 and x2 just for clarity.

    Yes, but, again, you should write down the equations in terms of x1 and x2 first.
     
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