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Springs with dashpot

  1. Sep 28, 2015 #1
    1. The problem statement, all variables and given/known data
    Capture.PNG

    2. Relevant equations
    ∑F=ma
    F=-kx
    x = Ae-γt/2cos(ωt + ∅)

    3. The attempt at a solution
    I have looked up this old thread but dont understand how to get the equation of motion from :
    ma1=-kx1-k(x1-x2) - bv
    ma2=-kx2-k(x2-x1) - bv

    https://www.physicsforums.com/threads/system-of-2-masses-3-springs.648400/
     
  2. jcsd
  3. Sep 29, 2015 #2

    haruspex

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    Is your problem what to do with v? How do you think that relates to x1 and x2?
    But there is something wrong with the signs. The forces the dashpot exerts on the two masses must be in opposite directions, so they can't both be -bv.
     
  4. Sep 30, 2015 #3
    What i got so far:
    ma1=-kx1-k(x1-x2)-bv <==> bv=-kx1-k(x1-x2)-ma1
    ma2=-kx2+k(x1-x2)-bv <==> bv=-kx2+k(x1-x2)-ma2
    Substitute bv
    -kx1-k(x1-x2)-ma1=-kx2+k(x1-x2)-ma2
    m(a2-a1)=-kx2+kx1+k(x1-x2)+k(x1-x2)
    m(a2-a1)=3k(x1-k2)

    This relation is only in terms of x1-x2 not x1+x2 and also the equation of motion should be ma1+ma2=m(a1+a2) but i cant get that since ma has the same sign on both sides of the equation. Does anyone have advice on how to get to b) ? thanks
     
  5. Oct 1, 2015 #4

    haruspex

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    You don't seem to have addressed either of my comments in post #2. Do I need to explain them more?
     
  6. Oct 1, 2015 #5

    rude man

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    The problem states that " .. v is the relative velocity of its two ends." By itself, this statement is of course ambiguous. But the additional statement that the dashpot force "opposes the motion", when considered for each mass separately, implies that F for the left mass (mass 1) has to be -b (d/dt)(x1 - x2) and for the #2 mass, -b (d/dt)(x2 - x1), thus resolving the ambiguity.
     
  7. Oct 1, 2015 #6

    haruspex

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    Ok, you are saying that the two v's are equal and opposite. But it does not look as though wololo understood that. (And it strikes me as a misleading way to have written the equations.)
     
  8. Oct 1, 2015 #7

    rude man

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    Agreed.
     
  9. Oct 1, 2015 #8
    I considered the force by the dashpot bv as being opposite for m1 and m2 but when I put them together to get the total force they cancel themselves. Also the force caused by the middle spring cancels itself (since it points on opposite direction at m1 and m2). The equation of motion I get is then solely expressed in terms of x1+x2 . I know that if if I considered the middle spring force as being equal for both m1 and m2, I would get an equation that depends on x1-x2, but I don't see how this could be true since a spring exerts a force in different direction at it's ends. Am I correct in assuming that in order to solve my two equations I need to put them together and find the total force exerted on the system? Perhaps the way I attempt to solve these equations is wrong and stops me from obtaining an equation that depends on both y1 and y2. Thanks for helping me clarify the way the dashpot works but I am still at a lost on how to successfully answer part b. Any ideas? Here is how I proceeded:

    ma1=-kx1-k(x1-x2)-b(d/dt)(x1-x2)
    ma2=-kx2+k(x1-x2)-b(d/dt)(x2-x1) <==> (rearrange) ma2=-kx2+k(x1-x2)+b(d/dt)(x1-x2)

    m(a1+a2)=-kx1-kx2-k(x1-x2)+k(x1-x2)-b(d/dt)(x1-x2)+b(d/dt)(x1-x2)
    Ftot=-k(x1+x2) (all the other terms cancel)
     
  10. Oct 1, 2015 #9

    haruspex

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    They only cancel because of the way you are combining the two equations.
    Assuming you corrected the sign (so that they are now opposite) if you add them the bv terms cancel. What else could you do?
    Please also answer my other question: what equation directly relates v to x1, x2 and time?
    Oh, and before all that, you need to define clearly what v means.
     
  11. Oct 1, 2015 #10

    rude man

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    These equations are correct, except why "rearrange" them? Solve for x1 and x2 for part (a).
    why are you adding the forces on m1 and m2???
    Do part (a); then part (b) and (c) follow without too much trouble.
     
  12. Oct 1, 2015 #11
    v is the relative speed betwen both ends of the dashpot, so v2-v1. Speed is equal to the derivate of position in regards to time, dx/dt, in this case dx will be the relative variation of x between both ends of the dashpot so bv=bd(x2-x1)/dt

    If I develop develop F1 and F2 seperately, assuming y1=x1+x2 and y2=x1-x2, I get:

    F1=-kx1-ky2-b(d/dt)(y2)
    F2=-kx2+ky2+b(d/dt)(y2)

    I dont see how I can express them in terms of y1

    Could it be that the dashpot only exerts a force on the system as a whole, such that

    Ftot=F1+F2-bv

    Which would give
    Ftot=(-kx1-ky2)+(-kx2+ky2)-b(d/dt)(y2)
    Ftot=-ky1-b(d/dt)(y2)
    ?
     
  13. Oct 1, 2015 #12

    rude man

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    No, as I pointed out before, the two v's are different for the two equations, and haruspex has also pointed out that they can't be the same for both equations.
    First, what are F1 and F2 in terms of x1 and x2?
    Then, try adding the two equations (in post 10) together.
    Then, try subtracting the second equation from the first.
     
    Last edited: Oct 2, 2015
  14. Oct 2, 2015 #13
    Post 8 (ma1=F1 and ma2=F2):
    I already added the two equations together in post 8:
    -kx1-k(x1-x2)-b(d/dt)(x1-x2)-(-kx2+k(x1-x2)-b(d/dt)(x2-x1))
    =-k(x1-x2)-2k(x1-x2)-2b(d/dt)(x1-x2)
    =-3k(x1-x2)-2b(d/dt)(x1-x2)

    I must be forgetting some relation since both you and Haruspex seem to be trying to point me towards something, though I evidently cant't tell what . Would one of you please be willing to point out what I am missing? I added and substracted the equations and still cant express them in terms of x1-x2 and x1+x2. I really am at a loss. Thanks!
     
  15. Oct 2, 2015 #14

    rude man

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    Again: what are F1 and F2 in terms of x1 and x2? Saying F1 = ma1 does not answer that question. This is the first thing you need to get right.
    Right, and you got F1 + F2 = -k(x1 + x2). So - one mo' time - what are F1 and F2 in terms of x1 and x2?
    EDIT: I just noticed, part (a) does not ask you to solve for x1 and x2. Good thing, because that's a real mess. But in part (b) you substitute y1 = x1 + x2 and y2 = x1 - x2, that makes solving for y1 very easy and then y2 is similarly easily obtained. Finally, solve for x1(t→∞) and x2(t→∞) from y1(t→∞) and y2(t→∞).
     
    Last edited: Oct 2, 2015
  16. Oct 2, 2015 #15

    rude man

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    Yes, that was good. Sorry i asked you why you did it. I was concentrating on solving for x1 and x2 separately which would be a big mess - see my previous post. You got F1 + F2 = -k(x1+x2) (all the other terms cancel) which is right. So can you replace x1+x2 with one of the two y's?
    equals what? Can you replace x1-x2 with the other y?
    see my previous post.
     
    Last edited: Oct 2, 2015
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