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Spring/Mass/Damper system

  1. Nov 1, 2014 #1
    1. The problem statement, all variables and given/known data
    The system is this:
    upload_2014-11-1_14-43-43.png
    I want to find the equation describing it. Assuming K = K and M = M, no friction.

    2. Relevant equations
    Nothing to post here.

    3. The attempt at a solution

    For left M: ##f(t) = Kx_1 + b \frac {d(x_1 -x_2)}{dt} + M \frac {d^2 x_1}{dt^2}##
    For right M: ## b \frac {d(x_1 -x_2)}{dt} = K x_2 + M \frac {d^2 x_2}{dt^2} ##
    Adding the equations: ##f(t) = Kx_1 +K x_2 + M \frac {d^2 x_2}{dt^2} + M \frac {d^2 x_1}{dt^2}##

    Is that the way to go here? I have no background in physics, this popped out in a homework question. Please let me know if I am on the right track.
     
  2. jcsd
  3. Nov 1, 2014 #2
    There is no single equation of motion, but rather two equations are required.

    What do you think gained by adding the two equations?
     
  4. Nov 1, 2014 #3
    I don't think I gained something. It's just that one variable is expressed in terms of two other variables, and I substituted... is that wrong? Anyway, are the two first equations ok? I am not sure about the damper's x variable. Should it be ##x_1 - x_2##?
     
  5. Nov 1, 2014 #4

    haruspex

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    No, I think it was a good move. Look at how x1 and x2 appear in your resulting equation. Notice anything?
    Btw, your equations are only valid while d/dt(x1-x2) > 0, right? In general, need some modulus function in there.
     
  6. Nov 1, 2014 #5
    ##x_1## and ##x_2## are added before multiplied to the spring constant, which is the same for both springs. Is not that what is happening? Once again, first time I saw this kind of problem was days ago in my homework. I was thinking that the resulting equation will equal the force applied, apparently is not what needs to be done.
    Yes, the equations are only valid for positive values of the distance difference. Will include absolute value, thank you.
    Besides that, are the first two equations correct? They look fine to me, but I won't trust myself on that.
     
  7. Nov 1, 2014 #6

    haruspex

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    Yes, and compare that with how they appear in the acceleration terms.
    Yes.
     
  8. Nov 1, 2014 #7
    I like what you did quite a bit. By adding the two equations together, you were able to uncouple them, so that you arrived at an equation in terms of the single variable x1+x2. If you subtract the two equations, you will get a second uncoupled equation in terms of the single variable x2-x1. You can solve each of these equations separately, and then combine the solutions in the end. Very nice.

    Chet
     
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