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Start-up time of a force

  1. Jan 1, 2016 #1
    1. The problem statement, all variables and given/known data
    given an oscillator with parameters k,\gamma,m acted upon by a force of the form f(t) = {0 for t<0, F*t/T for t<T, F for t>T} find the location of the mass as a function of time, what happens when T->0 (write x(t) in this case)?

    2. Relevant equations
    the equation for the oscillator: m*x''(t) + \gamma *x'(t) +k*x(t) = f(t)
    where f(t) = {0 for t<0, F*t/T for t<T, F for t>T}

    3. The attempt at a solution
    I managed to solve the first part using the double side Laplace transform ( given starting conditions of x(0) =0 and x'(0) =0 , but I have no idea how to approach the part about T->0, attached is the solution to the first part of the question
     

    Attached Files:

  2. jcsd
  3. Jan 1, 2016 #2

    TSny

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    Does your solution satisfy the initial conditions?
    You'll need to consider what happens to the solution as T approaches 0 for both t < T and t > T.
     
  4. Jan 1, 2016 #3
    yes my solution is correct, but the problem with T->0 is that the solution seems to diverge
     
  5. Jan 1, 2016 #4

    TSny

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    What do you get for y(0)? Do you get x(0) = 0?
    It shouldn't diverge in the limit as T -> 0. Can you show how you are taking this limit?
     
  6. Jan 2, 2016 #5
    y(0) isn't 0 but seeing that the total solution is multiplied by a step function ( 0 for t<=0 ,1 otherwise) I get x(0) = 0, and I haven't the slightest clue how to approach T-> 0 , my reasoning is that my solution is given by some expression z(t)*u(t) - z(t-T)*u(t-T), for the second term when T -> 0 I should get the derivative of that term but for the first term there's a point where I divide by T and I can't get the second term to balance that seeing that there will always be some time period as small as we want where the first term is non zero yet the second is
     
  7. Jan 2, 2016 #6

    gneill

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    Won't the applied force take the form of an impulse function as the width narrows towards zero? It has a particularly simple form in the Laplace domain.
     
  8. Jan 2, 2016 #7
    you sir are correct! , I just looked at the definition of limit again and it cleared up a lot of confusion I had with this question, thank you now this question is solved!
     
  9. Jan 2, 2016 #8

    TSny

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    I don't believe your expression for ##y(t)## is correct. Are you sure you wrote it down without any typographical errors? Your expression does not lead to ##x(0) = 0##, unless I keep making the same mistake when evaluating your expression for ##t = 0##

    Yes, taking the limit as ##T## approaches 0 will yield ##x'(t)\cdot T## where ##x(t)## is the solution for finite T when ##0<t<T##.
     
    Last edited: Jan 2, 2016
  10. Jan 2, 2016 #9

    TSny

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    For finite ##T## the impulse is ##\frac{1}{2}F_0 T##. So, in the limit as ##T## approaches 0 the impulse goes to zero. To get a Dirac delta impulse, the force ##F_0## would have to go to infinity as ##T \to 0##. But no mention is made of that in the statement of the problem.
     
  11. Jan 2, 2016 #10
    that's the raw output from matlab when calculating z(t) ( the final answer is z(t)*u(t) - z(t-T)*u(t-T)):
    (F*t)/(T*k) - (F*g)/(T*k^2) + (F*g*exp(-(g*t)/(2*m))*(cosh((t*(g^2/4 - k*m)^(1/2))/m) - (m*sinh((t*(g^2/4 - k*m)^(1/2))/m)*(g/(2*m) - (F*g^2 - F*k*m)/(F*g*m)))/(g^2/4 - k*m)^(1/2)))/(T*k^2) , so I might have missed a parenthesis somewhere, I'm rechecking it now

    I solved it by taking the limit of the force as T->0 , it's the derivative of F*t*u(t) where F is some constant. then I reanalysed in the Laplace domain, the confusion I had was that I was taking just a part of the expression to be the derivative and was confused as to what am I supposed to do with the other part, the force ends up being F*U(t)

    edit: I have rechecked the parenthesis and corrected y(t), now the expression goes to zero as it should
     
    Last edited: Jan 2, 2016
  12. Jan 2, 2016 #11

    gneill

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    The force was given to be F*t/T for t<T. That might be written as (F/T)*t. Then as T → 0 you'd have your infinite spike.
     
  13. Jan 2, 2016 #12
    actually it becomes F*U(t) + F*t*delta, t*delta=0 so we are left with F*U(t) but you are correct in that it was a derivative, anywho how do I mark this thread as solved?
     
  14. Jan 2, 2016 #13

    gneill

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    There's no "solved" state for a thread, so nothing needs to be done in that regard.
     
  15. Jan 2, 2016 #14

    TSny

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    The impulse, J, of the force over the interval from t = 0 to t = T is ##J = \frac{F}{T}\int_0^Tt dt = \frac{1}{2}FT## which goes to zero as T → 0.

    In the limit as T → 0 the problem becomes equivalent to applying a constant force F starting at t = 0. This is different than applying a sharp impulse at t = 0.
     
  16. Jan 2, 2016 #15

    TSny

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    Your solution for x(t) that you posted in the attachment in the first post cannot be correct. It does not satisfy the initial conditions.

    [EDIT: Sorry. I somehow overlooked your comments in post #10. Would you mind posting your final answers for both parts of the problem?]
     
    Last edited: Jan 2, 2016
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