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Classical mechanics differential equation F(x) = -kx

  1. Jun 23, 2017 #1
    1. The problem statement, all variables and given/known data
    A particle of mass m is subject to a force F (x) = -kx. The initial position is
    zero, and the initial speed is v0. Find x(t).

    2. Relevant equations
    F = m*v*dv/dx = -kx
    v = dx/dt

    3. The attempt at a solution
    I'm new to differential equations, so please excuse me if I make any amateurish mistakes.
    -kx = mv*dv/dx
    After integrating:
    -kx2/2 + c = mv2/2
    v2 = -kx2 + c

    This is the part I'm unsure about, the inital values:
    v2(0) = 0 + c = v02
    So c = v02

    v = (v02 - kx2/m)1/2

    dx/v(x) = dt
    After integrating: (a bit messy)
    t = (m/k)1/2*arcsin(x*(k/v02m)1/2) + c
    c=0
    Just wanted to know if I went wrong somewhere as I'm not convinced by the final answer.
     
  2. jcsd
  3. Jun 23, 2017 #2

    Orodruin

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    You are to find x as a function of t, it makes more sense to solve the differential equation in time. You can do it like that if you really want to complicate your life... just solve for x from your expression.
     
  4. Jun 23, 2017 #3
    How would I go about doing this? I'm not sure how to solve in time if the function is of position.
     
  5. Jun 23, 2017 #4

    Orodruin

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    What function is of position? There are only functions of time. The force is a function of position, which is a function of time. What you have to do is to solve the resulting differential equation.
     
  6. Jun 23, 2017 #5

    Orodruin

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    Also, the work-energy theorem is not really what you want to use. Newton 2 will suffice.
     
  7. Jun 23, 2017 #6
    Then I'd have dv/dt = -kx
    dv = -kx*dt
    I'm guessing that's not what you meant because the integral doesn't make sense.
     
  8. Jun 23, 2017 #7

    Orodruin

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    Of course it makes sense (the first expression is what you want to work with - apart from missing the mass), it is a differential equation. What is v in terms of x?
     
  9. Jun 23, 2017 #8
    Sorry but I just don't know how to solve it if the left side is with respect to time and the other side is position.
     
  10. Jun 23, 2017 #9

    scottdave

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    Net force = mass times acceleration, but you have F = (mass)*(velocity)*(dv / dx) {dv/dt is acceleration}
    Why have you set it up dv/dx?
    How about just this: Net force = (mass)*(dv/dt) which is (mass)*(d2x/dt2)
     
  11. Jun 23, 2017 #10
    To expand on what @scottdave said, you have the following 2nd order ordinary differential equation with constant coefficients:$$m\frac{d^2x}{dt^2}=-kx$$Do you know how to solve this equation for x as a function of t?
     
  12. Jun 23, 2017 #11

    Orodruin

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    The time derivative of velocity is the second derivative of position. This results in a linear second order differential equation with constant coefficients.
     
  13. Jun 23, 2017 #12
    I set it up as dv/dx so I could integrate both sides (with respect to position on one side and velocity on the other. v* dv/dx = dx/dt * dv/dx so the dx's cancel out to get dv/dt.
    The book I've been using (Morin) was introducing me to this and I haven't done second order DE's yet.
     
  14. Jun 23, 2017 #13
    I set it up as dv/dx so I could integrate both sides (with respect to position on one side and velocity on the other. v* dv/dx = dx/dt * dv/dx so the dx's cancel out to get dv/dt.
    No I've been using Morin's CM book and have only been introduced to first order DE's so far.
     
  15. Jun 23, 2017 #14
    Oh ok, I've been trying to solve it in first order because I haven't been introduced to second order yet.
     
  16. Jun 23, 2017 #15

    scottdave

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    Can you think of a function, which when you take the derivative of it twice, gives you back the negative of the function (multiplied by a constant)? Note that the behavior of the mass/spring is to oscillate back and forth.
     
  17. Jun 23, 2017 #16
    Oh yes I
    I yes I recognise it's cos and sin. Thanks all.
     
  18. Jun 23, 2017 #17

    haruspex

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    You did well to get to this, and it is quite easy from there. Just turn it around to make it x as a function of t:
    (t - c)(k/m)1/2=arcsin(x*(k/v02m)1/2)
    sin((t - c)(k/m)1/2)=x*(k/v02m)1/2
    etc.
     
  19. Jun 23, 2017 #18

    scottdave

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    Since it is messy, as you said, one good thing is to check dimensions, which may reveal possible errors
    From your formula, (m/k)1/2 needs to have the dimension of time. Knowing that Force = -kx, you can find that k has dimension of mass/time2. Working that through, it does indeed have dimension of time. Then argument of arcsine needs to be dimensionless.
    (x*(k/v02m)1/2) It indeed does turn out to be dimensionless.
     
  20. Jun 23, 2017 #19

    scottdave

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    Were you able to get the correct answer?
     
  21. Jun 24, 2017 #20
    I'm thinking it's:
    x = cos(kt)/k + v0sin(kt)/k -1/k
    So when t=0 , x=0 and v=v0
     
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