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Undamped motion: mass and spring system

  1. May 5, 2017 #1
    1. The problem statement, all variables and given/known data
    A mass of 10 kg is hanging from a spring with k=2500N/m that is attached to a roof (see pic).
    The length of the spring when it is not in tension is l0=0.5 m.
    At the time t=0, the mass has a speed of v0=0.5 m/s when it passes the system's equilibrium position.

    Determine

    a) The equilibrium position
    b) The natural frequency, fn
    c) The period, T
    d) The position of the mass as a function of time, t, measured from the roof - draw an approximate graph.

    2. Relevant equations
    F=ma

    Undamped motion
    Wn^2 = k/m
    fn=Wn/2pi
    T=1/fn
    x''+(Wn^2)x=C
    x(t)=Acos(Wn t) + Bsin(Wn t) + c/(Wn^2)

    3. The attempt at a solution
    I think I did this correctly. In part b I'm not really sure how to deal with the fact that the formula seems to be giving me Wn^2 = - k/m , but I just took the absolute value.

    cpoD1-b.jpg
    gUd4utp.jpg

    I have no idea why the images keep being uploaded sideways. Incredibly annoying but I can't seem to fix it.
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. May 5, 2017 #2

    rude man

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    First, the problem is not defined since the polarity of dx/dt(0) is not given. Maybe the missing pic does? ("Speed" is a magnitude term. What's needed is "velocity".)

    Irrespective of that, however, look at your equation for x(t) in part (d). What should x(0) be?
     
  4. May 5, 2017 #3

    haruspex

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    You have not been consistent about which direction is positive. You show the tension as kx up. If the displacement x is down then the force is up, but if the displacement is up the force is down. So the tension should be -kx.
     
  5. May 8, 2017 #4
    Oh yeah sorry, I forgot to draw in the velocity. The velocity is supposed to be negative - the mass is moving down at 0.5m/s. So then the only mistake is that v0 should be -0.5?
     
  6. May 8, 2017 #5
    Yeah I did. I don't really see what I did wrong with the force balance, I put kx as positive since it is pulling the weight up since we're saying that at t=0 it already passed the equilibrium position of l0. x(0) = -0.53924 no?
     
  7. May 8, 2017 #6

    haruspex

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    You need to measure displacement, force and acceleration as positive in the same direction. Up or down, pick one and stick to it.
    Either way, when the displacement is positive the force is negative, and vice versa. The force is -kx, not kx.
     
  8. May 8, 2017 #7
    Aaaah yeah, now I get it. Duh! Let me do some math and get back to you. Thanks.
     
  9. May 8, 2017 #8
    I ended up getting

    X(t) = 0.5 cos (15.811t) + 0.0316 sin (15.811 t) + 0.03924

    I decided just to call down positive and solve it like that instead. The results seem to agree with what it should be in the x(0) case so that's cool. Thanks for the help! Much appreciated.
     
  10. May 8, 2017 #9

    rude man

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    For future problems like part (d) , it is probably easier to set x=0 at the equilibrium point. The expression for x(t), and setting x'(0) to the initial condition, wil be simpler. Also, you can then quickly check for x=0 at t=0 and, if damping is present, at t=∞. But what you did was fine.
     
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