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Determine the particle's equation of motion

  1. Apr 25, 2008 #1
    1. The problem statement, all variables and given/known data
    A particle with a mass of 0.500 kg is attatched to a spring with a force constant of 50.0 N/m. At time t = 0 the particle has its maximum speed of 20.0 m/s and is moving to the left. (Use t as necessary.)
    (a) Determine the particle's equation of motion, specifying its position as a function of t, time.
    (b) Where in the motion is the potential energy three times the kinetic energy?
    (c) Find the length of a simple pendulum with the same period.
    (d) Find the minimum time interval required for the particle to move from x = 0 to x = 1.00 m.

    3. The attempt at a solution
    I only need help with the first part for now, but here is what I tried.

    [tex]\omega = \sqrt{\frac{k}{m}}[/tex]
    [tex]\omega = 10[/tex]

    [tex]v_{max} = A\omega[/tex]
    [tex]A = 2[/tex]
    So my equation of motion looked like this.
    [tex]x(t) = 2cos(10t)[/tex]

    but that was wrong and I am not entirely sure why
     
  2. jcsd
  3. Apr 25, 2008 #2

    Doc Al

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    Staff: Mentor

    You are off by a phase factor. Note that at t = 0 the particle is at max speed--so what's its position with respect to equilibrium?
     
  4. Apr 25, 2008 #3
    It's position, with respect to equilibrium, would be 0, which means the phase factor needs to be pi?
     
  5. Apr 25, 2008 #4

    Doc Al

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    Right.
    Write the general expression and see if that works.
     
  6. Apr 25, 2008 #5
    Ok that worked and I am on part D) now.

    x(t)= 2sin(10t + pi)

    I set x to 1 and solved for t but that didn't work. Can you not do that?
     
  7. Apr 25, 2008 #6

    Doc Al

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    That should work fine. What did you get? (Remember that you're working in radians.)
     
  8. Apr 25, 2008 #7
    [tex]\frac{sin^{-1}(.5)-\pi}{10}[/tex]


    that gives me a negative .261

    perhaps I am not doing my algebra right?
     
  9. Apr 25, 2008 #8

    Doc Al

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    Two things you need to do:
    (1) Choose a value of [itex]sin^{-1}(.5)[/itex] that gives you a positive time.
    (2) Figure out the times when x = 0. (t = 0 is one such time, but it's not the one you want. At t = 0 the particle is moving left; you want the time when it's moving right.)
     
  10. Apr 25, 2008 #9
    but my time = all that stuff, and sin of .5 - pi always comes out a negative number..
     
  11. Apr 25, 2008 #10

    Doc Al

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    Realize that there's more than one angle that satisfies [itex]\sin\theta = .5[/itex].
     
  12. Apr 25, 2008 #11
    For arcsin of 0.5 I get a value of 0.52359, so you are saying there is another value in addition to this?
     
  13. Apr 25, 2008 #12
    Can anyone that is still awake help me solve part D?
     
  14. Apr 26, 2008 #13

    Doc Al

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    I think you've figured it out by now, but sure:
    [tex]\sin\theta = \sin(\theta + 2\pi)[/tex]

    That's just one example (but it's the one you need). Look at a graph of a sine function and see all the places where it equals .5.
     
  15. Apr 26, 2008 #14
    I figured it out, but I still didn't understand it, so I am glad you posted again.
     
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