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Particle Motion

  1. Sep 10, 2005 #1
    Given a particle whose equation of motion is

    [tex] \frac {d^2x} {dt^2} = -Qx + R [/tex]

    I'm supposed to find a solution expressing x in terms of t, given that position is (5,0) at t=0 and speed=0. The next time the particle is at (5,0) it has travelled a distance of 12 units and t = Pi. I'm supposed to solve for Q and R as well and have x and t as the only 'variables'.

    The differential equation isn't that hard to work out - solving the characteristic eqn - leading to complex roots - I get

    [tex] x = A \sin{(\sqrt{Q}t)} + B \cos{(\sqrt{Q}t)}\ +\frac{R}{Q} [/tex]

    I believe I'm correct in saying A = 0 since taking [tex]\frac{dx}{dt}[/tex] and setting it equal to 0 with t=0 gives [tex]\sqrt(Q)*A = 0 [/tex]. From this point on I'm a little stuck - as nothing I do seems to be able to solve for B or R and Q - I don't even know how to bring in the 12 to help out. A little help would be appreciated.
  2. jcsd
  3. Sep 11, 2005 #2


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    Is this actually 2-d? what do you know about y?

    At t=0, x=5 = B + R/Q (presuming your soln is ok)
    at t=pi, x=5 = B cos(RootQpi) +R/Q
    subtract these to solve for Q
    find out when v=0 ; then, x=-1 (why?) R=?
    Last edited: Sep 11, 2005
  4. Sep 11, 2005 #3


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    At t=0, x=5 = B + R/Q (presuming your soln is ok)
    at t=pi, x=5 = B cos(RootQpi) +R/Q

    subtracting one from t'other,

    0 = B - Bcos(√Q.pi)
    B(1 - cos(√Q.pi)) = 0
    cos(√Q.pi) = 1
    √Q.pi = 0, 2pi, 4p, ...
    √Q = 0, 2, 4, ...
    Q = 0, 4
    Q = 4


    x = Bcos(2t) +R/Q
    at t = 0, x = 5
    at t = pi, x = 5 again, but the argument of the cos function is now 2pi, a movement of one cycle, or 4 times the amplitude
    .: movement, 12 units, = 4 times B
    B = 3

    Using again the expression,

    At t=0, x=5 = B + R/Q

    5 = 3 + R/4
    2 = R/4
    R = 8
    Last edited: Sep 11, 2005
  5. Sep 11, 2005 #4


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    |B| = 6 does not imply that B=6.
    There is no "choice" of Q - read the situation.

    While you're at it, read the sticky post:

    please DO NOT do someone's homework for them or post complete solutions to problems. Please give all the help you can, but DO NOT simply do the problem yourself and post the solution (at least not until the original poster has tried his/her very best).

    Members who:

    (1) attempt to get others to do their work for them, or
    (2) post the same question in multiple forums, or
    (3) prematurely post complete solutions to problems,

    are violating the spirit (and policy) of PF!
  6. Sep 11, 2005 #5


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    Hello lightrav,

    Reading those stickies was among the first things I did when I came to this board.

    If you read my earlier posts, you will see that I do try to adhere to the spirit (and policy) of this board.
  7. Sep 11, 2005 #6


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    Dearly Missed

    OP has done so much work already, and shown it, so I don't see that Fermat has done too much for him.
  8. Sep 11, 2005 #7
    Thanks a million folks :!!) - I didn't ask for the entire solution, only pointers, but at least now I have a way of checking my work :cool: .
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