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Simple harmonic motion of a particle

  1. Feb 13, 2008 #1


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    1. The problem statement, all variables and given/known data

    A particle moves along x axis It is initially at 0.270m moving with velocity of 0.140m/s
    and acceleration of -0.320m/s^2
    Suppose it moves with constant acceleration for 4.50s.

    Find it's:
    a) position and velocity at end of time interval

    next assume it moves with simple harmonic motin for 4.50s and x= 0 is it's equillibrium position.

    Find it's:

    b) position and velocity at end of this time interval

    2. Relevant equations
    [tex]x(t)= A cos (\omega t + \phi )[/tex]
    [tex]v(t)= - \omega A sin (\omega t + \phi ) [/tex]
    [tex]a(t)= - \omega ^2 A cos (\omega t + \phi) [/tex]

    3. The attempt at a solution

    well I know for the first situation that:

    [tex]x_o[/tex]= 0.270
    [tex]v_o[/tex]= 0.140m/s
    [tex]a_o[/tex]= -0.320m/s^2

    and the acceration remain constant for 4.50s

    to find a) the postiion and velocity at end of time interval....

    I was thinking that I would take the period (T) to equal T= 4.50s?
    and would I use 4.51 seconds and use the velocity as a function of time equation

    [tex]v(t)= - \omega A sin (\omega t + \phi ) [/tex]

    though I need the [tex] \phi [/tex]....hm..

    and I think I'm lost here.

    can someone Please help me out with this?
    Last edited: Feb 13, 2008
  2. jcsd
  3. Feb 13, 2008 #2
    so, a) it's no problem as is an 1 dimension problem, with movement equations...

    about b), it's a whole new problem:
    you will need to know the position and velocity at [tex]t_0=4.50[/tex]
    so, you can write: [tex]t_1=0[/tex] at [tex]x_1=0[/tex](equilibrium point)
    by a), you calculated velocity at [tex]t_0[/tex], so, it's the particles velocity at [tex]t_1[/tex]

    I guess this should help...
    Last edited: Feb 13, 2008
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