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Simple Harmonic Motion + Regular Kinematics

  1. Jan 23, 2012 #1
    1. The problem statement, all variables and given/known data
    A particle moves along the x axis. It is initially at the position 0.250 m,
    moving with velocity 0.200 m/s and
    acceleration -0.290 m/s2.
    Suppose it moves with constant acceleration for 5.80 s.

    Position of Particle after 5.80 Seconds - -3.4678 m.
    Velocity of Particle After 5.80 Seconds -1.482 m/s

    Next, assume it moves with simple harmonic motion for 5.80 s and x = 0 is its equilibrium position.

    What is its position at the end of this time interval?
    What is its velocity at the end of this time interval?


    2. Relevant equations


    Attempts :

    I attempted to use the formula X(t)= A cos (wt), with A being the amplitude 0.250 (because a spring can't stretch farther than its intial position) and w = angular velocity which I used 0.2 / .25.

    For the velocity I used the formula- v(t)=w A sin (t)

    From this point I was completely lost because I could not find a way to incorporate the constant acceleration.


    Simple Harmonic Motion equations

    if you could leave an explanation, it's be greatly appreciated!!
     
    Last edited: Jan 23, 2012
  2. jcsd
  3. Jan 23, 2012 #2

    vela

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    You need to make an attempt at solving the problem yourself and showing us what you did before you can receive assistant, per the forum rules.
     
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