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Hot Oscillation question

  1. May 2, 2007 #1
    A particle of mass 12 grams moves along the x axis. It has a restoring force F= -0.06 N/m. If it starts from x=10 cm with a speed of 20 cm/sec toward the equilibrium position, Find its amplitude, period, and frequency. Determine when the particle reaches the equilibrium point for the first time.


    Edit:


    I solved the question this way:

    F= -w^2*m -> w=1/2(sqroot)

    v=x*w
    20=x*1/2(sqrt)
    x=28.2

    --

    T=2pi/w -> T=8.8

    --

    a(accelaration) = -w^2*x
    a= -14,1

    Vf=Vi+a.t
    t=1,141 sec.

    --------------

    2nd way- and some one else solved the question with this way :

    Let w = angular frequency, A = amplitude, T = time period
    Acceleration = F / m = 0.06 / 0.012 = 5 m/s^2
    Acc = w^2 * x
    w^2 * 0.10 = 5
    Simplifying, w = 5 (sqrt 2)
    T = 2 pi / w
    m * v^2 + m *w^2* x^2 = m* w^2 * A^2
    0.04 + 0.5 = 50 * A ^2
    A = (sqrt 1.08) / 10 m = sqrt 1.08 * 10 cm




    --
    and I'M confused :(






    Thanks,..
     
    Last edited: May 2, 2007
  2. jcsd
  3. May 2, 2007 #2

    Mentz114

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    Gold Member

    To get homework help you must show some effort. What have you done towards solving this ?
     
  4. May 2, 2007 #3
    by the way thanks for reminding me Mentz, I'm new in the forum that's why.
     
  5. May 2, 2007 #4

    Mentz114

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    Gold Member

    Welcome to the forum. Assuming the particle will go into simple harmonic motion, we can use the force formula F = -kx to calculate k. The frequency is sqrt(k/m). Have a look here and you'll get some useful formulae -
    http://hyperphysics.phy-astr.gsu.edu/hbase/shm.html#c1
     
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