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Simple Harmonic Motion

  1. Nov 8, 2015 #1
    1. The problem statement, all variables and given/known data
    A particle of mass 4.00 kg is attached to a
    spring with a force constant of 100 N/m. It is oscillating
    on a frictionless, horizontal surface with an amplitude
    of 2.00 m. A 6.00-kg object is dropped vertically on top
    of the 4.00-kg object as it passes through its equilibrium
    point. The two objects stick together. What
    is the new amplitude of the vibrating system after the
    collision?



    2. Relevant equations


    3. The attempt at a solution
    Angular frequency before impact adding the block = (100/ 4)^1/2 = 5 s^-1

    Angular frequency after impact adding the block = (100/ 10)^1/2 = 3.16 s^-1

    Max acceleration before adding the block = Angular frequency^2 x amplitude = 25 x 2 = 50 ms^-2

    Force of max acceleration = 50 x 4 = 200N

    After addition of the block, the max acceleration will be 200N/ (6 + 4) = 20 ms^-2

    Max acceleration after adding the block = Angular frequency^2 x amplitude
    20 = 3.16^2 x amplitude
    amplitude = 2

    I know it is wrong but what is wrong, other than the fact my answer is exactly the same as the old amplitude.
     
  2. jcsd
  3. Nov 8, 2015 #2

    BvU

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    Did you do anything with the 'collision' when adding the 6 kg object ?
     
  4. Nov 8, 2015 #3
    thanks. I have got the answer!
     
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