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Simple Harmonic motion and collisions

  1. Apr 13, 2005 #1
    A block of mass 2kg oscillates on the end of a spring in SHM with a period of 20ms. The position of the block is given by: x=(1.0cm)cos(wt+2pi). Another block of mass 4kg slides toward the oscillating block with velocity of 6m/s. The two blocks undergo a completely inelatic collision at time t=5.0ms.

    a)what is the amplitude of oscllaltion after collision?
    I know momentum is conserved in the collision and time t=.005 secons, the velocity of the oscillating block is O. So I got V(final)=4m/s.
    But I am not sure how to account for this with the spring force to get the final amplitude.

    b) What is the frequency of the resulting SHM?
    without A, I dont think I can do this part, so I havent tried yet...
  2. jcsd
  3. Apr 14, 2005 #2


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    Are you sure it's cos(wt+2pi)? Might as well make it cos(wt).

    If the position is given by:
    x=(1.0cm)cos(wt+2pi) then, t=5.0ms corresponds to an angle of pi/2. The velocity equation is:
    v=-w(1.0cm)sin(wt+2pi). v does not equal 0 at t=5.0ms.

    I'd do part b first to get the new w... then use the fact that the magnitude of the maximum velocity = wA. Since you have the maximum velocity (you get it from the conservation of momentum etc...) you can get A (amplitude).


    You can calculate k, then solve for the new w using a mass of 6kg
  4. Apr 14, 2005 #3

    I meant (1/2)pi

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