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SHM question

  1. Dec 6, 2006 #1
    1. The problem statement, all variables and given/known data
    A block attached to a spring underneath oscillates vertically with a frequency of 4Hz and an amplitude of 7.00cm. A tiny bead is placed on top of the block jut as it reaches its lowest point. Assume the bead's mass is so small that its effect on the motion of the block is negligible. At what distance from the block's equilibrium position does the bead lose contact with the block

    2. Relevant equations
    Hooke's Law?

    3. The attempt at a solution
    My first instict was that the bead would launch up until the maximum amplitude when the spring starts to move downward. Then I thought about Hooke's Law. Am I supposed to sum up the forces and say the bead is launched when there is no normal (restoring) force? This would mean the bead loses contact at equilibrium. Am I right about this?
  2. jcsd
  3. Dec 6, 2006 #2
    Ive found a better way of explaning my thoughts. Will spring accelerate the block and the bead upward until the sum of the forces is zero, or until the normal (restoring) force is zero? Im pretty sure its the former, but without the mass of the block or bead, can I get a numerical answer?
    Last edited: Dec 6, 2006
  4. Dec 6, 2006 #3
    I'm think it is until the normal force is zero. Since after the bead is launched into the air, the only force acting on it would be the force of gravity.
  5. Dec 7, 2006 #4


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    That's the right idea. The forces on the bead are its weight and the reaction force from the block. You know the motion of the bead (the same as the block) so you can find its acceleration and therefore the resultant force on it. The mass of the bead will cancel out in the equations.

    The force between the bead and block can only act in one direction. The bead isn't glued to the block, so the block can only "push" not "pull".

    You could also work in terms of acceleration not force. The maximum downwards acceleration the bead can have is g (caused by gravity). When the block has a bigger acceleration than g downwards, the bead will lose contact with it.
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