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Simple Harmonic Motion formula help

  1. Jan 15, 2010 #1
    1. The problem statement, all variables and given/known data

    A small bock that has a mass equal to M rest on a piston that is vibrating vertically with simply harmonic motion described by the formula y = Asin(wt)

    a) Show that the block will leave the piston if w2A > g.

    b) If w2A = 3g and A = 15 cm, at what time will the block leave the piston?


    2. Relevant equations
    F = m*a (maybe...)

    3. The attempt at a solution
    The only thing I did was find that the equation for the acceleration is,

    a = -w2Asin(wt)
     
  2. jcsd
  3. Jan 15, 2010 #2

    vela

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    What forces are on the block?
     
  4. Jan 15, 2010 #3
    The force of gravity (Mg) and the force of the piston pushing up on the block (Normal Force). Although in this case I'm not too sure how to write out normal force.

    And actually I think the way I wrote this is only if it's sitting still so actually maybe it would be M(g+a) because the push from the piston would make it feel heavier. I may be wrong about that.

    Actually I'm kind of thinking that...
    Fn - Mg = Ma
    Fn = M(g+a)

    So if the acceleration on the piston is ever greater than g there will be no normal force exerted on the block. Meaning it's not touching it. Is this right?
     
    Last edited: Jan 15, 2010
  5. Jan 15, 2010 #4

    vela

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    You're right about the two forces. Unlike when the block is sitting at rest, the two forces generally won't be equal in magnitude, so there will be a net force on the block that causes it to accelerate. When it's accelerating upward, the normal force will be bigger than its weight. When it's accelerating downward, the block's weight will be bigger than the normal force.

    Can you say anything about the forces on the block when it's right on the verge of losing contact with the piston?
     
  6. Jan 15, 2010 #5
    The only thing I can really think of is that the normal force might be zero and the only force acting on the block would be Mg. Although I think that would only be if it were in the air already so I'm probably wrong about this.

    So maybe on the way down it would be...
    Fn + mg = ma
    Fn = m(a - g)
     
  7. Jan 15, 2010 #6

    vela

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    You're right. The normal force can only push up on the block. It can get as big as necessary to accelerate the block upward, but the smallest it can get is zero. So what's the maximum downward acceleration the block can have? And how does this tell you when the block and piston separate?
     
  8. Jan 15, 2010 #7
    I believe the maximum downward acceleration it can feel it g. So when the block feels this max acceleration it is about to separate.

    I'm not sure if I'm right on this completely but thanks for your help.
     
    Last edited: Jan 15, 2010
  9. Jan 15, 2010 #8

    vela

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    You're right again. (You need to have more confidence!)

    What's the piston's acceleration at the moment of separation?
     
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