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Homework Help: Standard harmonic motion question

  1. Jan 5, 2010 #1
    A block rides on a piston that is moving vertically with simple harmonic motion. (a) If the SHM has period 1.0 s, at what amplitude of motion will the block and piston separate? (b) If the piston has an amplitude of 5.0 cm, what is the maximum frequency for which the block and piston will be in contact continuously?
     
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  3. Jan 5, 2010 #2

    vela

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    Show us what you've got so far.
     
  4. Jan 5, 2010 #3
    Thanks for responding. Well, I am guessing that I need to find the maximum acceleration using what I am given, but I don't know how to.
     
  5. Jan 5, 2010 #4

    vela

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    Start with the equations you have that describe simple harmonic motion.
     
  6. Jan 5, 2010 #5
    Well i tried using a(t)=-[tex]\omega[/tex]2xcos([tex]\omega[/tex]t+[tex]\phi[/tex]) but i came out with two variables. I had a=-4[tex]\pi[/tex]2xcos(2[tex]\pi[/tex]t)
     
  7. Jan 5, 2010 #6

    berkeman

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    Not sure what the 2nd equation means, but the first one us reasonable.

    Now step back for a second. For the block to separate from the piston at the top of the travel, what acceleration must it have? What is pulling the block downward through the whole range of motion?
     
  8. Jan 5, 2010 #7
    the second equation is when i plugged in values. I plugged in w, which i got using w=2[tex]\pi[/tex]f. And do i need to account for gravity and the spring?
     
  9. Jan 5, 2010 #8

    berkeman

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    Can you answer my "step back" question first?
     
  10. Jan 5, 2010 #9
    i said gravity and the spring
     
  11. Jan 5, 2010 #10

    berkeman

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    Who said there's any spring? There are other ways to drive the piston in SHM, and in this problem, the driving mechanism would appear to be irrelevant, I believe.
     
  12. Jan 5, 2010 #11
    sorry, i meant to say piston. so gravity pulls down and the piston pushes up. i dont have a mass, though. could i use F=-m[tex]\omega[/tex]2x and Fg=mg and set them equal?
     
  13. Jan 5, 2010 #12

    berkeman

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    I don't think I'd use forces. Just stick with accelerations. At the top of the piston's motion, if the mass separates just slightly from the piston as the piston starts downward, what is the downward acceleration of the separated block? So the piston's acceleration at that same moment has to be what?
     
  14. Jan 5, 2010 #13
    they must be equal?
     
  15. Jan 5, 2010 #14

    berkeman

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    Yes. That would be the point where there just barely could be some separation. If the piston accelerates downward even a but faster than that, there definitely will be separation.

    So you had a correct equation earlier for the acceleration of the piston in terms of omega. You can use omega = 2*pi*f as you said, and write the equation in terms of the piston frequency (or period). I assume in your equation that "x" was the maximum single-sided amplitude of the piston's motion, right? That part will be important in the part (b) of the question as well.

    So go ahead and solve parts (a) and (b), and show us your work and the solutions.
     
  16. Jan 5, 2010 #15
    should i use -9.8m/s2 for my acceleration? and should i use the period for my time?
     
  17. Jan 5, 2010 #16

    berkeman

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    Yes on the acceleration. On the time -- try thinking more intuitively that that. What is the max acceleration of the piston? When does it occur in its motion?

    They gave you the period mostly so that you could figure out omega, which along with the amplitude leads you to the max acceleration...

    I going to be away from computers for a few hours, but will check back in later tonight. Just relax and think about what is going on physically, and write out the equations. (And be sure to check your units.)
     
  18. Jan 6, 2010 #17
    i solved it with another equation. it was amax=[tex]\omega[/tex]2xmax. my teacher says i got the right answer. thanks a lot.
     
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