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A mass and vertical spring

  1. Jan 3, 2009 #1
    1. The problem statement, all variables and given/known data

    A 0,824kg mass attached to a vertical spring of a force constant 162N/m oscillates with a maximum speed of 0,372 m/s. Calculate the maximum acceleration of the mass.

    3. The attempt at a solution

    amax=vmax*[tex]\omega[/tex]

    [tex]\omega[/tex]=[tex]\sqrt{\frac{k}{m}}[/tex]

    Is that correct?
     
  2. jcsd
  3. Jan 3, 2009 #2

    Doc Al

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    Staff: Mentor

    Yes, that will work.
     
  4. Jan 27, 2009 #3
    I am a bit confused on how to calculate the acceleration of the mass. Are the accelerations at the bottom and the top the same or different? at eq we have kx=mg. when you displace the mass downward and release, ma=kx-mg. when the mass reaches the top, kx+mg=ma. since x is the same in both cases, the two a's need to be different. what is the flaw in my line of thinking?
     
  5. Jan 27, 2009 #4

    Doc Al

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    Staff: Mentor

    Realize that the mass oscillates about the equilibrium point, where kx0 = mg.

    At the lowest point, the net force is: k(x0 + A) - mg = kA (upward)
    At the highest point, the net force is: k(x0 - A) - mg = -kA (downward)
     
  6. Jan 27, 2009 #5
    i still dont understand. can u please explain further?
     
  7. Jan 27, 2009 #6
    so this means that the force due to the spring at the top and the bottom are different? while if the system was horizontal, it'd be the same at both extremes?
     
  8. Jan 28, 2009 #7

    Doc Al

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    Staff: Mentor

    Right.
    If the system was horizontal, the equilibrium point would be at the unstretched position, x = 0. Thus at one extreme the spring force would be kA and at the other it would be -kA.
     
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