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Simple harmonic oscilator

  1. Jul 18, 2009 #1
    Hallo,

    Does the velocity i simple harmonic oscillator is zero in equilibrium points? if it's true

    how does it make sense with the fact that i suppose to get a maximum kinetic

    Energy in those points (stable ones)

    i would realy appriciate if someone could clear this issue for me.

    Thanks,

    Omri
     
  2. jcsd
  3. Jul 18, 2009 #2

    diazona

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    The velocity of a simple harmonic oscillator is a maximum when it passes through its equilibrium point. Is that what you were asking?
     
  4. Jul 18, 2009 #3
    yes,

    but what about the non stable equilibrium point (maximum points)?

    thanks
    omri
     
  5. Jul 18, 2009 #4
    Simple Harmonic Scillators have stable equilibria (or they wouldn't oscillate).

    The extremes in position are not equilibrium points.
     
  6. Jul 18, 2009 #5
    sorry for that... i ment that if i have a potential with maximum points does the velocity
    is zero there?
     
  7. Jul 19, 2009 #6
    In general, the total energy of a pendulum is T + V (kinetic plus potential energy). T + V = constant. One is a maximum when the other is minimum, and vice versa. There is one other quasistable equilibrium point for a pendulum, when the pendulum is exactly upside down. In principle, it should stay there forever, barring vibration and air currents. In actually, it is unstable, because of the Heisenberg Uncertainty Principle. If the tip of the pendulum has a momentum uncertainty Δ p, and the uncertainty in position is Δ x, then the product is Δp Δx <=h/2 pi.

    Based on this uncertainty (and barring friction), the pendulum will begin swinging (as I recall) in a few seconds.

    Γ Δ Θ Λ Ξ Π Σ Φ Ψ Ω
     
  8. Jul 21, 2009 #7
    If I'm understanding your question right, YES. PE is a maximum when KE is zero, and vice versa.

    The farthest extent of position is the maximum of potential energy, and V is zero.
     
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