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Homework Help: Harmonic Motion

  1. May 10, 2010 #1
    1. The problem statement, all variables and given/known data
    A 130 g ball attached to a spring with spring constant 3.0 N/m oscillates horizontally on a frictionless table. Its velocity is 17 cm/s when x = 4.2 cm

    Find the:
    -Amplitude of Oscillation
    -Max acceleration
    -Speed of the ball when x = 2.8cm

    2. Relevant equations
    T = 2*pi*(m/k)1/2
    Sinusoidal equations for position and velocity of a particle in SHM

    3. The attempt at a solution

    I'm not entirely sure where to start. I solved the velocity and position equations for t and set them equal to solve for A, but finding an intersection for that is beyond me. Not sure how else to work the equation.

    I figured if I could find A then I could take the derivative of the velocity equation and try to solve for the maximum acceleration, but that still leaves the phase shift and doesn't give an acceleration value to set it equal to.

    Not sure where to begin on this one at all.
  2. jcsd
  3. May 10, 2010 #2
    I think using energy equations is the easiest way to solve this. The total energy is given by [tex]1/2kA^2 = 1/2kx^2 + 1/2mv^2[/tex]

    Solve for amplitude A.

    As for the max acceleration, this will occur intuitively when all energy is potential and being converted into kinetic. Use [tex]F=ma[/tex] where[tex]F=-kx[/tex].

    Hope that helps.
    Last edited: May 10, 2010
  4. May 10, 2010 #3
    I hadn't thought about the energy perspective. Thanks for that one.

    But about finding the max acceleration...

    Since the potential of a spring is given by (1/2)(k)(x^2), I figure it was intuitive that the max acceleration would be when the spring is stretch to its max distance, its amplitude.

    But (1/2)(3)(5.5^2) = (.13)(a) gives me a huge value for acceleration, which is being spit back. So I'm wondering what I'm missing. (yes, 5.5cm is the right amplitude).
  5. May 10, 2010 #4
    I don't think I made myself clear before. I meant to solve the equation [tex]kx = ma[/tex] (where x is the amplitude A) by equating forces. Your intuition is right, though.
    Last edited: May 11, 2010
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