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SHM Calculus Twister

  1. Jul 22, 2004 #1


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    I saw this in an old, junior-level, classical mechanics
    textbook and haven't been able to figure it out.

    A particle undergoing simple harmonic motion has a velocity:


    when the displacement is:


    and a velocity


    when the displacement is:


    What is the angular frequency and the amplitude of the motion in terms of the given quantities?

    I know the solution to the SHM wave equation is:

    x(t) = A \cdot sin( \omega t + \phi )\end{equation}[/tex]

    And that:

    dx(t)/dt = A \omega \cdot cos( \omega t + \phi )\end{equation}[/tex]

    But can't see how to express omega or A in these terms.
  2. jcsd
  3. Jul 23, 2004 #2
    Ain't this same problem answered somewhere else on this same forum?
  4. Aug 26, 2004 #3


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    You need to write an equation relating displacement to velocity, which doesn't have any mention of time. Then, by plugging in the two pairs of displacement/velocity values, you get yourself two simultaneous equations in two unknowns (you can take the phase constant to be zero without loss of generality).
  5. Aug 30, 2004 #4


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    The way to relate position to velocity without mention of time is through conservation of energy. Recall that

    1/2 k x^2 + 1/2 m v^2 = 1/2 k A^2

    where A is the amplitude and v, x are the velocity (or speed) and position at any time (I mean they must be taken at the same value of "t" but "t" can be anything).

    Divinding by 1/2 and by m and using k/m = omega^2, you get

    omega^2 x^2 + v^2 = omega^2 A^2

    So knowing v and x at two different times allows you to find A and omega (they are positive values by definition so there is only one root allowed).

  6. Sep 10, 2004 #5
    The angular frequency is equal to omega, and the ammplitude is equal to A.
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