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Simple Harmonic Motion Brain Teaser

  1. Jul 22, 2004 #1

    cj

    User Avatar

    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:

    [tex]\frac{dx_1}{dt}[/tex]

    when the displacement is:

    [tex]x_1[/tex]

    and a velocity

    [tex]\frac{dx_2}{dt}[/tex]

    when the displacement is:

    [tex]x_2[/tex]

    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:

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

    And that:

    [tex]\begin{equation}
    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.
     
    Last edited: Jul 22, 2004
  2. jcsd
  3. Jul 22, 2004 #2
    If x = A.sin(w.t) then

    x1 = A sin(w.t1)
    x2 = A.sin(w.t2)
    dx/dt 1= A.w.cos(w.t1)
    dx/dt 2 = A.w.cos(w.t2)

    I am not going to do it but there appears to be enough information , to eliminate t1,t2 and get A and w.
    For instance a) and c) can eliminate t1 , b) and d) eliminate t2 .
     
    Last edited: Jul 22, 2004
  4. Jul 22, 2004 #3

    Doc Al

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    Staff: Mentor

    Apply what you know. I'll get you started. You are given: At time [itex]t_1[/itex] the displacement is [itex]x_1[/itex] and the speed is [itex]v_1[/itex]. (I didn't like your notation, so I changed it. :smile: )

    So... just plug into your SHM equations:
    [itex]x_1 = A sin(\omega t_1)[/itex]
    [itex]v_1 = A \omega cos(\omega t_1)[/itex]
    Combine these equations to get a relationship between [itex]\omega[/itex] and A.

    Now do the same for time [itex]t_2[/itex], and then you should be able to solve for [itex]\omega[/itex] and A.
     
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