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Quick Harmonic Motion problem (got most of it solved)

  1. Nov 4, 2004 #1
    To start, I already know the answer.. but I can't seem to get it the "hard" way, i.e. through solving the 2nd order ODE.

    Redundant, but it's a block on water, and it's oscillating after a mass is removed from it.
    There's an initial displacement, and no
    [itex]
    \begin{align*}
    y(0) &= -0.025m \\
    y'(0) &= 0 \\
    \omega^2 &= 39.2 s^{-2} \\
    y'' + \omega^2 y &= -g = -9.8 N/kg \\
    y_h &= c_1 \cos(\omega t) + c_2 \sin(\omega t) \\
    y_p = \frac{g}{\omega^2} &= -0.25m\\
    y'(0) = 0 &\implies c_2 = 0 \\
    y(t) &= y_h + y_p \\
    y(t) &= c_1 \cos(39.2^{0.5} t) - 0.25 \\
    y'(0) = -0.025 &\implies c_1 = 0.225 ?
    \end{align*}
    [/itex]

    Now, I'm looking for the amplitude, that should be c1 right? But by logic, the amplitude is 0.025m.

    What am I doing wrong?

    Am I supposed to use the initial conditions to solve the homogeneous solution and find c1 like that?

    I've already solved the problem the other way (just assume amplitude is the initial displacement), but I don't like to assume things.

    Thanks.
     
  2. jcsd
  3. Nov 5, 2004 #2
    check your algebra

    [tex]y_{p}=-\frac{g}{\omega^2}[/tex]
     
    Last edited: Nov 5, 2004
  4. Nov 5, 2004 #3
    Well.. whatever the particular solution is, it makes c1 != y(0), because it will always be off by Yp.

    But the answer(the amplitude) is clearly just the initial displacement (y(0)).

    Or am I missing something ?

    Yp can't be 0, can it ?
     
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