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Verifying that a critical damped oscillator approaches zero the fastest.

  1. Jan 7, 2009 #1
    1. The problem statement, all variables and given/known data

    Not actually a homework problem, just something from my book I'm trying to verify.

    2. Relevant equations

    The general form of the equation for damped oscillations...

    [tex]\ x(t) = e^{-\gamma t}(Ae^{\sqrt{\gamma^{2}-\omega^{2}}t}+Be^{-\sqrt{\gamma^{2}-\omega^{2}}t})[/tex]

    Here gamma = b/m, where b is the constant associated with the strength of a velocity dependent damping force. (A and B are constants determined by the initial conditions)

    and omega is the angular frequency, root (k/m).

    And, in the critically damped case, the solution is

    [tex]\ x(t) = e^{-\gamma t}(A+Bt)[/tex]

    3. The attempt at a solution

    It's obvious that it b<k, that is gamma<omega, then when the oscillation is critically damped (gamma = omega), it will reach zero more quickly. When the oscillation is underdamped, then the exponents in the larger expression in parenthesis are imaginary, so we have simple harmonic motion falling off as e^-gamma(t), and since gamma is less than omega, this expression reaches zero faster than in the critically damped case, which falls off as e^-omega(t).

    However when the oscillation is overdamped, and we have b>k, gamma>omega, how can we show that the exponents (which are then positive) are still less than omega? That is, how can we show

  2. jcsd
  3. Jan 7, 2009 #2
    Rearranging the inequality etc.:

    [tex]\gamma -\omega \leq \sqrt{\gamma^2-\omega^2} [/tex]

    But we know that: [tex]\gamma-\omega = \sqrt{\gamma-\omega}\sqrt{\gamma-\omega}[/tex].

    So we have:

    [tex]\sqrt{\gamma-\omega}\sqrt{\gamma-\omega} \leq \sqrt{\gamma-\omega}\sqrt{\gamma+\omega} [/tex]

    Wich is obviously true, since [tex]\gamma > \omega[/tex]
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