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Motion equation of moving particle

  1. Nov 26, 2008 #1
    Hi,
    assume that motion equation of moving particle are given by:
    [tex]\ddot{x} + \omega^2 x = \frac{f}{m} \cos ( \gamma t + \beta )[/tex]​
    the solution is of course [tex]x = a \cos (\omega t + \alpha) + \tfrac{f}{m(\omega^2 - \gamma^2)} \cos ( \gamma t + \beta)[/tex].

    My question is - what exactly happens when [tex]\omega = \gamma[/tex] ?

    As [tex]\gamma \to \omega[/tex] but [tex]\omega \not= \gamma[/tex] the amplitude of oscillations grows rapidly, right?
    We might guess that when [tex]\omega = \gamma[/tex] aplitude is infinite! But... let's right [tex]x[/tex] as:
    [tex]x = b \cos (\omega t + \alpha) + \frac{f}{m(\omega^2 - \gamma^2)} \left\{ \cos ( \gamma t + \beta) - \cos (\omega t + \beta) \right\}[/tex]​
    .
    (b is a new const). In the second term of the sum we have [tex]0/0[/tex] symbol and using de L'Hospital rule we obtain
    [tex]x = b \cos (\omega t + \alpha) + \frac{f}{2m \omega} t \sin (\omega t + \beta)[/tex]​
    This function grows to infinity as [tex]t \to \infty[/tex] but not so fast and it of course has a finite amplitude.

    So again - what exactly happens when [tex]\omega = \gamma[/tex] ?
     
  2. jcsd
  3. Nov 26, 2008 #2

    Dale

    Staff: Mentor

    Re: Resonance

    Why don't you try to set them equal in the original equation and solve that.
     
  4. Nov 26, 2008 #3
    Re: Resonance

    I did that and got solution of the form [tex]x = b \cos (\omega t + \alpha) + \frac{f}{2m \omega} t \sin (\omega t + \beta)[/tex] again.
    But I still wonder why when [tex]\gamma \to \omega[/tex] the amplitude can be very very large and when [tex]\gamma = \omega[/tex] it is totally different.
     
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