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Initial Conditions of an Undamped Forced Harmonic Oscillator

  1. May 2, 2005 #1
    The equation of motion of an undamped harmonic oscillator with driving force [tex]F=F_ocos(\omega*t)[/tex] is

    [tex]x(t) = Acos(\omega_0*t) + Bsin(\omega_0*t) + \frac{F_0}{m}\frac{cos(\omega*t)}{\omega_0^2-\omega^2}[/tex]

    I am to determine the initial conditions such that the undamped oscillator begins steady state motion immediately. Is steady state motion simply when [tex]\frac{F_0}{m}\frac{cos(\omega*t)}{\omega_0^2-\omega^2} = Acos(\omega*t-\theta)? [/tex]

    I really have no idea how to approach this problem. Any help would be appreciated. No answers, just hints. Thanks.
  2. jcsd
  3. May 2, 2005 #2


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    My first impression is that a couple of derivatives of x(t) with respect to t would be helpful.
  4. May 2, 2005 #3
    it appears that one gets the correct answer if you set A and B equal to zero. if i do so, i get [tex]x(0) = \frac{F_0}{m}\frac{1}{\omega_0^2 - \omega^2}[/tex] and the derivative produces a sin term which would make v(0)=0, which is correct as well. Are the first two terms (Acoswt and B sinwt) the damping or some sort of interfering expression that goes away when the motion becomes steady state?
  5. May 2, 2005 #4


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    Some justification for setting those terms equal to zero would be good. Those terms represent the motion of the system at its natural frequency. They are solutions to the homgeneous differential equation of a harmonic oscillator. If there were no driving force and you set the system in motion, those terms would be the ones to keep, with A and B established by the initial displacement and velocity of the system.

    I think the satement of the problem is weak. If the oscillator is truly undamped, then "steady state" is a misnomer. There really would not be a decay to steady state motion in that case. The natural frequency oscillations would last forever if they were ever excited. If there were any damping it would show up as a decaying exponential in front of those first two terms. So when they say steady state, what they mean is the steady state of a lightly damped oscillator where you gradually remove the damping as steady state is approached. In any case, what they are calling "steady state" is achieved when those first two terms are gone. Damping would kill them eventually if they were present. You can also kill them with the intial conditions making A = B = 0 as you have done.
  6. May 2, 2005 #5
    Yeah, I recall dealing with such cases of harmonic oscillators in Differential Equations. I think I got this now. Thanks Dan.
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