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Undamped HO transfer function

  1. Sep 23, 2010 #1
    Undamped Harmoni Oscillator differential equation:
    mx''=-kx
    (solution (to IC: x(0)=A, x'(0)=0) : x=Acos(wt) with w2=k/m)

    I need to find out the solution to the forced case with (Initial Condition zero):
    mx''+kx=F (with F=AFcos(wFt) ) evaluating with the transfer function. But I m not sure this transfer function does exists, or is limied.

    Second question) Is the solution to the Undamped HO forced sinusoidally stable ?
    I suppose not, because without energy dissipation, the energy that enter is never consumed and just adds up to the system. But I m not sure, I have not time to calculate the solution which i haven't found.

    Third ) In case of damped HO forced with a sinusoid, in the resonant band of frequencies, where there is amplification, the response function is higher than the input forcing sinusoid. Is a way to see the mathematic of energy transfer from input to output ? There is no creation of energy in the system , just the amplitude of the oscillation of the input is amplified, but this doesn't mean that the power of the output is higher than the input. Isn't it?
     
  2. jcsd
  3. Sep 23, 2010 #2

    K^2

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    Not really. The transfer function here is not well-defined.

    Working backwards from solution. (I'm not in the mood to derive all this. You can find it in any ODE text.)

    [tex]x(t) = \frac{2 A_F}{m(\omega_F^2 - \omega_0^2)} sin(\frac{\omega_F + \omega}{2} t) sin(\frac{\omega_F - \omega_0}{2} t), \omega_F \neq \omega_0[/tex]

    [tex]x(t) = \frac{A_F}{2 m} t sin(\omega_0 t), \omega_F = \omega_0[/tex]

    (You might want to check the amplitudes, I scribbled them down rather hastily.)

    You'll have to treat the two cases where omegas are different or same (resonance) separately. Lets take a look at the first case.

    Input:
    [tex]X(s) = L[A_F cos(\omega_F t)] = \frac{\delta(s - i \omega_F) + \delta(s + i \omega_F)}{2}[/tex]

    Output:
    [tex]Y(s) = L[x(t)] = \frac{A_F}{m(\omega_F^2 - \omega_0^2)} \frac{\delta(s + i \omega_0 + \delta(s - i \omega_0) - \delta(s + i \omega_F) - \delta(s - i\omega_F)}{2}[/tex]

    And, of course, the transfer function H(s) must be such that Y(s) = H(s)X(s). This works everywhere except s= ±(iω0).

    This solution is stable, of course, and the resonance solution has divergent Y(s), so obviously, there is no transfer function.

    As far as power goes, the oscillator does no work, so there is no output power. The applied force does work on the oscillator if the phases match and against it if they don't. So the input power fluctuates between positive and negative if there is no resonance. In resonance, it's always positive, and energy stored in oscillator increases without bounds.
     
  4. Sep 24, 2010 #3
    Really thankyou for the math solution. This is exactly what I was looking for (and I already plotted some examples)! I need it for evaluating a system without damping factor, or without vibration absorber, just a metal frame joined with bolts very fixed. (I have a lot of cases and a lot of mathematic examples with the damping formula of SDOF model mx''+cx'+kx=F (where F=Afcos(wf*t)), but few formulas for the case of no damping c=0.)

    There are still some question in my mind, without damping: in case wf is not the resonant freq w0, the response is clearly divergent; in the other case the solution is a sinusoid modulated with another sinusoid (or you can see it as a sinusoid (wf-w0) transmitted at higher freq (wf+w0) like in AM radio) (and I suppose the same energy of the input). Now consider the two case from a thermodynamic point of view, in the first case you are adding energy without dissipation, so energy adds up to the system and the oscillation becomes more and more large; but in the second case it looks like a periodic wave, which does not diverge, while I was expecting it to become more and more high. I am missing something, I know, but cannot figure it out.
     
  5. Sep 24, 2010 #4
    So you mean in resonance the oscillator can store/absorbe energy, while not in resonance, the oscillator is somewhat transparent to the input vibration ?
     
  6. Sep 26, 2010 #5

    K^2

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    It stores energy from input either way. However, in non-resonance case, the input power isn't always positive. When the driving force pushes with oscillation, the work is positive. When it pushes against, it's negative. So your driving force isn't always supplying power to the oscillator. Half the time it is taking energy away from the oscillator.

    Using the above solutions, try plotting the total energy of the oscillator as a function of time and comparing it to the plot of the oscillations themselves. Maybe it will make a little more sense.
     
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