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Forced damping

  1. Jun 26, 2012 #1
    Does changing the angular frequency change the amplitude of a forced oscillation? If so, I don't understand how that can be. I guess that the angular frequency is based only on the springs and dampers? So assuming those are the same then the angular frequency will always be the same?

    Are there any equations for solving the amplitude of a forced oscilliating system without knowing the angluar frequency? I know everything else except the angular frequency and the amplitude. It is strongly dampered.
     
  2. jcsd
  3. Jun 26, 2012 #2

    HallsofIvy

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    No, in general changing the frequency of an oscillation does NOT change the amplitude. Of course it is possible to make a change in the situation that changes both frequency and oscillation but one is not dependent on the other.
     
  4. Jun 26, 2012 #3
    But I guess that this equation links amplitude and the frequency?


    [tex]\begin{array}{l}
    A = \frac{P}{{\sqrt {{{(k - m{\Omega ^2})}^2} + {r^2}{\Omega ^2}} }}\\
    \\
    k = stiffness\\
    m = mass\\
    \Omega = frequency\\
    r = damper\\
    P = force
    \end{array}[/tex]

    So, an increase in the angular frequency should decrease the amplitude.
     
  5. Jun 26, 2012 #4
    A linear 2nd order system response to sinusoidal excitation in complex exponential form is

    [tex]A(\omega)e^{j(\omega t + \theta)}[/tex] where [tex]|A(\omega)|^2[/tex] is frequency response and [tex]\theta(\omega)[/tex] is phase response and [tex]-\frac{d \theta}{d\omega}[/tex] is group delay. In general all of them are function of excitation frequency ω.

    Note that the frequency response usually contains terms of [tex]\omega_{d}^{2} \quad and \quad \omega^2 \quad where \quad \omega_{d} \quad is [/tex] damped frequency which is only dependent of system characteristics, i.e. the mass, stiffness, damper.


     
    Last edited: Jun 26, 2012
  6. Jun 26, 2012 #5
    Sorry, not sure if I follow. That equation I posted says that it is dependent on the angular frequency too, doesn't it?
     
  7. Jun 26, 2012 #6
    You are correct if you meant the angular frequency of the sinusoidal driving force. Your formula is in terms of m, r, and Ω. It can be also written in terms of mass, damping ratio ζ and natural frequency of the system. Natural frequency is not dependent of angular frequency of the driving force.

    Your problem is basically
    [tex]my''+ry'+ky=f[/tex]
    The amplitude of steady-state sinusoidal excitation can be solved by setting driving force [tex]f(j\omega t)=Fe^{j\omega t}[/tex] and let solution be [tex]y(j\omega t)= Ae^{j\theta}e^{j\omega t}[/tex]. Plugging y and f into the ODE, canceling the complex exponential and equating the amplitude of both side, then you will get
    [tex]A(\omega)=\dfrac{F}{\sqrt{(m\omega^2-k)^2-\omega^2r^2}}[/tex]

    You can now see that the ω in the formula is the angular frequency of the driving force. In other context, you may see A in terms of damping ratio and natural frequency where
    [tex]\zeta=\frac{r}{2\sqrt{km}} \quad \quad and \quad \quad \omega_n=\sqrt{\frac{k}{m}}[/tex]

    Note that the angular natural frequency only depends of k and m.


     
    Last edited: Jun 26, 2012
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