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Forced Vibration

  1. Jun 24, 2005 #1
    Questions:
    (a) At what resonance ([itex]\omega = \omega_0[/itex]), what is the value of the phase angle [itex]\phi[/itex]?
    (b) What, then, is the displacement at a time when the driving force [itex]F_{ext}[/itex] is a maximum, and at a time when [itex]F_{ext} = 0[/itex]?
    (c) What is the phase difference (in degrees) between the driving force and the displacement in this case?

    Equations related to this problem:

    [tex]F_{ext} = F_0\cos{\omega t}[/tex]

    [tex]x = A_0\sin{(\omega t + \phi_0)}[/tex]

    [tex]\phi_0 = \tan^{-1}\frac{\omega_0^2 - \omega^2}{\omega(b/m)}[/tex]

    My Answers:
    (a) Since [itex]\omega = \omega_0[/itex], [itex]\phi_0 = \tan^{-1}0[/itex] which means [itex]\phi_0 = k\pi[/itex] for some non-negative integer k.
    (b) [itex]F_{ext}[/itex] has its maximum value when [itex]\omega t = 2j\pi[/itex] for some non-negative integer j. The displacement is then [itex]x = A_0\sin{(2j\pi + k\pi)} = 0[/itex]. [itex]F_{ext} = 0[/itex] implies that [itex]\omega t = l\pi/2[/itex] where l is some odd positive integer.The displacement is then [itex]x = A_0\sin{(l\pi/2 + k\pi)}[/itex], so x = A0 or -A0.
    (c) This question I don't understand well. I'm guess the difference is [itex]\pi/2 + \phi_0[/itex] because the driving force is a cosine function and the displacement is a sine function with a phase angle.

    Is this right?
     
  2. jcsd
  3. Jun 25, 2005 #2

    OlderDan

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    Homework Helper

    (c) What is the phase difference (in degrees) between the driving force and the displacement in this case?

    Assuming your x equation is correct, and using

    [tex] \sin{\alpha} = \cos{(\alpha - 90)} [/tex]

    you have

    [tex]x = A_0\sin{(\omega t + \phi_0)} = A_0\cos{(\omega t + \phi_0 - 90)}[/tex]
     
  4. Jun 25, 2005 #3
    Ah, so the phase difference is just [itex]\phi_0 - \pi/2[/itex] radians. Thanks.
     
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