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Vibration periods

  1. May 20, 2005 #1
    hello guys getting a bit stuck with this problem? was wondering if anyone could help me please? how would one work out the maximum vibration period during harmonic oscillation of a an object moving up and down with amplitude 50mm if it is constantly losing mass? :cry:
     
  2. jcsd
  3. May 20, 2005 #2
    I'd try to start from from NII like this:

    [tex] \frac {dp} {dt} = \dot m \dot x + m \ddot x = C \dot x + m \ddot x =-kx[/tex] (1)

    You also know that the derivative of mass wrt time is the constant C so you can solve the simple DE for m(t) and plug it into (1). Manipulate and you should get a 2nd degree DE for x. Look for oscillating solutions and see how the period developes with time.
     
  4. May 20, 2005 #3
    cheers mate i will try to give it a whirl :biggrin:
     
  5. May 20, 2005 #4
    thanks mate i will give it a go
     
  6. May 20, 2005 #5

    OlderDan

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    The motion of the object depends on how the mass is being lost. If the mass is simply detaching itself from the object, then it still carries momentum and the remaining mass will experience the full force of the spring and nothing else. With less mass the object will experience a greater acceleration and the period of oscillation will change, but so might the amplitude. If the mass comes off the object with some different velocity, then there will be a force applied to the ramaining object in addition to the spring force. I don't see that happening in this problem, so I don't think there is any additional term. It seems to me you just have

    [tex] (M_0 - Ct) \ddot x =-kx[/tex]

    with [tex]m(t) = (M_0 - Ct)[/tex] where C is the constant rate of change of mass.

    The only deviation I see possible from this is if the given amplitude is required to be maintained and the mass is required to leave the object with altered velocity to make that happen. I'm not sure what exit velocity would be required to maintain the amplitude, but I think you can rule out simple detachment as a possibility. Simple detachment should conserve total energy, so the lost mass will be carrying away energy that will reduce the amplitude of the motion. To maintain amplitude, the leaving mass would have to provide propulsion to the remaining object.

    Constant amplitude would require that no kinetic energy be lost by the object because the mass was leaving. You would need

    [tex] (M - Ct) {\dot x}^2 + kx^2 = kA^2[/tex]

    [tex] \frac{d}{dt}(M - Ct) {\dot x}^2 + \frac{d}{dt}kx^2 = 0[/tex]

    [tex] - C {\dot x}^2 + 2(M - Ct)\dot x \ddot x + 2kx \dot x= 0[/tex]

    With that said, I don't see any solution to this problem other than that the period is a maximum when the mass is maximum. Any loss of mass is going to reduce the period of oscillation of a mass on a spring, unless it is deliberately expelled in such a way as to oppose the force of the spring. Since no such thing was stated, there is no justification for making such an assumption. Maybe trying to write the correct DE is overkill.
     
    Last edited: May 20, 2005
  7. May 20, 2005 #6

    arildno

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    This is just nonsense.

    You need to know the ejection velocity of the ejected mass, as well as the mass flux if you are to solve this problem properly.

    (Note: this was posted before I had the chance to read OlderDan's reply. I'll get to that now)
     
  8. May 20, 2005 #7

    OlderDan

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    We certainly agree about that. Maybe it is nonsense to try to salvage the problem by speculating about how the mass could be expelled.
     
  9. May 20, 2005 #8

    arildno

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    Well, you made a good try on it anyways, so I'll give my best shot here as well.
    However, it IS rather speculative, and thus in the final analysis, possibly meaningless.

    As I see it, it is practically impossible to say something definite if we let the mass be expelled parallell to the motion.
    However, let us suppose that the spring resists sideways motion perfectly, and let the mass be ejected sideways.
    In that case, the force acting upon the object through ejection of mass is counteracted by the resistance from the spring.
    Thus, the only interesting component of Newton's 2.law is along the line of motion, and it reduces to:
    [tex](M-Ct)\ddot{x}=-kx[/tex]
    which coincides with your first answer..
     
    Last edited: May 20, 2005
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