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LC Oscillations how are they even possible?

  1. Dec 30, 2007 #1
    If i connect an inductor and a charged capacitor as the only circuital elements in a circuit [other than lead wires], something called 'LC Oscillations' happen. According to what I've read, due to the charge on the capacitor, there is a potential difference across the ends across which the capacitor is connected. So, the capacitor undergoes discharge [i.e. current starts flowing in the circuit].

    Since the inductor is there in the circuit, current flows through the inductor too... and hence causes a back emf in the inductor. Which basically means the ends across the inductor also have an emf across them, opposite to that caused by the capacitor and hence a current flows backward from the inductor and charges the capacitor, but this time the polarity of the capacitor is different from the previous one.

    My question is that, assuming that this oscillation is not dampened, then won't there be a time when the voltage across the ends of the capacitor undergoing discharge and the voltage across the ends of the inductor giving a back e.m.f become equal and opposite in direction. At this time, there will be no e.m.f in the circuit i.e. no potential difference across any two points in the circuit. As such, the current should stop to flow and hence, no further charge/discharge of the capacitor should take place.

    Once a capacitor is connected, the e.m.f. it causes across it begins to decrease and the back e.m.f. of the inductor continues to increase. Therefore, the point where these e.m.f's become equal should happen in the first cycle itself and hence, oscillations are just not possible, not even theoretically.

    But we all know, that they do happen.. so where am I going wrong?
     
  2. jcsd
  3. Dec 31, 2007 #2
    Here is an analogous case with a mass on a spring, you should be able to spot the error.

    If you have a mass on a spring, and you displace it and let it go, it will freely oscillate. But, at a certain time, it passes through the equilibrium position. When it is at the equilibrium position, there is no force acting on the mass! Does this imply that the mass will not oscillate any more?
     
  4. Dec 31, 2007 #3

    Defennder

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    Homework Helper

    Well I think you have to note it is misleading to say that at some point the induced emf in the inductor and the potential difference across the capacitor will be the same. In fact, at all times, the potential difference across the capacitor and the induced emf across the inductor are always the same. This follows because they are connected parallel to one another.

    Note what happens when the capacitor fully discharges; at this point the voltage across it and the inductor is zero. But the current in the circuit is not zero, because the presence of the inductor prevents the current from going to zero immediately when the pd across the capacitor reaches zero.

    It is important to note that the current in the circuit, due to the presence of the inductor, does not start out at the a maximum value and then declines exponentially, as in the RC circuit. In fact it is the opposite, it starts out at 0, then increases to a maximum. This is because the inductor resists the change in current in the circuit, so to speak. The maximum value of the current in the circuit is attained when the capacitor is fully discharged.

    So what happens when the current is at a maximum and the capacitor is discharged fully and the potential difference across it as well as the inductor is 0? Now note that the emf of the inductor is in the same direction as that of the current, and the inductor would then act to maintain the current at at this value. What happens is that that the current would then decrease at a slower rate, until the current becomes zero, and the magnetic field in the inductor has been discharged completely. Note this is when the capacitor has been charged fully, albeit to its opposite polarity.
     
  5. Jan 13, 2008 #4
    In that case, I'd like to ask: How are oscillations even possible? When the object is at the equillibrium position, it is at rest. Also, there is no force acting on it. Newton's first law of motion says that, "A body continues to be in it's state of motion or rest unless acted upon by an external force." In that case, we can say that Newton's laws are not valid here. Which means that in the case of an oscillatory device, we are dealing with a non-inertial frame. Is it true?

    also, how can i decide whether a given frame is inertial or not. I want to decide it mathematically or by logic, not instinct.
     
  6. Jan 13, 2008 #5

    dlgoff

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    Here you go. This should give you a good mathematical read on harmonic oscillators.
     
  7. Jan 13, 2008 #6

    mda

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    Absolutely not. Just because it is at the equilibrium position does not mean it is at equilibrium... it depends on the previous history of the object. Think about the direction of the force before it reaches this point... what does this do to the velocity?
     
  8. Jan 13, 2008 #7
    The 'equilibrium position' is simply a position. Not a state. A car flying down the road crosses an intersection of another road. In one instant it is in a position, or more correctly stated crosses a position, but it most certainly is not at rest.
     
  9. Jan 13, 2008 #8
    How do we define being 'at rest'? afaik, it is the state when the velocity of the body is zero. In that case following from:

    [tex]
    x = Asin(\omega t)
    [/tex]
    [tex]
    v = A \omega cos(\omega t)
    [/tex]

    When [itex]x = 0, \omega t = 0[/itex] and hence [itex]v \neq 0[/itex]

    So, i get how i was wrong here. But at the maximum offset, we have [itex]x = A; \omega t = \frac{\pi}{2}[/itex]. And hence, we have: [itex]v = 0[/itex].

    But then does the motion continue because [itex]a \neq 0[/itex] as:

    [tex]
    a = A \omega^2sin(\omega t)
    [/tex]

    And obviously, we have maximum acceleration over here. That'd be the only reason i can state...

    So, do we define rest as the state in a particular frame where the [itex]n^{th}[/itex] derivative of displacement with respect to time is zero in that frame for all [itex]n \in N[/itex]
     
  10. Jan 14, 2008 #9
    This is right.
    No, rest is the state in which the velocity of an object is 0 with respect to the reference frame...as you showed, the object is at rest momentarily when [itex]\omega t = \frac{\pi}{2}[/itex] (with respect to the frame of the coordinate system). It doesn't necessarily mean that all time-derivatives of the object's position are zero.
     
    Last edited: Jan 14, 2008
  11. Jan 14, 2008 #10
    k.. thanks a lot.
     
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