# Re: Standing Wave LC Resonance

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• Subject: Re: Standing Wave LC Resonance
• From: "Tesla list" <tesla@xxxxxxxxxx>
• Date: Sun, 18 Sep 2005 11:53:57 -0600
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• Resent-date: Sun, 18 Sep 2005 11:58:51 -0600 (MDT)
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`Original poster: Jared E Dwarshuis <jdwarshui@xxxxxxxxx>`

```For an LC system the energy is described as U = Ub +Ue  or 1/2 L Isqrd
+1/2C Qsqrd
This classic equation unfortunately gives no hint about energy storage
within an LC system under conditions of standing wave resonance.```

```For this relationship we can turn to the energy equations of Jacob and
Daniel Bernoulli used to describe a mechanical linear array of coupled
harmonic oscillators.```

`We will modify the Bernoulli equations here to describe an LC system.`

```We can write the inductance of an air cored inductor as L = u (2 pi r
N)sqrd / 4 pi l
Or: L = u (wire length)sqrd / 4 pi l```

```We can conclude that under conditions of dynamic equilibrium, that
both charge and current are distributed along the entire length of the
inductors wire.```

```We define elements of charge (q) as being distributed, where (q) is a
displacement from an equilibrium value.```

`Thus we can write:`

`Ub = L/2 ( dq/dt1 +dq/dt2 +dqdt3 + + + dq/dtn)`

```The capacitance requirements for standing wave resonance (can) rely on
both capacitance internal to the inductor and capacitance external to
the inductor. But under conditions of standing wave resonance we can
examine the effects of the combined capacitance on the charge
distribution along the length of the inductor.```

`Ue = 1/2C ( (q1)sqrd + (q2-q1)sqrd + + + (qn ­(qn-1))sqrd + (qn) sqrd )`

```Thus the Lagrangian function: La = 1/2 the sum to k [ Ldq/dt k ­ 1/C (
q k+1 ­ q k )]```

```Becomes: L d2q/dt2 = -1/C( (qk ­ qk-1) + 1/C( qk+1 ­ qk) )   Where k =
1,2,3?.n```

```If the number of charge is made very large and the distance between
charges is very small we can write:```

```(partial derivative) d2q/dt2 = (velocity squared) X (d2q/dx2) where q
is a sinusoidal
function of position and time.  (a linear function).```

```Thus we can add two waves of equal amplitude traveling in opposite
direction and get:```

```  q = 2A sin (2 pi x/ wavelength) cos wt      Where:  w = 2 pi freq. /
wavelength```

```The implications are that at intervals of wavelength/2 we will find
current nodes and at intervals of wavelength/4 we will find voltage
nodes.```

```There are definite implications for impedance that fall from such
analysis, I will leave this as an ?exercise for the reader?```

Sincerely: Jared Dwarshuis