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*To*: tesla@xxxxxxxxxx*Subject*: Re: Frequency splitting...... Food For Thought*From*: "Tesla list" <tesla@xxxxxxxxxx>*Date*: Wed, 24 May 2006 22:37:32 -0600*Delivered-to*: testla@xxxxxxxxxx*Delivered-to*: tesla@xxxxxxxxxx*Old-return-path*: <vardan01@xxxxxxxxxxxxxxxxxxxxxxx>*Resent-date*: Wed, 24 May 2006 22:22:08 -0600 (MDT)*Resent-from*: tesla@xxxxxxxxxx*Resent-message-id*: <qGIqipRRqCL.A.O1B.wDTdEB@chip1>*Resent-sender*: tesla-request@xxxxxxxxxx

Original poster: Jared E Dwarshuis <jdwarshui@xxxxxxxxx> We can actually use a mechanical model of a tuned coupled oscillator to gain insight into the nature of the phenomena of "Frequency Splitting" We can have a spring stuck to a wall with a spring constant k and a mass m . Then we add a coupling spring (k prime) then an identical mass followed by another duplicate spring k (which we attach to the adjacent wall). Now we need a relevant starting point. Let us hold the left mass stationary at its natural resting point where displacement equals zero. Let us now displace the right mass an arbitrary unit of 1 to the right of its natural resting point. Now let us examine the predicted frequency of the left mass at the instant where time equals zero. It is simply: w = sqrt (k/m). But what is the instantaneous frequency at time equals zero of the right side's mass that is displaced one unit. It is: w = sqrt(( k plus Kprime)/m) Now the interesting part, is that given a duration of time we will find that the mass on the left has been displaced by -1 unit, and the mass on the right will have zero displacement. Thus at this instant in time, they have switched frequencies, where the right side now has the left sides old frequency, and the left side now has the right sides old frequency. Implication: The frequency for a given side is constantly changing between sqrt (k/m) and the sqrt of ( k plus k prime)/ m.

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