[Phys-L] Q for band width is not valid. Was: Re: finding Q from band width?

[Bernard Cleyet <bernardcleyet@redshift.com>]

On 2013, Oct 07, , at 20:12, Bernard Cleyet <bernard@cleyet.org> wrote:

> I'm attempting to find the Q of a clock pendulum while free from the escapement.  (crutch and crown removed, so only the spring suspension in use) using the bandwidth method, i.e. Q = "natural" or resonance frequency / FWHPower, no?  

I very stupidly thought the simple formula [ frequency / half power bandwidth] was valid in the case of a quadratically damped oscillator.  Obviously not as the Q is not constant, but is a function of the amplitude.  But can one not derive a valid one from the particular amplitude solution of the appropriate ODE?

I've found many sources for the complimentary solution of the driven ODE, but not the particular one.  

Any maths mavens out there who can show me?

bc maths declined, so won't even try a trial solution.
  
> I'm driving it E-magnetically using a PM waxed the the edge of the bob (lenticular) and a coil waxed to the case opposite the PM.  Not only is  this asymmetrical, but also the force is a strong (4th power?) function of the separation (near field).  Therefore, I vary the drive to maintain the same amplitude, which is measured by the blocking time of a photogate at ~ BDC, as I manually change the drive frequency in 0.001Hz steps.  I assume the drive power is the square of the applied EMF.   The  undriven frequency agrees w/ the minE drive frequency (~ 5%)  [0.565Hz]  
>
>
> So if the min drive E is 0.3V^2 then the diff between the two frequencies at 0.6V^2 is the FWHP, yes??
>
> The result is ~ 1k, which agrees ~ with the ring down method.  
>
> bc
> _______________________________________________