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

[brian whatcott <betwys1@sbcglobal.net>]

You now appear to accept that losses vary in a non-linear way with 
amplitude.
Though a math solution to an experimental difficulty is forlorn (in my 
view) I can sketch an experimental approach which may be helpful.
It is  wise to recall that the drag loss is not the only loss, and that 
there may be other losses acting in a  non-linear way,  in addition to 
the flexure loss which is also amplitude dependent.
If one applies a narrow magnetic impulse to the pendulum at the bottom 
dead center position to find the frequency for maximal amplitude for 
constant power, then the variation in losses may then be mapped for 
varied energy values and pendulum amplitudes at  the SAME driven frequency.

Given an experimental protocol which can then relate pendulum amplitude 
to total losses;  varying the drive frequency to establish the two 
frequencies which result in  the experimentally derived  amplitude for 
half input power will provide you with an improved procedure for 
estimating the Q at a specific amplitude.

Sincerely

Brian Whatcott     Altus OK

On 10/20/2013 5:19 PM, Bernard Cleyet wrote:
> 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
> > _____________