Re: [Phys-L] Trinity Pendulum Q

["Donald Polvani" <dgpolvani@verizon.net>]

On May 22, 2015 at 11:55 PM, Brian Wattcott wrote:

"It was tantalizing to see the responses to your pointer for data on a Trinity clock restart for October 2009, in connection with seeking a value for Q.
 Though the various determinations of a restart Q were scattered (from~1000 to 2000) they all were remarkably lower than the Q values found earlier for free decay of the pendulum."

With this wide disparity in Q results, I redid my determination of the exponential fit constant gamma and Q.  I used the same data I had used previously (clock restart of March 29, 2009 using times t = 600 s to 9300 s after start-up for the fit (where the data points changed smoothly and, apparently, in a (rising) exponential manner).  I took the amplitude at t = 10500 s of 53.318653 mrad as the "final" amplitude for large time.  Time and amplitude shifting the data, as in my previous email, I ended up fitting to a now decaying exponential function:

theta_shift = theta_0_shift * exp (-gamma*t_shift)

I now took the natural log of this function and did linear regression on the equation:

ln(theta_shift) = -gamma*t_shift + ln(theta_0_shift)

I determined gamma and ln(theta_0_shift) using three different linear regression implementations:
1) Using the Excel linear regression function I got:

gamma = 0.0005264 rad/s, ln(theta_0_shift) = -6.923, so theta_0_shift = 0.9849 mrad

2) Manually coding the least squares formulas in Excel and got:

gamma =0 .0005264 rad/s, ln(theta_0_shift) = -6.923, so theta_0_shift = 0.9849 mrad

3) Using linear regression in MATLAB and got:

gamma = 0.0005264 rad/s, ln(theta_0_shift) = -6.923, so theta_0_shift = 0.9849 mrad

All three calculations agree.  So, if you do an exponential fit to the March 29 data specified above, then gamma = 0.0005264 rad/s seems a good value to me.

The next (and more controversial) step is to calculate Q.  I used Q = w/(2*gamma) where I take w to be the radial clock running frequency (w = 2*pi/T and T = 3 s)

This gives Q = pi/(gamma*T) = 1989.

I can see where we might disagree on the value of Q, since we seem to be using different definitions and/or different ways of calculating it.  However, a gamma value of 0.0005264 rad/s seems quite consistent with an exponential fit to the data and doesn't, in my opinion, permit much latitude in computing its value.  The R^2 value for the linear fit is 0.8997 with most of the residuals coming from the last 3300 s of the fit.  Had I fit from 600 to only 6000 s after start-up, I believe the fit would have had an even higher R^2 value, although the gamma value may not have changed much.  I say this just from looking at the plots of actual and fitted data where the fitted line appears to go right through the early points and deviate significantly from the data only from 6000 s on.

Don
Dr. Donald Polvani
Adjunct Faculty, Physics
Anne Arundel Community College