Who's Ringing? Wobbly Hydrogen?

From the post "Beijing Mask Opera and Wobbling Hydrogen",

$F_{ t }=m_{ e }a=\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o }r^{ 2 } } cos(\theta )$ --- (1)

When $v=c$,

$r=\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o }m_{ e }c^{ 2 } } sin^{ 2 }(\theta )=\cfrac { x }{ cos(\theta ) }$ --- (2)

$x=\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o }m_{ e }c^{ 2 } } sin^{ 2 }(\theta )cos(\theta )$

$\cfrac { dx }{ d\theta  } =-\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o }m_{ e }c^{ 2 } } (2{ \cos ^{ 2 } \left( \theta  \right) \sin  \left( \theta  \right) -\sin ^{ 3 } \left( \theta  \right)  })$

The negative sign is the result of $x$ decreasing as $\theta$ increases.

$\cfrac { d^2x }{ d\theta^2  } =-\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o }m_{ e }c^{ 2 } }( 2cos^3(\theta)-7cos(\theta)sin^2(\theta))$

$\cfrac { d^{ 2 }x }{ d\theta ^{ 2 } } \cfrac { d\theta ^{ 2 } }{ d^{ 2 }t } =\cfrac { dx^{ 2 } }{ dt^{ 2 } } =-\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o }m_{ e }c^{ 2 } } (2cos^{ 3 }(\theta )-7cos(\theta)sin^{ 2 }(\theta))\cfrac { d\theta ^{ 2 } }{ d^{ 2 }t }$ --- (*)

Substitute (2) into  (1)

$m_{ e }a=\cfrac { 4\pi \varepsilon _{ o }m^{ 2 }_{ e }c^{ 4 } }{ q^{ 2 } } \cfrac { cos(\theta ) }{ sin^{ 4 }(\theta ) }$

$a=\cfrac { 4\pi \varepsilon _{ o }m_{ e }c^{ 4 } }{ q^{ 2 } } \cfrac { cos(\theta ) }{ sin^{ 4 }(\theta ) } =\cfrac { dx^{ 2 } }{ dt^{ 2 } }$

From  (*),

$\cfrac { 4\pi \varepsilon _{ o }m_{ e }c^{ 4 } }{ q^{ 2 } } \cfrac { cos(\theta ) }{ sin^{ 4 }(\theta ) } =-\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o }m_{ e }c^{ 2 } } (2cos^{ 3 }(\theta )-7cos(x)sin^{ 2 }(\theta))\cfrac { d\theta ^{ 2 } }{ d^{ 2 }t }$

$\cfrac { 16\pi ^{ 2 }\varepsilon ^{ 2 }_{ o }m^{ 2 }_{ e }c^{ 6 } }{ q^{ 4 } } =-sin^{ 4 }(\theta )(2cos^{ 2 }(\theta )-7sin^{ 2 }(\theta))\cfrac { d\theta ^{ 2 } }{ d^{ 2 }t }$

$\cfrac { 16\pi ^{ 2 }\varepsilon ^{ 2 }_{ o }m^{ 2 }_{ e }c^{ 6 } }{ q^{ 4 } } \iint _{ 0 }^{ T }{ d^{ 2 }t } =-\iint _{ 0.9553 }^{ \frac { \pi  }{ 2 }  }{ sin^{ 4 }(\theta )(2cos^{ 2 }(\theta )-7sin^{ 2 }(\theta)) } d\theta ^{ 2 }$

This was solved on the web,

$-\int _{ 0.955 }^{ \frac { \pi  }{ 2 }  } \int  \sin ^{ 4 } \left( \theta \right) \left( 2\cos ^{ 2 } \left( \theta \right) -7\sin ^{ 2 } \left( \theta \right)  \right) d\theta d\theta\: =\: \cfrac { 2358870+3 }{ 1280000 } -\cfrac { 33\pi ^{ 2 } }{ 128 } +\cfrac { 103\cos  \left( \cfrac { 191 }{ 100 }  \right)  }{ 128 } -\cfrac { 23\cos  \left( \cfrac { 191 }{ 50 }  \right)  }{ 256 } +\cfrac { \cos  \left( \cfrac { 573 }{ 100 }  \right)  }{ 128 }$

$\cfrac { 8\pi ^{ 2 }\varepsilon ^{ 2 }_{ o }m^{ 2 }_{ e }c^{ 6 } }{ q^{ 4 } } { T }^{ 2 }$=0.8927

And the period of this oscillation is $4T$,

$T=\sqrt{0.8927\cfrac { q^{ 4 } }{ 8\pi ^{ 2 }\varepsilon ^{ 2 }_{ o }m^{ 2 }_{ e }c^{ 6 } }}$

$T=\sqrt{\cfrac{0.8927}{8}}\cfrac { q^{ 2 } }{ \pi \varepsilon _{ o }m_{ e }c^{ 3 } }$

$T_p = 4*T$  = 4*( 0.8927/8)^(1/2)*(1.602176565e-19)^2/(pi*(8.854187817e-12)*(9.10938291e-31)*(299792458)^3)

$T_p$ = 5.02388e-23 s

And the resonance frequency is,

$f$ = 1.9905e22 Hz

Comparing this to the previous estimate, this value is fair.  It is expected that the oscillation frequency be higher because the system is not linear ($\frac{1}{r^2}$) as was the first estimate.  The correction factor is pi/4*(8/0.8927)^(1/2) = 2.3511.