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.
