Twirl Plus SHM, Spinning Coin

It is possible that the negative particle is in both circular motion and simple harmonic motion,

The total energy of the particle in circular motion is a constant, both KE and PE are constants.  Adding a constant to the simple harmonic equation does not introduce any changes.  Combined, the path of the negative particle is a sin wave wrap around in circular motion, wider at the top that is further from the center of the positive particle.

The path due to SHM marked by the double arrows when extrapolated passes through the center of the positive particle, $O$.

This motion is like a spot on the rim of a coin or thin disc spinning around on its circular edge, at an low incline to the horizontal plane.  Except the electron's path is in a wider circle further from center of the positive particle.  In the diagram above, the crown opens up on top.

The equations are

$F_{SHM}=m\cfrac{d^2x}{dt^2}=-2{ mc^{ 2 } }ln(cosh(\cfrac { G }{ \sqrt { 2{ mc^{ 2 } } }  }x))+\psi_d$

where $x$ is the displacement from the point $x=x_z$ along the line $OP$, where $\psi(x_z)=\psi_{max}$ on the energy density graph, $\psi(x)$ along $OP$ of the containing particle.  Furthermore,

$\psi_d=\psi_{max}-\psi_n$

We have assumed here that the negative particle has energy density $\psi_n$ when it enters into the $\psi$ cloud.

$F_{cir}=m\cfrac{v^2_{cir}}{(x+x_z)cos(\theta)}$

$v_{shm}$ along $OP$ and $v_{cir}$ are at an angle $\angle \theta$. The force on the particle is,

$F_{p}=\psi(x)-\psi_{n}=F_{SHM}$ along $OP$

where the negative particle moves into region of higher $\psi$.

$F_{cir}=F_pcos(\theta)=-2{ mc^{ 2 } }cos(\theta)ln(cosh(\cfrac { G }{ \sqrt { 2{ mc^{ 2 } } }  }x))+\psi_d$

So,

$m\cfrac{v^2_{cir}}{(x+x_z)cos(\theta)}=-2{ mc^{ 2 } }cos(\theta)ln(cosh(\cfrac { G }{ \sqrt { 2{ mc^{ 2 } } }  }x))+\psi_d$

$v^{ 2 }_{ cir }=cos(\theta )(x+x_{ z })\left\{ -2{ c^{ 2 } }cos(\theta )ln(cosh(\cfrac { G }{ \sqrt { 2{ mc^{ 2 } } }  } x))+\cfrac { \psi _{ d } }{ m }  \right\}$

where $x=0$ is not at $O$ but at $OP=x_z$.  And

$v^{ 2 }_{ SHM }=sin(\theta )(x+x_{ z })\left\{ -2{ c^{ 2 } }cos(\theta )ln(cosh(\cfrac { G }{ \sqrt { 2{ mc^{ 2 } } }  } x))+\cfrac { \psi _{ d } }{ m }  \right\}$

$v_{SHM}$ is the velocity along $OP$.

When the particle passes $x=0$ with velocity $v_{cir,\,x=0}$,

$v^{ 2 }_{ cir,\,x=0}=\cfrac { \psi _{ d } }{ m } x_{ z }cos(\theta )$

From,

$F=-\psi$

here $\psi$ has the unit of a force, $Nm$, and the expression is unit dimension consistent.

A illustrative plot of this graph is shown below,

When the particle is in periodic motion, $v(x)$ should remain analytical (ie infinitely differentiable).  $v(x)$ remains analytical only when at the turn points, $\cfrac{dv}{dx}=0$.  This requires that $v$ oscillates between its two extrema.  If we consider only positive real values for $v_{cir}$, we have a further requirement on $\psi_d$ for a double root at $x=-x_z$,

$\cfrac { \psi _{ d } }{ m } =2{ c^{ 2 } }cos(\theta )ln(cosh(\cfrac { G }{ \sqrt { 2{ mc^{ 2 } } }  } x_z))$

This is a very strict condition.

If however, we allow for complex values of $v_{cir}$, in which case as $v_{cir}$ approaches zero, the radius of  its circular motion is also zero.  At that point the particle set off in a direction perpendicular to its original approach direction and increases.  At the extrema point, the particle retraces along the $v(x)$ curves and decreases.  When it once again attain a zero value at zero radius, the particle set off in a new direction perpendicular to its approach to zero.

In both cases, the amplitude of of oscillation about $x=0$, along the line joining the particle to the center of the containing particle, $A_{osc}$ is the distance along $x$ between the two extrema in the $v^2(x)$ plot.

The particle attains zero circular motion radius at a point before the center of the containing particle or at the center of the containing particle.