From the previous post "No Solution But Exit Velocity Anyway",
\(\cfrac { d^{ 2 }x }{ dt^{ 2 } } =\cfrac { 1 }{ m } \psi _{ c } -{ \cfrac { u }{ \psi _{ c }+u } }\cfrac { 1 }{ e^{ u }\left( { e^{ 2u }-1 } \right) ^{ 1/2 } } (\cfrac { dx }{ dt } )^{ 2 }\)
Obviously,
\(\cfrac { d^{ 2 }x }{ dt^{ 2 } } =\cfrac { 1 }{ m } \psi _{ c } \)
when
\(\cfrac { dx }{ dt }=v=0\)
irrespective of the value for \(x\).
This is the maximum acceleration of the oscillating particle when \(v=0\).
