Nice To Know Instantaneously

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 $v=0$.