From this reason, both \(\small{v_{cir}}\) and \(\small{v_{shm}}\) are negative at the same time. Where, from the post "Twirl Plus SHM, Spinning Coin" dated 17 Jul 2015,
\(v^{ 2 }_{ cir }=cos(\theta )(x+x_{ z })\left\{ -2{ c^{ 2 } }cos(\theta )ln(cosh(\cfrac { G }{ \sqrt { 2{ mc^{ 2 } } } } x))+\cfrac { \psi _{ d } }{ m } \right\}\)
and
\(v^{ 2 }_{ shm }=sin(\theta )(x+x_{ z })\left\{ -2{ c^{ 2 } }cos(\theta )ln(cosh(\cfrac { G }{ \sqrt { 2{ mc^{ 2 } } } } x))+\cfrac { \psi _{ d } }{ m } \right\}\)
their signs depend on the common term,
\((x+x_{ z })\left\{ -2{ c^{ 2 } }cos(\theta )ln(cosh(\cfrac { G }{ \sqrt { 2{ mc^{ 2 } } } } x))+\cfrac { \psi _{ d } }{ m } \right\}\)
When
\(v_{cir}=v_{shm}=c\)
\(x_d=x_v\)
circular motion is in a plane whose normal at the center of the circular path passes above \(O\), the center of oscillation. This occurs instantaneously and does not alter the analysis presented.
Have a nice day.
