A particle \(P\) of mass \(m\) is attached to one end of a light inextensible string of length \(a\). The other end of the string is attached to a fixed point \(O\). The particle is held so that the string is taut, with \(O P\) horizontal. The particle is projected downwards with speed \(\sqrt { } \left( \frac { 2 } { 5 } a g \right)\) and begins to move in a vertical circle. The string breaks when its tension is equal to \(\frac { 11 } { 5 } m g\).

1. Show that the string breaks when \(O P\) makes an angle \(\theta\) with the downward vertical through \(O\), where \(\cos \theta = \frac { 3 } { 5 }\). Find the speed of \(P\) at this instant.
2. For the subsequent motion after the string breaks, find the distance \(O P\) when the particle \(P\) is vertically below \(O\).