A light elastic string has modulus of elasticity \(\frac { 3 } { 2 } m g\) and natural length \(a\). A particle of mass \(m\) is attached to one end of the string. The other end of the string is attached to a fixed point \(A\). The particle is released from rest at \(A\). Show that when the particle has fallen a distance \(k a\) from \(A\), where \(k > 1\), its kinetic energy is $$\frac { 1 } { 4 } m g a \left( 10 k - 3 - 3 k ^ { 2 } \right) .$$ Show that the particle first comes to instantaneous rest at the point \(B\) which is at a distance \(3 a\) vertically below \(A\). Show that the time taken by the particle to travel from \(A\) to \(B\) is $$\sqrt { } \left( \frac { 2 a } { g } \right) + \frac { 2 \pi } { 3 } \sqrt { } \left( \frac { 2 a } { 3 g } \right)$$