2 A light elastic string AB has stiffness \(k\). The end A is attached to a fixed point and a particle of mass \(m\) is attached at the end B . With the string vertical, the particle is released from rest from a point at a distance \(a\) below its equilibrium position. At time \(t\), the displacement of the particle below the equilibrium position is \(x\) and the velocity of the particle is \(v\).

1. Show that $$m v \frac { \mathrm {~d} v } { \mathrm {~d} x } = - k x$$
2. Show that, while the particle is moving upwards and the string is taut, $$v = - \sqrt { \frac { k } { m } \left( a ^ { 2 } - x ^ { 2 } \right) }$$
3. Hence use integration to find an expression for \(x\) at time \(t\) while the particle is moving upwards and the string is taut.