5. A light elastic string, of natural length \(2 a\) and modulus of elasticity \(m g\), has a particle \(P\) of mass \(m\) attached to its mid-point. One end of the string is attached to a fixed point \(A\) and the other end is attached to a fixed point \(B\) which is at a distance \(4 a\) vertically below \(A\).

1. Show that \(P\) hangs in equilibrium at the point \(E\) where \(A E = \frac { 5 } { 2 } a\). The particle \(P\) is held at a distance \(3 a\) vertically below \(A\) and is released from rest at time \(t = 0\). When the speed of the particle is \(v\), there is a resistance to motion of magnitude \(2 m k v\), where \(k = \sqrt { } \left( \frac { g } { a } \right)\). At time \(t\) the particle is at a distance \(\left( \frac { 5 } { 2 } a + x \right)\) from \(A\).
2. Show that $$\frac { \mathrm { d } ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } + 2 k \frac { \mathrm {~d} x } { \mathrm {~d} t } + 2 k ^ { 2 } x = 0$$
3. Hence find \(x\) in terms of \(t\).