5. A particle \(P\) of mass \(m\) is attached to one end of a light elastic string, of natural length \(a\) and modulus of elasticity \(2 m a k ^ { 2 }\), where \(k\) is a positive constant. The other end of the string is attached to a fixed point \(A\). At time \(t = 0 , P\) is released from rest from a point which is a distance \(2 a\) vertically below \(A\). When \(P\) is moving with speed \(v\), the air resistance has magnitude \(2 m k v\). At time \(t\), the extension of the string is \(x\).

1. Show that, while the string is taut, $$\frac { \mathrm { d } ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } + 2 k \frac { \mathrm {~d} x } { \mathrm {~d} t } + 2 k ^ { 2 } x = g$$ You are given that the general solution of this differential equation is $$x = \mathrm { e } ^ { - k t } ( C \sin k t + D \cos k t ) + \frac { g } { 2 k ^ { 2 } } , \quad \text { where } C \text { and } D \text { are constants. }$$
2. Find the value of \(C\) and the value of \(D\). Assuming that the string remains taut,
3. find the value of \(t\) when \(P\) first comes to rest,
4. show that \(2 k ^ { 2 } a < g \left( 1 + \mathrm { e } ^ { \pi } \right)\).