4. A particle \(P\) of mass \(m\) is suspended from a fixed point by a light elastic spring. The spring has natural length \(a\) and modulus of elasticity \(2 m \omega ^ { 2 } a\), where \(\omega\) is a positive constant. At time \(t = 0\) the particle is projected vertically downwards with speed \(U\) from its equilibrium position. The motion of the particle is resisted by a force of magnitude \(2 m \omega v\), where \(v\) is the speed of the particle. At time \(t\), the displacement of \(P\) downwards from its equilibrium position is \(x\).

1. Show that \(\frac { \mathrm { d } ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } + 2 \omega \frac { \mathrm {~d} x } { \mathrm {~d} t } + 2 \omega ^ { 2 } x = 0\). Given that the solution of this differential equation is \(x = \mathrm { e } ^ { - \omega t } ( A \cos \omega t + B \sin \omega t )\), where \(A\) and \(B\) are constants,
2. find \(A\) and \(B\).
3. Find an expression for the time at which \(P\) first comes to rest.
   (3)