Sine displacement of an oscillating particle, verifying the motion equation and deriving its velocity–displacement relation

A particle oscillates along a horizontal straight line between two end points, and the fixed point \(O\) lies midway between them. At time \(t\) seconds, the displacement of the particle from \(O\) is \(x\) metres, where \(x=4\sin(2t+\frac\pi3)\) and \(0\leq t\leq\pi\).

1. Determine
   1. the amplitude of the motion;[1]
   2. the displacement of the particle when \(t=0\);[2]
   3. the time at which the particle reaches \(O\) for the first time.[2]
2. In general, a particle moving along a straight line has displacement \(s=A\sin(nt+b)\) metres at time \(t\) seconds, where \(t\geq0\), \(A>0\), \(n>0\) and \(-\pi\leq b\leq\pi\).
   Its velocity is \(v=\frac{\mathrm{d}s}{\mathrm{d}t}\) and its acceleration is \(a=\frac{\mathrm{d}v}{\mathrm{d}t}\).
   Obtain \(\frac{\mathrm{d}s}{\mathrm{d}t}\) and \(\frac{\mathrm{d}^2s}{\mathrm{d}t^2}\), and hence show that \(a=-n^2s\).[2]
3. 1. Using the chain rule, show that \(a=v\frac{\mathrm{d}v}{\mathrm{d}s}\), where \(v\) denotes the velocity.[1]
   2. When \(s=A\), the particle is momentarily at rest. Separate the variables in \(v\frac{\mathrm{d}v}{\mathrm{d}s}=-n^2s\) to show that \(v^2=n^2(A^2-s^2)\).[5]
   3. Hence, or otherwise, find the greatest speed of the particle.[2]