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 \(2m\omega^2a\), 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 \(2m\omega v\), where \(v\) is the speed of the particle. At time \(t\), the displacement of \(P\) downwards from its equilibrium position is \(x\).
- Show that \(\frac{\mathrm{d}^2x}{\mathrm{d}t^2} + 2\omega \frac{\mathrm{d}x}{\mathrm{d}t} + 2\omega^2x = 0\).
[5]
Given that the solution of this differential equation is \(x = e^{-\omega t}(A \cos \omega t + B \sin \omega t)\), where \(A\) and \(B\) are constants,
- find \(A\) and \(B\).
[4]
- Find an expression for the time at which \(P\) first comes to rest.
[3]