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A spring of natural length l with modulus of elasticity λ has one end fixed to the ceiling. A particle of mass m is attached to the other end of the spring and is left to hang in its equilibrium position under the influence of gravity.

(i) Find the extension $x_E$ of the spring in the equilibrium position.

(ii) The particle is now lowered a distance $a$ below its equilibrium position and released from rest. Working from N2, show that if $z$ denotes the vertical displacement of the particle below its equilibrium position and we neglect theeffects of air resistance, then

$\ddot{z}+ω^2z = 0$,

where $ω^2 = λ/(lm)$. Deduce that $z(t) = a cos(ωt)$.

(iii) Suppose now that the particle is initially at the equilibrium position and is given a positive downward initial velocity $\dot{z}(0) = b$. Show that in this case,

$z(t) = \frac{b}{ω} sin ωt$,

and hence find the first time when the particle comes to rest."

I'm really stuck on this one. I know the position vector of the particle is $\vec{r}=(l+x)\vec{k}$

and I think $T=\frac{\lambda x_E}{l}$ so $x_E=\frac{Tlm}{\lambda}$ but not sure if thats way off.

Thanks