A light elastic spring hanging vertically with a mass attached to its lower end. A light elastic spring has a natural length $a$ and modulus of elasticity $λ$. The energy stored in the spring when it is stretched is
$$\frac{λx^2}{2a}$$
where $x$ is the extension. A light elastic spring of natural length $0.2m$ and modulus of elasticity $50N$ hangs vertically with one end attached to a fixed point and with a particle of mass $2Kg$ attached to the lower end.
(i) Calculate the extension of the spring when the particle is in equilibrium.
(ii) The particle is pulled down below its equilibrium position until the total extension of the spring is $0.2m$ and it is then released from rest in this position. Calculate the speed of the particle when it passes the equilibrium position, and find the maximum compression of the spring in the resulting motion.
Answers so far:
(i) In equilibrium, $b$ is the static extension and $T$ is the tension in the spring (Hooke's Law):
$$T=\frac{λb}{a}$$
Resolving vertically:
$$T=mg$$
so
$$b=\frac{mga}{λ}=\frac{(2)(10)(0.2)}{50}=0.08m$$
(ii) The total extension of the spring is $(b+x)=0.2m(b+x)=0.2m$, and the spring is released from rest in this position. In the extended position:
$$T=\frac{λ(b+x)}{a}=\frac{(50)(0.2)}{0.2}=50N$$
The velocity should be $0.6 m/s$ (according to the book, which now turns out to be wrong - see the comments below). I have done another edit in this section as well to give the correct value for $ω$ and the velocity of the particle as it passes the equilibrium position.
Using:
$$v^2=ω^2(a^2−x^2)$$
with $ω=√125$ $/s$, $a=0.12m$ and $x=0 m$, as we want the velocity as the mass passes the equilibrium position, which is at $x=0m$, so we have:
$$v^2 = (√125)^2.(0.12^2 - 0^2) = 1.8$$
and $$v=√(1.8) = 1.3m/s$$
The $ω$ comes from the differential equation of the system, which is:
$$\frac{d^2 x}{dt^2}=-125x$$
For the maximum compression of the spring the answer is quoted as $3N$, but should this be in $m$ not $N$ as we can talk about a compression of a spring in terms of its length?
How was the $3N$ for the maximum compression of the spring worked out?

Your value of $\omega$ is incorrect, and you should have $$\omega^2=\frac{\lambda}{ml}=\frac{50}{2\times0.2}=125$$
The particle performs simple harmonic motion with centre at the equilibrium position, and the amplitude is $a=0.2-0.08=0.12$.
The particle attains its maximum speed as it passes the equilibrium position, and $$v_{max}=\omega a=\sqrt{125}\times0.12$$
The maximum compression occurs when the particle is displaced by one amplitude from the centre, so is $0.12-0.08=0.04$
The $3N$ must be a misprint
