Answer: The equilibrium position is \(\boxed{AP=0.75\ \mathrm{m}}\), the period is \(\boxed{\pi/5\ \mathrm{s}}\), the speed at equilibrium is \(\boxed{0.5\ \mathrm{m\ s^{-1}}}\), and the speed when the acceleration is half its maximum value is \(\boxed{0.433\ \mathrm{m\ s^{-1}}}\).
(i) Let \(AP=x\) at equilibrium. Then the extension of the spring attached to \(A\) is \(x-0.6\).
Also \(PB=1.2-x\), so the extension of the spring attached to \(B\) is
\[(1.2-x)-0.4=0.8-x.\]
At equilibrium, the two tensions are equal:
\[\frac{10(x-0.6)}{0.6}=\frac{20(0.8-x)}{0.4}.\]
Simplify:
\[\frac{50}{3}(x-0.6)=50(0.8-x).\]
Multiply by \(3/50\):
\[x-0.6=2.4-3x.\]
So
\[4x=3,\]
and therefore
\[\boxed{x=0.75}.\]
Hence \(AP=0.75\ \mathrm{m}\) at equilibrium.
(ii) Let \(x\) now be the displacement of \(P\) from equilibrium away from \(A\), towards \(B\). Then \(AP=0.75+x\).
The extension of the spring attached to \(A\) is
\[0.75+x-0.6=0.15+x.\]
The extension of the spring attached to \(B\) is
\[(1.2-(0.75+x))-0.4=0.05-x.\]
Taking the direction towards \(B\) as positive, Newton's second law gives
\[\frac23\frac{d^2x}{dt^2}=-\frac{10(0.15+x)}{0.6}+\frac{20(0.05-x)}{0.4}.\]
Now
\[\frac{10(0.15+x)}{0.6}=2.5+\frac{50}{3}x,\]
and
\[\frac{20(0.05-x)}{0.4}=2.5-50x.\]
Therefore
\[\frac23\frac{d^2x}{dt^2}=-(2.5+\frac{50}{3}x)+(2.5-50x)=-\frac{200}{3}x.\]
Multiplying by \(\dfrac32\),
\[\frac{d^2x}{dt^2}=-100x.\]
This is the standard SHM form \(x''=-\omega^2x\), so
\[\omega^2=100,\qquad \omega=10.\]
The period is
\[T=\frac{2\pi}{\omega}=\frac{2\pi}{10}=\boxed{\frac\pi5\ \mathrm{s}}.\]
(iii) The particle is released from rest after being displaced \(0.05\ \mathrm{m}\), so the amplitude is
\[A=0.05.\]
The maximum speed occurs at the equilibrium position:
\[v_{\max}=\omega A=10(0.05)=\boxed{0.5\ \mathrm{m\ s^{-1}}}.\]
(iv) In SHM, the magnitude of the acceleration is
\[|a|=\omega^2|x|.\]
The maximum acceleration occurs when \(|x|=A\). Half of the maximum acceleration therefore occurs when
\[|x|=\frac A2=0.025.\]
Use the SHM speed formula
\[v^2=\omega^2(A^2-x^2).\]
Then
\[v=10\sqrt{0.05^2-0.025^2}=0.433\ldots.\]
So the speed is
\[\boxed{0.433\ \mathrm{m\ s^{-1}}}\quad\text{to 3 significant figures}.\]
