Answer:

(a)

\(x=2\ln\!\left(\cos\left(t-\frac{\pi}{4}\right)\right)+\ln2.\)

(b)

\(t=\frac{\pi}{2}\text{ s}.\)

Take the positive direction to be the initial direction of motion of \(P\). Since the acceleration is in the direction \(PO\), it is towards \(O\), so while \(P\) is moving away from \(O\),

\(\frac{dv}{dt}=-\frac12(v^2+4).\)

Separate variables:

\(\frac{dv}{v^2+4}=-\frac12\,dt.\)

Integrating,

\(\frac12\tan^{-1}\left(\frac{v}{2}\right)=-\frac12t+C.\)

Initially \(t=0\) and \(v=2\), so

\(C=\frac12\tan^{-1}(1)=\frac{\pi}{8}.\)

Thus

\(\frac12\tan^{-1}\left(\frac{v}{2}\right)=-\frac12t+\frac{\pi}{8},\)

so

\(\tan^{-1}\left(\frac{v}{2}\right)=\frac{\pi}{4}-t.\)

Hence

\(v=2\tan\left(\frac{\pi}{4}-t\right).\)

Since \(v=\frac{dx}{dt}\),

\(\frac{dx}{dt}=2\tan\left(\frac{\pi}{4}-t\right) =-2\tan\left(t-\frac{\pi}{4}\right).\)

Therefore

\(x=2\ln\!\left(\cos\left(t-\frac{\pi}{4}\right)\right)+B.\)

Using \(x=0\) when \(t=0\),

\(0=2\ln\left(\cos\left(-\frac{\pi}{4}\right)\right)+B =2\ln\left(\frac{1}{\sqrt2}\right)+B.\)

So \(B=\ln2\), and

\(x=2\ln\!\left(\cos\left(t-\frac{\pi}{4}\right)\right)+\ln2.\)

For \(P\) to pass through \(O\) again, set \(x=0\):

\(2\ln\!\left(\cos\left(t-\frac{\pi}{4}\right)\right)+\ln2=0.\)

Hence

\(\ln\!\left(2\cos^2\left(t-\frac{\pi}{4}\right)\right)=0,\)

so

\(2\cos^2\left(t-\frac{\pi}{4}\right)=1.\)

The next positive time after \(t=0\) is

\(t=\frac{\pi}{2}\text{ s}.\)