b
by using formula of time period

\(T=2 \pi \sqrt{\frac{m}{k}}\)

\(0.2=2 \pi \sqrt{\frac{0.5}{k}}\)

\(k=50 \pi^{2}\)

\(\approx 500\)

\(x=A\, \sin (\omega t+\phi)\)

\(=5\, c m\, \sin \left(\frac{\omega t}{4}+0\right)\)

\(=5\, {cm} \sin \left(\frac{\pi}{2}\right)\)

\(=5 \,{cm}\)

\(P E=\frac{1}{2} k x^{2}\)

\(=\frac{1}{2}(500)\left(\frac{5}{100}\right)^{2}\)

\(=0.6255\)