Applying linear momentum conservation,

$0.5 \times 3=(1+0.5) v \text { or } v =1 m / s$

By conversation of energy,

After collision

$\frac{1}{2}(1+0.5) v ^2=\frac{1}{2} kA ^2$

$\Rightarrow A =\sqrt{\frac{1.5}{ k }} \times v$

$\Rightarrow A =\sqrt{\frac{1.5}{600}} \times 1=\frac{1}{20} m =0.05 m$

$A =5 cm$

Time period of oscillation,

$T =2 \pi \sqrt{\frac{ m _1+ m _2}{ k }}=2 \pi \sqrt{\frac{1.5}{600}}=\frac{2 \pi}{20}=\frac{\pi}{10} s$