Explanation

(b) $m_{1} = 1\ kg,\ extension\ l_{1} = 5\ cm = 5 \times 10^{2}\ m$
$\therefore$ $m_{1}g = kl_{1}$
$k =$ force constant of the spring
$k = \frac{m_{1}g}{l_{1}} = \frac{1 \times 10}{5 \times 10^{- 2}} = 200\ Nm^{- 1}$
Time period of the block of mass 2 kg.
$T = 2\pi\sqrt{\frac{m}{k}} = 2\pi\sqrt{\frac{2}{200}} = 2\pi \times \frac{1}{10} = \frac{\pi}{5}$ s
Maximum velocity $v_{\max} = A\omega$
where $A =$ Amplitude
$= 10\ cm = 10 \times 10^{- 2}\ m$
$v_{\max} = A \times \frac{2\pi}{T} = 10 \times 10^{- 2} \times \frac{2\pi}{\frac{\pi}{5}}$
$= 10^{- 1} \times 2 \times 5 = 1\ ms^{- 1}$