b
(b) Initially when \(1\, kg\) mass is suspended then by using \(F = kx\)

\( \Rightarrow mg = kx\)

\( \Rightarrow k = \frac{{mg}}{x} = \frac{{1 \times 10}}{{5 \times {{10}^{ - 2}}}} = 200\frac{N}{m}\)

Further, the angular frequency of oscillation of \(2\, kg\) mass is

\(\omega = \sqrt {\frac{k}{M}} = \sqrt {\frac{{200}}{2}} = 10\,rad/sec\)

Hence, \({v_{\max }} = a\omega = (10 \times {10^{ - 2}}) \times 10 = 1\,m/s\)