and \(v = \frac{{dx}}{{dt}} = - a\omega \sin (\omega \,t + \theta )\) ….(ii)

Given at \(t = 0\), \(x = 1\,cm\) and \(v = \pi \) and \(\omega = \pi \)

Putting these values in equation (i) and (ii) we will get \(\sin \theta = \frac{{ - 1}}{a}\) and \(\cos \theta = \frac{1}{a}\)

==> \({\sin ^2}\theta + {\cos ^2}\theta = {\left( { - \frac{1}{a}} \right)^2} + {\left( {\frac{1}{a}} \right)^2}\)

==> \(a = \sqrt 2 \,cm\)