Solution:
Distance travelled by the mass sideways is given as $a=2.0 \mathrm{~cm}$
Angular frequency of oscillation can be calculated as,
$\omega=\sqrt{k} / \mathrm{m}$
$=\sqrt{1200 / 3}$
$=\sqrt{400}$
$=20 \mathrm{rad} \mathrm{s}^{-1}$
(a) As time is noted from the mean position,
So, using
$x=a \sin \omega t$
We have,
$x=2 \sin 20 t$
(b) At maximum stretched position, the body is at the extreme right position, with an initial phase of $\pi /2 rad$. Then,
$x=a \sin (\omega t+m / 2)$
$=\mathrm{a} \cos \omega \mathrm{t}$
$=2 \cos 20 \mathrm{t}$