Answer

In simple harmonic motion,
acceleration $\quad a=-\omega^2 x\quad\quad...(1)$
$\quad$$\quad$(where $\omega$ is constant)
and velocity $\quad v=\frac{d x}{d t}\quad\quad...(2)$
(i) When $t=0.3$ $sec$ and $x$ is negative.
The slope of the $x-t$ graph is negative from which $v$ is also negative.
From equation (1) the acceleration a is positive.
(ii) When $t =1.2$ $sec$ then $x$ is positive.
The slope of the $x-t$ graph is positive from which $v$ is also positive.
From equation (1) the acceleration a is negative.
(iii) when $t =-1.2$ $sec$ them $x$ is negative.
The slope of the $x-t$ graph is positive so that $v$ is positive.
From equation (1) the acceleration $a$ is positive.

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