The figure shows the $x-t$ plot of a particle executing one-dimensional simple harmonic motion. Determine the signs of the position,velocity,and acceleration variables of the particle at $t = 0.3 \; s$,$t = 1.2 \; s$,and $t = -1.2 \; s$.

(N/A) For a particle in simple harmonic motion $(SHM)$,the acceleration $(a)$ is related to position $(x)$ by the equation:
$a = -\omega^2 x$ ... $(i)$
where $\omega$ is the angular frequency.
$1$. At $t = 0.3 \; s$:
The particle is in the region where $x < 0$ (negative position). The slope of the $x-t$ graph,which represents velocity $(v = dx/dt)$,is also negative. Using equation $(i)$,since $x$ is negative,$a = -\omega^2 (-|x|) = +\omega^2 |x|$,so acceleration is positive.
Signs: Position: Negative,Velocity: Negative,Acceleration: Positive.
$2$. At $t = 1.2 \; s$:
The particle is in the region where $x > 0$ (positive position). The slope of the $x-t$ graph is positive,so velocity is positive. Using equation $(i)$,since $x$ is positive,$a = -\omega^2 (+|x|) = -\omega^2 |x|$,so acceleration is negative.
Signs: Position: Positive,Velocity: Positive,Acceleration: Negative.
$3$. At $t = -1.2 \; s$:
The particle is in the region where $x < 0$ (negative position). The slope of the $x-t$ graph at this point is positive (the curve is rising),so velocity is positive. Using equation $(i)$,since $x$ is negative,$a = -\omega^2 (-|x|) = +\omega^2 |x|$,so acceleration is positive.
Signs: Position: Negative,Velocity: Positive,Acceleration: Positive.