$A$ particle is in linear simple harmonic motion between two points,$A$ and $B$,$10 \; cm$ apart. Take the direction from $A$ to $B$ as the positive direction and give the signs of velocity,acceleration,and force on the particle when it is:
$(a)$ at the end $A$.
$(b)$ at the end $B$.
$(c)$ at the mid-point of $AB$ going towards $A$.
$(d)$ at $2 \; cm$ away from $B$ going towards $A$.
$(e)$ at $3 \; cm$ away from $A$ going towards $B$.
$(f)$ at $4 \; cm$ away from $B$ going towards $A$.

(N/A) The mean position $O$ is the midpoint of $AB$. $A$ is at $x = -5 \; cm$ and $B$ is at $x = +5 \; cm$. The positive direction is $A \to B$. In $SHM$,acceleration $a = -\omega^2 x$ and force $F = ma = -m\omega^2 x$.
$(a)$ At $A$ $(x = -5 \; cm)$: Velocity $v = 0$ (extreme point). Acceleration $a = -\omega^2(-5) > 0$ (positive). Force $F$ is positive.
$(b)$ At $B$ $(x = +5 \; cm)$: Velocity $v = 0$ (extreme point). Acceleration $a = -\omega^2(+5) < 0$ (negative). Force $F$ is negative.
$(c)$ At $O$ $(x = 0)$ going towards $A$: Velocity $v < 0$ (moving left). Acceleration $a = 0$. Force $F = 0$.
$(d)$ At $2 \; cm$ from $B$ towards $A$ $(x = +3 \; cm)$: Velocity $v < 0$ (moving left). Acceleration $a = -\omega^2(+3) < 0$ (negative). Force $F$ is negative.
$(e)$ At $3 \; cm$ from $A$ towards $B$ $(x = -2 \; cm)$: Velocity $v > 0$ (moving right). Acceleration $a = -\omega^2(-2) > 0$ (positive). Force $F$ is positive.
$(f)$ At $4 \; cm$ from $B$ towards $A$ $(x = +1 \; cm)$: Velocity $v < 0$ (moving left). Acceleration $a = -\omega^2(+1) < 0$ (negative). Force $F$ is negative.