Q.Figure 2.13 gives the $x$-$t$ plot of a particle executing one-dimensional simple harmonic motion. (You will learn about this motion in more detail in Chapter 13). Give the signs of position, velocity and acceleration variables of the particle at $t = 0.3$ s, $1.2$ s, $-1.2$ s.

At $t = 0.3\,$s: the phase is $\pi(0.3) = 0.3\pi$ rad $= 54^\circ$, which lies between $0^\circ$ and $90^\circ$ (the particle is in its first swing away from the origin, not yet at the far end). Here $\sin(0.3\pi) > 0$, so $x = -A\sin(0.3\pi) < 0$. Also $\cos(0.3\pi) > 0$, so $v = -A\pi\cos(0.3\pi) < 0$ (still moving away from the origin, in the negative direction). Since $x<0$, $a = -\pi^2 x > 0$ (accelerating back towards $x=0$).

At $t = 1.2\,$s: the phase is $1.2\pi$ rad $= 216^\circ$, which lies between $180^\circ$ and $270^\circ$ (third quadrant), where both $\sin$ and $\cos$ are negative. So $x = -A\sin(1.2\pi) = -A(\text{negative}) > 0$, $v = -A\pi\cos(1.2\pi) = -A\pi(\text{negative}) > 0$, and since $x>0$, $a = -\pi^2 x < 0$. …

Setting up the motion

The curve is zero at $t=0$ and negative just after, so write

Then

The last relation, $a = -\omega^2 x$, is the defining property of SHM: acceleration is always opposite in sign to the position.

Evaluating the signs

At $t = 0.3\,$s ($0.3\pi = 54^\circ$):

• $x = -A\sin 54^\circ = -0.81A < 0$ → negative
• $v = -A\pi\cos 54^\circ = -0.59A\pi < 0$ → negative
• $a = -\omega^2 x > 0$ → positive

At $t = 1.2\,$s ($1.2\pi = 216^\circ$):

• $x = -A\sin 216^\circ = -A(-0.59) = +0.59A > 0$ → positive
• $v = -A\pi\cos 216^\circ = -A\pi(-0.81) = +0.81A\pi > 0$ → positive
• $a = -\omega^2 x < 0$ → negative …

Concept: Reading Signs Directly Off the Sketched Curve

Method: Direct Curve-Reading by Region (slope and concavity, no explicit trigonometric formula)

Rather than writing $x(t)=-A\sin(\pi t)$ and differentiating it twice, this method reads every sign directly off the described shape of the curve — using only two general facts valid for any smooth, periodic up-down curve: (i) inside any trough (a stretch between two zero-crossings that dips below the axis), the curve is concave up everywhere, and inside any crest, concave down everywhere;

(ii) within a trough or crest, the velocity sign flips exactly once, at the extremum itself — negative (falling) before it, positive (rising) after it, for a trough (and the reverse for a crest).

Setting up the regions from the description

• Trough (dips below axis): between $t=0$ and $t=1$ (bottom near $t=0.5$), and again between $t=-2$ and $t=-1$ (bottom near $t=-1.5$), by the stated period of $2\ \text{s}$.
• Crest (rises above axis): between $t=-1$ and $t=0$ (top near $t=-0.5$), and between $t=1$ and $t=2$ (top near $t=1.5$).

Evaluating each instant

1. $t=0.3\ \text{s}$ lies inside the trough $(0,1)$, before its bottom at $t=0.5$: position is below the axis ($x<0$); the curve is still descending toward the bottom, so velocity is still falling ($v<0$); concavity is upward throughout this entire trough ($a>0$).

   → $x<0,\ v<0,\ a>0$

2. $t=1.2\ \text{s}$ lies inside the crest $(1,2)$, before its top at $t=1.5$: position is above the axis ($x>0$); the curve is still rising toward the top, so velocity is still rising ($v>0$); concavity is downward throughout this entire crest ($a<0$).

   → $x>0,\ v>0,\ a<0$ …