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Q2.16: NCERT Physics - Part 1 MOTION IN A STRAIGHT LINE 2.16 Solution: Figure 2.13 gives the x-t plot of a particle executing…
⚠️ Common Exam Traps & Conceptual Pitfalls
Students often read the curvature of the x‑t graph and assign the acceleration sign directly, forgetting that in simple harmonic motion $a=-\omega^2 x$, so the acceleration sign is always opposite the position sign, not the slope sign. Always check the displacement first, then apply $a=-\omega^2 x$ to determine the correct sign.
Figure 2.13 gives the x-t plot of a particle executing one-dimensional simple harmonic motion. Give the signs of position, velocity and acceleration variables of the particle at t = 0.3 s, 1.2 s, � 1.2 s.
Step-by-Step Verified Solution
**Answer:** At t = 0.3 s: x > 0, v < 0, a < 0. At t = 1.2 s: x < 0, v > 0, a > 0. At t = -1.2 s: x < 0, v > 0, a > 0.
**Step-by-Step Explanation:**
From the x-t graph of SHM: at t = 0.3 s, position is positive, slope is negative (v < 0), and acceleration is opposite to displacement (a < 0). At t = 1.2 s, position is negative, slope is positive (v > 0), and acceleration is opposite to displacement (a > 0). At t = -1.2 s (same as t = 1.2 s due to periodicity), same signs. [citation:19]
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