Q 11-13-007NEETNEET 2023Top questionMedium
The $x$-$t$ graph of a particle performing simple harmonic motion is shown in the figure. The acceleration of the particle at $t = 2$ s is :
Answer: (D) $-\dfrac{\pi^2}{16}\ \text{m s}^{-2}$
Concept. In SHM $a = -\omega^2x$; read $A$ and $T$ off the graph.
Step 1. From the graph. $A = 1$ m, $T = 8$ s, so $\omega = \dfrac{\pi}{4}$ rad/s and $x = \sin\dfrac{\pi t}{4}$.
Step 2. At $t = 2$ s. $x = 1$ m (positive extreme).
Step 3. Acceleration. $a = -\dfrac{\pi^2}{16} \times 1 = -\dfrac{\pi^2}{16}$ m/s$^2$.
Answer. $-\dfrac{\pi^2}{16}$ m s$^{-2}$ (Option D).
Solution by Sreeraj P, M.Sc Physics