The displacement-time ($x-t$) graph of a particle performing simple harmonic motion is shown in the figure. Th

Question

The displacement-time ($x-t$) graph of a particle performing simple harmonic motion is shown in the figure. The acceleration of the particle at $t = 2$ s is:

Accepted Answer

1.  **Analyze the Graph Data:** Although the figure is not explicitly provided in the text, this is a standard NEET 2023 question. The graph depicts a sine wave starting from the origin ($x=0$ at $t=0$).
    *   **Time Period ($T$):** The graph completes one full cycle at $t = 8$ s. Thus, $T = 8$ s.
    *   **Amplitude ($A$):** The maximum displacement shown is $1$ m.
2.  **Calculate Angular Frequency ($\omega$):**
    $$ \omega = \frac{2\pi}{T} = \frac{2\pi}{8} = \frac{\pi}{4} 	ext{ rad/s} $$
3.  **Determine Displacement at $t = 2$ s:**
    The equation of motion for a particle starting from the mean position is $x(t) = A \sin(\omega t)$.
    At $t = 2$ s:
    $$ x = 1 \sin\left(\frac{\pi}{4} 	imes 2ight) = \sin\left(\frac{\pi}{2}ight) = 1 	ext{ m} $$
    (Alternatively, from the graph, at $t = 2$ s (which is $T/4$), the particle is at the positive extreme position, so $x = +A = +1$ m).
4.  **Calculate Acceleration ($a$):**
    The acceleration in SHM is given by $a = -\omega^2 x$.
    $$ a = -\left(\frac{\pi}{4}ight)^2 (1) = -\frac{\pi^2}{16} 	ext{ ms}^{-2} $$

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