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NEET PHYSICSEasy

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

A

−π216 ms−2-\frac{\pi^2}{16} \text{ ms}^{-2}

B

π28 ms−2\frac{\pi^2}{8} \text{ ms}^{-2}

C

−π28 ms−2-\frac{\pi^2}{8} \text{ ms}^{-2}

D

π216 ms−2\frac{\pi^2}{16} \text{ ms}^{-2}

Step-by-Step Solution

  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=0x=0 at t=0t=0).
  • Time Period (TT): The graph completes one full cycle at t=8t = 8 s. Thus, T=8T = 8 s.
  • Amplitude (AA): The maximum displacement shown is 11 m.
  1. Calculate Angular Frequency (ω\omega): ω=2πT=2π8=π4 rad/s\omega = \frac{2\pi}{T} = \frac{2\pi}{8} = \frac{\pi}{4} \text{ rad/s}
  2. Determine Displacement at t=2t = 2 s: The equation of motion for a particle starting from the mean position is x(t)=Asin⁡(ωt)x(t) = A \sin(\omega t). At t=2t = 2 s: x=1sin⁡(π4×2)=sin⁡(π2)=1 mx = 1 \sin\left(\frac{\pi}{4} \times 2\right) = \sin\left(\frac{\pi}{2}\right) = 1 \text{ m} (Alternatively, from the graph, at t=2t = 2 s (which is T/4T/4), the particle is at the positive extreme position, so x=+A=+1x = +A = +1 m).
  3. Calculate Acceleration (aa): The acceleration in SHM is given by a=−ω2xa = -\omega^2 x. a=−(π4)2(1)=−π216 ms−2a = -\left(\frac{\pi}{4}\right)^2 (1) = -\frac{\pi^2}{16} \text{ ms}^{-2}
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