NEETPhysicsOscillationsMCQ+4 / −1
The - graph of a particle performing simple harmonic motion is shown in the figure. The acceleration of the particle at is :

- A
- B
- C
- D
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Correct answer: C
$$ \begin{aligned} & \mathrm{x}=\mathrm{A} \sin (\omega \mathrm{t}) \\ & \frac{\mathrm{dx}}{\mathrm{dt}}=\mathrm{v}=\mathrm{A} \omega \cos (\omega \mathrm{t}) \\ & \frac{\mathrm{dv}}{\mathrm{dt}}=\mathrm{a}=-\omega^{2} \mathrm{~A} \sin (\omega \mathrm{t}) \\ & \mathrm{a}=-\left(\frac{2 \pi}{8}\right)^{2} \times 1 \sin \left(\frac{2 \pi}{8} \times 2\right) \\ & \Rightarrow \mathrm{a}=-\frac{\pi^{2}}{16} \times \sin \left(\frac{\pi}{2}\right) \\ & \therefore \mathrm{a}=\frac{-\pi^{2}}{16} \mathrm{~m} / \mathrm{s}^{2} \end{aligned} $$
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