JEE Advanced · Physics · 15. Oscillations
The \(x-t\) graph of a particle undergoing simple harmonic motion is shown below. The acceleration of the particle at \(t=\frac{4}{3} \mathrm{~s}\) is

- A \(\frac{\sqrt{3}}{32} \pi^2 \mathrm{cms}^{-2}\)
- B \(\frac{-\pi^2}{32} \mathrm{cms}^{-2}\)
- C \(\frac{\pi^2}{32} \mathrm{cms}^{-2}\)
- D \(-\frac{\sqrt{3}}{32} \pi^2 \mathrm{cms}^{-2}\)
Answer & Solution
Correct Answer
(D) \(-\frac{\sqrt{3}}{32} \pi^2 \mathrm{cms}^{-2}\)
Step-by-step Solution
Detailed explanation
\(T=8 \mathrm{~s}, \omega=\frac{2 \pi}{T}=\left(\frac{\pi}{4}\right) \operatorname{rads}^{-1}\)
\(x =A \sin \omega t\)
\(\therefore a=-\omega^2 x =-\left(\frac{\pi^2}{16}\right) \sin \left(\frac{\pi}{4} t\right)\)
Substituting \(t=\frac{4}{3} \mathrm{~s}\), we get
\(
a=-\left(\frac{\sqrt{3}}{32} \pi^2\right) \mathrm{cms}^{-2}
\)
\(x =A \sin \omega t\)
\(\therefore a=-\omega^2 x =-\left(\frac{\pi^2}{16}\right) \sin \left(\frac{\pi}{4} t\right)\)
Substituting \(t=\frac{4}{3} \mathrm{~s}\), we get
\(
a=-\left(\frac{\sqrt{3}}{32} \pi^2\right) \mathrm{cms}^{-2}
\)
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