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JEE Advanced 2009Paper 1
Exam
JEE Advanced
Year
2009
Paper
Paper 1
Subject
Physics
Chapter
Oscillations

JEE Advanced 2009 · Paper 1 · Physics · Oscillations

The x-t graph of a particle undergoing simple harmonic motion is...

The x−tx-t graph of a particle undergoing simple harmonic motion is shown below. The acceleration of the particle at t=43 st=\frac{4}{3} \mathrm{~s} is The x-t graph of a particle undergoing simple harmonic motion is shown below. The acceleration of the particle at

Options

  1. A332π2cms−2\frac{\sqrt{3}}{32} \pi^2 \mathrm{cms}^{-2}
  2. B−π232cms−2\frac{-\pi^2}{32} \mathrm{cms}^{-2}
  3. Cπ232cms−2\frac{\pi^2}{32} \mathrm{cms}^{-2}
  4. D−332π2cms−2-\frac{\sqrt{3}}{32} \pi^2 \mathrm{cms}^{-2}

Correct answer

D. −332π2cms−2-\frac{\sqrt{3}}{32} \pi^2 \mathrm{cms}^{-2}

Step-by-step solution

T=8 \mathrm{~s}, \omega=\frac{2 \pi}{T}=\left(\frac{\pi}{4}\right) \operatorname{rads}^{-1} x=Asin⁡\omegat∴a=−ω2x=−(π216)sin⁡(π4t)\begin{aligned}x & =A \sin \omegat \\ \therefore \quad a=-\omega^2 x & =-\left(\frac{\pi^2}{16}\right) \sin \left(\frac{\pi}{4}t\right) \end{aligned} Substituting t=43 st=\frac{4}{3} \mathrm{~s}, we get a=-\left(\frac{\sqrt{3}}{32} \pi^2\right) cms−2\mathrm{cms}^{-2}
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