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MHT CET 2026 (12 May, Shift 1) · Periodic Motion & SHM Basics

Periodic Motion & SHM Basics question from MHT CET 2026 (12 May, Shift 1)

Q11MHT CET 2026 · 12 MAY · SHIFT 1Medium

Periodic Motion & SHM Basics

A particle performing S.H.M., has a period of 1212 s. The particle starts from a mean position. The ratio of distances travelled by the particle in the first and second second is {cos⁡(π/6)=32}\left\{\cos(\pi/6) = \frac{\sqrt{3}}{2}\right\}

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Answer: (d)

  1. ω=2π12=π6\omega = \frac{2\pi}{12} = \frac{\pi}{6}, x=Asin⁡πt6x = A\sin\frac{\pi t}{6}
  2. x(1)=A2x(1) = \frac{A}{2}, x(2)=3A2x(2) = \frac{\sqrt{3}A}{2}
  3. 1st second: A2\frac{A}{2}; 2nd second: (3−1)A2\frac{(\sqrt{3} - 1)A}{2}
  4. Ratio =13−1= \frac{1}{\sqrt{3} - 1}
  5. So the answer is (d).
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