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MHT CET 2025 (22 Apr, Shift 2) · Periodic Motion & SHM Basics

Periodic Motion & SHM Basics question from MHT CET 2025 (22 Apr, Shift 2)

Q22MHT CET 2025 · 22 APR · SHIFT 2Medium

Periodic Motion & SHM Basics

A particle performing linear S.H.M. has period 8 seconds. At time t=0t = 0, it is in the mean position. The ratio of the distances travelled by the particle in the 11st and 22nd second is (cos⁡45∘=1/2)\left(\cos 45^\circ = 1/\sqrt{2}\right)

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Answer: (a)

  1. ω=π4\omega = \frac{\pi}{4}, x=Asin⁡πt4x = A\sin\frac{\pi t}{4}
  2. 1st second: A2\frac{A}{\sqrt{2}}; 2nd second: A−A2=A(2−1)2A - \frac{A}{\sqrt{2}} = \frac{A(\sqrt{2} - 1)}{\sqrt{2}}
  3. Ratio =1:(2−1)= 1 : (\sqrt{2} - 1)
  4. So the answer is (a).
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