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MHT CET 2026 (19 Apr, Shift 1) · Periodic Motion & SHM Basics

Periodic Motion & SHM Basics question from MHT CET 2026 (19 Apr, Shift 1)

Q12MHT CET 2026 · 19 APR · SHIFT 1Medium

Periodic Motion & SHM Basics

A particle executes a simple harmonic motion with a periodic time 88 second. At time t=0t = 0, it is at a mean position. The ratio of the distance traveled by a particle in the 2nd and that in the 1st second of its motion is

(sin⁡45∘=cos⁡45∘=12, sin⁡90∘=cos⁡0∘=1\sin 45^\circ = \cos 45^\circ = \frac{1}{\sqrt{2}},\ \sin 90^\circ = \cos 0^\circ = 1)

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Answer: (c)

  1. ω=2π8=π4\omega = \frac{2\pi}{8} = \frac{\pi}{4}, x=Asin⁡πt4x = A\sin\frac{\pi t}{4}
  2. x(1)=A2x(1) = \frac{A}{\sqrt{2}}, x(2)=Ax(2) = A
  3. 1st second: A2\frac{A}{\sqrt{2}}; 2nd second: A(1−12)A\left(1 - \frac{1}{\sqrt{2}}\right)
  4. Ratio (2nd : 1st) =2−1= \sqrt{2} - 1
  5. So the answer is (c) 2−1\sqrt{2} - 1.
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