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MHT CET 2025 (20 Apr, Shift 1) · Periodic Motion & SHM Basics

Periodic Motion & SHM Basics question from MHT CET 2025 (20 Apr, Shift 1)

Q23MHT CET 2025 · 20 APR · SHIFT 1Medium

Periodic Motion & SHM Basics

A particle oscillates in straight line simple harmonically with period 8 second and amplitude 42 m4\sqrt{2}\text{ m}. Particle starts from mean position. The ratio of the distance travelled by it in 11st second of its motion to that in 22nd second is (sin⁡45∘=1/2,sin⁡π2=1)\left(\sin 45^\circ = 1/\sqrt{2}, \sin\frac{\pi}{2} = 1\right)

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Answer: (d)

  1. ω=π4\omega = \frac{\pi}{4}, x=42sin⁡πt4x = 4\sqrt{2}\sin\frac{\pi t}{4}
  2. x(1)=4 mx(1) = 4\text{ m}, x(2)=42 mx(2) = 4\sqrt{2}\text{ m}
  3. 1st second: 44; 2nd second: 42−4=4(2−1)4\sqrt{2} - 4 = 4(\sqrt{2} - 1)
  4. So the answer is (d) 1:(2−1)1 : (\sqrt{2} - 1).
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