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JEE Main 2025 (24 Jan, Shift 1) · Periodic Motion & SHM Basics

Periodic Motion & SHM Basics question from JEE Main 2025 (24 Jan, Shift 1)

Q24JEE Main 2025 · 24 JAN · SHIFT 1Medium

Periodic Motion & SHM Basics

A particle is executing simple harmonic motion with time period 2 s and amplitude 1 cm. If D and d are the total distance and displacement covered by the particle in 12.5 s, then Dd\frac{D}{d} is

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Answer: (c)

  1. 12.5 s=6T+T412.5\text{ s} = 6T + \frac{T}{4} (taking start at the mean position).
  2. Distance: 6×4A+A=25A=25 cm6\times 4A + A = 25A = 25\text{ cm}
  3. In 6T6T the particle returns; in the extra T4\frac{T}{4} it goes mean to extreme, so d=A=1 cmd = A = 1\text{ cm}.
  4. Dd=25\frac{D}{d} = 25
  5. So the answer is (c) 25.
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