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JEE Main 2026 (6 Apr, Shift 1) · Periodic Motion & SHM Basics

Periodic Motion & SHM Basics question from JEE Main 2026 (6 Apr, Shift 1)

Q14JEE Main 2026 · 6 APR · SHIFT 1Medium

Periodic Motion & SHM Basics

A particle is executing simple harmonic motion. Its amplitude is AA and time period is 5 sec. The time required by it to move from x=Ax = A to x=A2x = \frac{A}{\sqrt{2}} is ______ sec.

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Answer: (c)

  1. From the extreme: x=Acos⁡ωt=A2⇒ωt=π4x = A\cos\omega t = \frac{A}{\sqrt{2}} \Rightarrow \omega t = \frac{\pi}{4}
  2. t=T8=58 st = \frac{T}{8} = \frac{5}{8}\text{ s}
  3. So the answer is (c) 58\frac{5}{8} s.
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