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JEE Main 2022 (27 Jun, Shift 2) · Periodic Motion & SHM Basics

Periodic Motion & SHM Basics question from JEE Main 2022 (27 Jun, Shift 2)

Q43JEE Main 2022 · 27 JUN · SHIFT 2Medium

Periodic Motion & SHM Basics

A particle executes simple harmonic motion. Its amplitude is 8 cm and time period is 6 s. The time it will take to travel from its position of maximum displacement to the point corresponding to half of its amplitude, is ___________ s.

Numerical answer type. Work it out, then open the answer.

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Answer: 1

  1. From the extreme: Acos⁡ωt=A2⇒ωt=π3⇒t=T6A\cos\omega t = \frac{A}{2} \Rightarrow \omega t = \frac{\pi}{3} \Rightarrow t = \frac{T}{6}
  2. t=66=1 st = \frac{6}{6} = 1\text{ s}
  3. So the answer is 1 s.
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