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JEE Main 2024 (29 Jan, Shift 2) · Periodic Motion & SHM Basics

Periodic Motion & SHM Basics question from JEE Main 2024 (29 Jan, Shift 2)

Q30JEE Main 2024 · 29 JAN · SHIFT 2Medium

Periodic Motion & SHM Basics

A simple harmonic oscillator has an amplitude AA and time period 6π6\pi second. Assuming the oscillation starts from its mean position, the time required by it to travel from x=x = A to x=32x = \frac{\sqrt{3}}{2} A will be πx s\frac{\pi}{x}\text{ s}, where x=x = _________.

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

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Answer: 2

  1. From the extreme: Acos⁡ωt=32A⇒ωt=π6⇒t=T12A\cos\omega t = \frac{\sqrt{3}}{2}A \Rightarrow \omega t = \frac{\pi}{6} \Rightarrow t = \frac{T}{12}
  2. t=6π12=π2 st = \frac{6\pi}{12} = \frac{\pi}{2}\text{ s}
  3. So x=2x = 2.
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