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JEE Main 2021 (25 Feb, Shift 2) · Periodic Motion & SHM Basics

Periodic Motion & SHM Basics question from JEE Main 2021 (25 Feb, Shift 2)

Q49JEE Main 2021 · 25 FEB · SHIFT 2Medium

Periodic Motion & SHM Basics

The point A moves with a uniform speed along the circumference of a circle of radius 0.36 m and covers 30∘^\circ in 0.1 s. The perpendicular projection 'P' from 'A' on the diameter MN represents the simple harmonic motion of 'P'. The restoration force per unit mass when P touches M will be:

Figure

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Answer: (a)

  1. ω=π/60.1=5π3 rad/s\omega = \frac{\pi/6}{0.1} = \frac{5\pi}{3}\text{ rad/s}
  2. At M, PP is at the extreme, ∣x∣=R=0.36 m|x| = R = 0.36\text{ m}.
  3. Force per unit mass =ω2R=25π29×0.36=π2≈9.87= \omega^2 R = \frac{25\pi^2}{9}\times 0.36 = \pi^2 \approx 9.87
  4. So the answer is (a) 9.87 (unit is really N/kg).
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