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MHT CET 2026 (18 Apr, Shift 1) · Velocity & Acceleration in SHM

Velocity & Acceleration in SHM question from MHT CET 2026 (18 Apr, Shift 1)

Q3MHT CET 2026 · 18 APR · SHIFT 1Easy

Velocity & Acceleration in SHM

A particle is performing simple harmonic motion about x=0x = 0 with an amplitude 'aa' and periodic time T. The speed of the particle at x=a3x = \frac{a}{3} will be

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Answer: (c)

  1. v=ωa2−x2v = \omega\sqrt{a^2 - x^2} with ω=2πT\omega = \frac{2\pi}{T}
  2. a2−a29=22a3\sqrt{a^2 - \frac{a^2}{9}} = \frac{2\sqrt{2}a}{3}
  3. v=2πT⋅22a3=42πa3Tv = \frac{2\pi}{T}\cdot\frac{2\sqrt{2}a}{3} = \frac{4\sqrt{2}\pi a}{3T}
  4. So the answer is (c) 42πa3T\frac{4\sqrt{2}\pi a}{3T}.
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