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MHT CET 2023 (12 May, Shift 1) · Velocity & Acceleration in SHM

Velocity & Acceleration in SHM question from MHT CET 2023 (12 May, Shift 1)

Q40MHT CET 2023 · 12 MAY · SHIFT 1Easy

Velocity & Acceleration in SHM

A simple pendulum performs simple harmonic motion about x=0x = 0 with an amplitude 'aa' and time period 'TT'. The speed of the pendulum at x=a2x = \frac{a}{2} is

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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−a24=3a2\sqrt{a^2 - \frac{a^2}{4}} = \frac{\sqrt{3}a}{2}
  3. v=2πT⋅3a2=πa3Tv = \frac{2\pi}{T}\cdot\frac{\sqrt{3}a}{2} = \frac{\pi a\sqrt{3}}{T}
  4. So the answer is (c) πa3T\frac{\pi a\sqrt{3}}{T}.
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