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MHT CET 2023 (14 May, Shift 1) · Velocity & Acceleration in SHM

Velocity & Acceleration in SHM question from MHT CET 2023 (14 May, Shift 1)

Q39MHT CET 2023 · 14 MAY · SHIFT 1Easy

Velocity & Acceleration in SHM

The displacement of a particle executing S.H.M. is x=asin⁡(ωt−ϕ)x = a\sin(\omega t - \phi). Velocity of the particle at time t=ϕωt = \frac{\phi}{\omega} is (cos⁡0∘=1)(\cos 0^\circ = 1)

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Answer: (b)

  1. v=dxdt=aωcos⁡(ωt−ϕ)v = \frac{dx}{dt} = a\omega\cos(\omega t - \phi)
  2. At t=ϕωt = \frac{\phi}{\omega}: ωt−ϕ=0\omega t - \phi = 0, so v=aωcos⁡0∘=aωv = a\omega\cos 0^\circ = a\omega
  3. So the answer is (b) aωa\omega.
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