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MHT CET 2024 (2 May, Shift 2) · Periodic Motion & SHM Basics

Periodic Motion & SHM Basics question from MHT CET 2024 (2 May, Shift 2)

Q29MHT CET 2024 · 2 MAY · SHIFT 2Medium

Periodic Motion & SHM Basics

The displacement of a particle performing S.H.M. is given by Y=Acos⁡[π(t+ϕ)]Y = A\cos[\pi(t + \phi)]. If at t=0t = 0, the displacement is y=2 cmy = 2\text{ cm} and velocity is 2π cm/s2\pi\text{ cm/s}, the value of amplitude AA in cm is

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Answer: (c)

  1. Let θ=πϕ\theta = \pi\phi. At t=0t = 0: Acos⁡θ=2A\cos\theta = 2.
  2. v=−πAsin⁡θ=2π⇒Asin⁡θ=−2v = -\pi A\sin\theta = 2\pi \Rightarrow A\sin\theta = -2
  3. A=22+22=22 cmA = \sqrt{2^2 + 2^2} = 2\sqrt{2}\text{ cm}
  4. So the answer is (c) 222\sqrt{2}.
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