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JEE Main 2003 (AIEEE) · Periodic Motion & SHM Basics

Periodic Motion & SHM Basics question from JEE Main 2003 (AIEEE)

Q80JEE Main 2003 · AIEEEMedium

Periodic Motion & SHM Basics

The displacement of particle varies according to the relation

x=4(cos⁡πt+sin⁡πt).x = 4(\cos \pi t + \sin \pi t). The amplitude of the particle is

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Answer: (c)

  1. x=4cos⁡πt+4sin⁡πtx = 4\cos\pi t + 4\sin\pi t
  2. Amplitude =42+42=42= \sqrt{4^2 + 4^2} = 4\sqrt{2}
  3. So the answer is (c) 424\sqrt{2}.
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