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Periodic Motion & SHM Basics question from NEET 2024

Q27NEET 2024Easy

Periodic Motion & SHM Basics

If x=5sin⁡(πt+π3)mx = 5\sin\left(\pi t + \frac{\pi}{3}\right)\text{m} represents the motion of a particle executing simple harmonic motion, the amplitude and time period of motion, respectively, are

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Answer: (b)

  1. Compare with x=Asin⁡(ωt+ϕ)x = A\sin(\omega t + \phi): A=5 mA = 5\ \text{m}, ω=π rad/s\omega = \pi\ \text{rad/s}.
  2. T=2πω=2ππ=2 sT = \frac{2\pi}{\omega} = \frac{2\pi}{\pi} = 2\ \text{s}.
  3. So the answer is (b) 5 m, 2 s.
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