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PhysicsNEET 2024Simple harmonic motion: equations, energy

Oscillations

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

  1. A.5 m, 2 s✓
  2. B.5 cm, 1 s
  3. C.5 m, 1 s
  4. D.5 cm, 2 s

Solution

Comparing with x=Asin⁡(ωt+ϕ)x = A\sin(\omega t + \phi): amplitude A=5A = 5 m and ω=π\omega = \pi rad/s, so T=2πω=2T = \frac{2\pi}{\omega} = 2 s.

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