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ExamsJEE MainPhysics

A particle of mass 0.1 kg performs simple harmonic motion with amplitude 0.1 m. As it crosses the equilibrium position, its kinetic energy is 8 × 10⁻³ J. Find the equation of motion of the particle, given that the initial phase is 45°.

  1. y = 0.1 sin (4t + π/4)
  2. y = 0.2 sin (4t + π/4)
  3. y = 0.1 sin (2t + π/4)
  4. y = 0.2 sin (2t + π/4)

Correct answer: y = 0.1 sin (4t + π/4)

Solution

(1/2)*0.1*w^2*(0.1)^2 = 8*10^-3 -> w^2 = 16 -> w = 4 rad/s. Amplitude is 0.1 m, so y = 0.1 sin(4t + pi/4).

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