A particle of mass 0.1 kg performs simple harmonic motion with amplitude 0.1 m. At the mean position, its kinetic energy is 18 × 10⁻³ J. If the initial phase of the oscillation is 45°, the particle’s equation of motion is given by:

Question

Accepted Answer

x = 0.1 sin(6t + π/4). KE at mean = (1/2)m*omega^2*A^2 = 18e-3 gives (1/2)(0.1)(0.01)omega^2 = 0.018, so omega^2 = 36, omega = 6 rad/s. With initial phase 45 deg, x = 0.1 sin(6t + pi/4).

A particle of mass 0.1 kg performs simple harmonic motion with amplitude 0.1 m. At the mean position, its kinetic energy is 18 × 10⁻³ J. If the initial phase of the oscillation is 45°, the particle’s equation of motion is given by:

Solution

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