Question:medium
A point particle of mass 0.1kg is executing S.H.M. of amplitude 0.1m. When the particle passes through the mean position, its kinetic energy is 8×10⁻3J. Obtain the equation of motion of this particle if its initial phase of oscillation is \(45^\circ\).
A point particle of mass 0.1kg is executing S.H.M. of amplitude 0.1m. When the particle passes through the mean position, its kinetic energy is 8×10⁻3J. Obtain the equation of motion of this particle if its initial phase of oscillation is \(45^\circ\).
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Maximum kinetic energy in SHM occurs at mean position.
Updated On: Apr 2, 2026
- \(y=0.1\sin(4t+\tfrac{\pi}{4})\)
- \(y=0.2\sin(4t+\tfrac{\pi}{4})\)
- \(y=0.1\sin(2t+\tfrac{\pi}{4})\)
- y=0.2sin(2t+(π)/(4))
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The Correct Option is A
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