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^∘.
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^∘.
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Maximum kinetic energy in SHM:
Kmax=(1)/(2)mω²A²
Always use mean-position condition to find ω.
Updated On: Mar 19, 2026
- \(y=0.1\sin\!\left(-4t+\dfrac{\pi}{4}\right)\)
- \(y=0.2\sin\!\left(-4t+\dfrac{\pi}{4}\right)\)
- \(y=0.1\sin\!\left(-2t+\dfrac{\pi}{4}\right)\)
- y=0.2sin(-2t+(π)/(4))
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The Correct Option is A
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