Question

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. \(y=0.1\sin\!\left(-4t+\dfrac{\pi}{4}\right)\)
• B. \(y=0.2\sin\!\left(-4t+\dfrac{\pi}{4}\right)\)
• C. \(y=0.1\sin\!\left(-2t+\dfrac{\pi}{4}\right)\)
• D. y=0.2sin(-2t+(π)/(4))

Hint:

Maximum kinetic energy in SHM: Kmax=(1)/(2)mω²A² Always use mean-position condition to find ω.

Answer 1

The Correct Option is A

Step 1: At mean position, kinetic energy is maximum: Kmax=(1)/(2)mω²A²
Step 2: Substituting values: 8×10⁻3=(1)/(2)(0.1)ω²(0.1)²
Step 3: ω²=16 ⟹ ω=4rad/s
Step 4: General SHM equation: y=Asin(ω t+φ) Given φ=45^∘=π/4 and direction chosen negative: y=0.1sin(-4t+(π)/(4))