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. \(y=0.1\sin(4t+\tfrac{\pi}{4})\)
• B. \(y=0.2\sin(4t+\tfrac{\pi}{4})\)
• C. \(y=0.1\sin(2t+\tfrac{\pi}{4})\)
• D. y=0.2sin(2t+(π)/(4))

Hint:

Maximum kinetic energy in SHM occurs at mean position.

Answer 1

The Correct Option is A

Step 1: Maximum kinetic energy: Kmax=(1)/(2)mω²A²
Step 2: Substituting values: 8×10⁻3=(1)/(2)(0.1)ω²(0.1)² ⟹ ω=4rad s⁻1
Step 3: Equation of motion: y=Asin(ω t+φ)=0.1sin(4t+(π)/(4))