Question Detail
Question
The periodic time of a S.H.O. oscillating about a fixed point is 2 s. After what time will the kinetic energy of the oscillator become 25% of its total energy?
Correct Answer
Option B is the correct answer.
Detailed Explanation
The total energy E = ½kA². Kinetic energy K = ½k(A² - x²). K = E/4 => ½k(A² - x²) = (1/4)½kA² => A² - x² = A²/4 => x² = 3A²/4 => x = ±(√3/2)A. In SHM, x = A cos(ωt). So cos(ωt) = √3/2. ω = 2π/T = 2π/2 = π rad/s. ωt = π/6 => t = (π/6)/π = 1/6 s.
Hint
Use the relation K = E sin²(ωt) or K = E cos²(ωt).