Question Detail
Question
The periodic time of a SHO is 2 s. After what time will kinetic energy become 25% of total energy?
Correct Answer
Option D is the correct answer.
Detailed Explanation
KE = ¼ E_total → PE = ¾ E_total → x = (√3/2)A. For SHO, x = A sin(ωt) → sin(ωt)=√3/2 → ωt = π/3, 2π/3, etc. Smallest t = (π/3)/ω = (π/3)/(2π/T) = T/6 = 2/6 = 1/3 s? That gives 0.333 s. Option D is 1/3 s. Option B is 1/6 s (0.167 s). Which is correct? At t=T/6, sin(60°)=0.866, KE=½mv², v=ωA cos(ωt)=ωA×0.5 → KE=¼×½mω²A²=¼E_total. Yes t=T/6=1/3 s? T=2s, T/6=0.333 s = 1/3 s. So D.
Hint
KE = E_total cos²(ωt).