Starting from the origin, a body oscillates simple harmonically with a period of 2 s . After what time will its kinetic energy be 75\% of the total en - Physics MCQ AIEEE 2006 2026 | ExamTest.live

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Starting from the origin, a body oscillates simple harmonically with a period of 2 s . After what time will its kinetic energy be 75% of the total energy?

Solution & Step-by-step Explanation

Total Energy E = \frac{1}{2} m A^{2} ω^{2} .Kinetic Energy K E = \frac{1}{2} m v^{2} = \frac{1}{2} m A^{2} ω^{2} cos^{2} (ω t) .Given K E = 0.75 E : \frac{1}{2} m A^{2} ω^{2} cos^{2} (ω t) = \frac{3}{4} (\frac{1}{2} m A^{2} ω^{2}) cos^{2} (ω t) = \frac{3}{4} ⟹ cos (ω t) = \frac{3}{2} ω t = \frac{π}{6} Since T = 2 s, ω = \frac{2 π}{T} = π . π t = \frac{π}{6} ⟹ t = \frac{1}{6} s

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Starting from the origin, a body oscillates simple harmonically with a period of 2 s . After what time will its kinetic energy be 75% of the total energy?

1/12 s

1/6 s

1/4 s

1/3 s

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