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When the potential energy of a particle executing simple harmonic motion is one-fourth of the maximum value during the oscillation, its displacement from the equilibrium position in terms of amplitude ' $a$ ' is(a) $a / 4$ (b) $a / 3$ (c) $a / 2$ (d) $2 a / 3$ When the displacement of a particle executing simple harmonic motion is half its amplitude, the ratio of its kinetic energy to potential energy is ______. When the displacement of a particle executing simple harmonic motion is half its amplitude, the ratio of its kinetic energy to potential energy is 3 : 1. Explanation: x = A cos ωt x = `"A"/2` we get ωt = 60° `"KE"/"PE" = (1/2 "m"ω^2"A"^2 sin^2 ω"t")/(1/2 "m"ω^2"A"^2 cos ω"t")` = tan2 ωt = tan2 60° = 3 Concept: Energy in Simple Harmonic Motion Is there an error in this question or solution? |