A particle starts oscillating simple harmonically from its equilibrium position with time period T. The ratio of KE and PE of the particle at the…

Question

A particle starts oscillating simple harmonically from its equilibrium position with time period T. The ratio of KE and PE of the particle at the t=T/12 is

MARKS Content Team · Accepted Answer

Correct answer: (C) 3:1. When t=T/12, then x=Asin⁡(2π)/T×T/12=A/2 KE=1/2 mv^2=1/2 mω^2(A^2-x^2) = 1/2 mω^2A^2-(A^2)/4 = 3/41/2 mω^2A^2 PE=1/2 mω^2x^2=1/41/2 mω^2A^2 KE/PE=3/1

A particle starts oscillating simple harmonically from its equilibrium position with time period $T$ . The ratio of KE and PE of the particle at the $t = T / 12$ is

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