QJEEM19Ph5372: JEE Main 2019 Physics - Class 11 JEEM19Ph5372 Solution: A particle undergoing simple harmonic motion has time dependent …

K = ${1 \over 2}m\omega^2 A^2 \cos^2 \omega t$

U = ${1 \over 2}m\omega^2 A^2 \sin^2 \omega t$

${K \over U} = \cot^2 \omega t = \cot^2 \left( {{\pi \times 210} \over {90}} \right) = \cot^2 \left( {{7\pi} \over 3} \right) = \cot^2 (60^\circ) = {1 \over 3}$

Note: Mathematically, the ratio of kinetic to potential energy is ${1 \over 3}$ . However, in the original JEE Main 2019 (11 Jan Shift 1) paper, ${1 \over 3}$ was not provided as an option, while 3 was. 3 is considered the most appropriate selection for this MCQ.