Oscillations

JEE Main level, single correct, medium

94 of 194

Question 94 of 194

A particle undergoing simple harmonic motion has time

dependent displacement given by x(t)=Asin⁡πt90x(t)=\mathrm{A} \sin \frac{\pi t}{90}. The ratio of kinetic to potential energy of this particle at t=210 s\mathrm{t}=210 \mathrm{~s} will be

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Answer: (D) 13\frac{1}{3}

Kinetic energy, k=12 mω2 A2cos⁡2ωt\mathrm{k}=\frac{1}{2} \mathrm{~m} \omega^{2} \mathrm{~A}^{2} \cos ^{2} \omega \mathrm{t}

Potential energy, U=12mω2A2sin⁡2ωtU=\frac{1}{2} m \omega^{2} A^{2} \sin ^{2} \omega t

kU=cot⁡2ωt=cot⁡2π90(210)=13\frac{k}{U}=\cot ^{2} \omega t=\cot ^{2} \frac{\pi}{90}(210)=\frac{1}{3}

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