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MHT CET 2024 (10 May, Shift 2) · Energy in SHM

Energy in SHM question from MHT CET 2024 (10 May, Shift 2)

Q34MHT CET 2024 · 10 MAY · SHIFT 2Medium

Energy in SHM

A particle starts oscillating simple harmonically from its equilibrium position with time period ' T '. What is the ratio of potential energy to kinetic energy of the particle at time t=T12t = \frac{T}{12}? (sin⁡(π6)=12)\left(\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}\right)

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Answer: (c)

  1. At t=T12t = \frac{T}{12}, ωt=2πT⋅T12=π6\omega t = \frac{2\pi}{T}\cdot\frac{T}{12} = \frac{\pi}{6}.
  2. UK=sin⁡2ωtcos⁡2ωt=tan⁡230∘=13\frac{U}{K} = \frac{\sin^2\omega t}{\cos^2\omega t} = \tan^2 30^\circ = \frac{1}{3}
  3. So the answer is (c) 1:31 : 3.
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