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MHT CET 2024 (15 May, Shift 1) · Energy in SHM

Energy in SHM question from MHT CET 2024 (15 May, Shift 1)

Q33MHT CET 2024 · 15 MAY · SHIFT 1Medium

Energy in SHM

Starting from mean position, a body oscillates simple harmonically with a period ' TT '. After what time will its kinetic energy be 75%75\% of the total energy? (sin⁡30∘=0.5)(\sin 30^\circ = 0.5)

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Answer: (b)

  1. K=75%K = 75\% of EE means U=25%U = 25\% of EE: x2A2=14⇒x=A2\frac{x^2}{A^2} = \frac{1}{4} \Rightarrow x = \frac{A}{2}.
  2. x=Asin⁡ωt⇒sin⁡ωt=12⇒ωt=π6x = A\sin\omega t \Rightarrow \sin\omega t = \frac{1}{2} \Rightarrow \omega t = \frac{\pi}{6}.
  3. t=π/62π/T=T12t = \frac{\pi/6}{2\pi/T} = \frac{T}{12}
  4. So the answer is (b) T12\frac{T}{12}.
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