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NEET PHYSICSOscillationsMedium

Question

A particle executing simple harmonic motion has a kinetic energy of K=K0cos2(ωt)K = K_0 \cos^2(\omega t). The values of the maximum potential energy and the total energy are, respectively:

A

00 and 2K02K_0

B

K0/2K_0/2 and K0K_0

C

K0K_0 and 2K02K_0

D

K0K_0 and K0K_0

Step-by-Step Solution

  1. Analyze Kinetic Energy: The given kinetic energy is K=K0cos2(ωt)K = K_0 \cos^2(\omega t). The maximum value of cos2(ωt)\cos^2(\omega t) is 1. Therefore, the maximum kinetic energy is Kmax=K0K_{max} = K_0.
  2. Total Energy (EE): In simple harmonic motion, the total mechanical energy is conserved and is equal to the maximum kinetic energy (at the mean position) or the maximum potential energy (at the extreme position) . Thus, Total Energy E=Kmax=K0E = K_{max} = K_0.
  3. Maximum Potential Energy (UmaxU_{max}): Since the total energy is the sum of kinetic and potential energies (E=K+UE = K + U), the potential energy is maximum when the kinetic energy is minimum (zero). At this point, the entire energy is potential. Therefore, Umax=E=K0U_{max} = E = K_0.
  4. Conclusion: Both the maximum potential energy and the total energy are equal to K0K_0.

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from Oscillations. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSOscillationsparticleexecutingsimpleharmonicmotion

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