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Q.At which position, the total energy of a particle executing linear S.H.M. is purely potential?

Maharashtra MsbshseMaharashtra HSC (MSBSHSE) Board 2020Subjective· 1mImportance★★★★★
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At extremes, v = 0 so KE = 0, hence the (constant) total energy is entirely potential.

Total energy E=KE+PE=12mω2(A2−x2)+12mω2x2=12mω2A2E = \text{KE} + \text{PE} = \tfrac{1}{2}m\omega^2(A^2-x^2) + \tfrac{1}{2}m\omega^2x^2 = \tfrac{1}{2}m\omega^2A^2 (constant). At the extreme positions x=±Ax = \pm A, velocity v=ωA2−x2=0v = \omega\sqrt{A^2-x^2} = 0, so KE =0=0 and the entire energy 12mω2A2\tfrac{1}{2}m\omega^2A^2 is potential.

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