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# The potential energy function for a particle executing linear simple harmonic motion is given by U(x)=12 kx2, where k is the force constant. For k=0.5Nm−1,the graph of U(x) versus x is shown in figure. Show that a particle of total energy 1 J moving under this potential ′turns back′ when it reaches x=± 2m.

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#### At extreme oh SHM, KE=0Total energy =potential energy =12kx21=12(0.5)x24=x2x=±2

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