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gyluncn,Iitiszeroatorigin.F-5. The potential energy function for a particle executing linear simple harmonic motion is given by Uoo2where k is the force constant. For k = 0.5 N m the graph of U(x) versus x 's shown in figure. Show that aparticle of total energy 1 J m oving under this potential turns back, when it reaches x = ± 2m.U(x)2-2m-1mx=0+1m+2m |
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