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Sahay Sir > Question Answers > NCERT Exemplar Class 11 > Physics > Chapter 5 - Work, Energy and Power > The potential energy function for a particle executing linear simple harmonic motion is given by V(x) = kx 2 /2, where k is the force constant of the oscillator. For k = 0.5 Nm^−1 , the graph of V(x) versus x is shown in Fig. Show that a particle of total energy 1 J moving under this potential must turn back when it reaches x = ± 2 m.

The potential energy function for a particle executing linear simple harmonic motion is given by V(x) = kx 2 /2, where k is the force constant of the oscillator. For k = 0.5 Nm^−1 , the graph of V(x) versus x is shown in Fig. Show that a particle of total energy 1 J moving under this potential must turn back when it reaches x = ± 2 m.

the graph of V(x) versus x is shown in Fig. Show that a particle of total energy 1 J moving under this potential must turn back when it reaches x = ± 2 m. The potential energy function for a particle executing linear simple harmonic motion is given by V(x) = kx 2 /2 where k is the force constant of the oscillator. For k = 0.5 Nm^−1