Question
The potential energy function for a particle executing linear simple harmonic motion is given by V(x) =kx2/2, where k is the force constant of the oscillator. For k = 0.5 N m-1, the graph of V(x) versus x is shown in Fig. 6.12. Show that a particle of total energy 1 J moving under this potential must ‘turn back’ when it reaches x = ± 2 m.
Solution
Given,
Potential energy for a particle executing linear simple harmonic motion is,
where,
Total energy of particle E = 1J
Since at extreme position total energy is potential energy,
∴ 
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Hence the results.
