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The potential energy function for a particle executing linear simple harmonic motion is given byV(x) = \(\frac{kx^2}{2}\), where k is the force constant of the oscillator, For k = 0.5N nm-1, the graph of V(x) versus x is shown. Show that a particle of total energy 1 J moving under this potential must ‘turn back’ when it reaches x = ± 2m. |
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Answer» We know that maximum potential energy = total energy ∴ (\(\frac{1}{2}\)kx2) max = 1 joule or \(\frac{1}{2}\) × 0.5 × (x2)max = 1 or (x2)max = 4 or (x)max = ± 2m. |
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