The potential energy function for a particle executing linear simple harmonic motion is given by \(V(x)=\frac{kx^2}{2}\), where \(k\) is the force constant of the oscillator. For \(k=0.5\) N/m, the graph of \(V(x)\) versus \(x\) is shown in the figure. A particle of total energy \(1\) J moving under this potential must turn back when it reaches:

      
1. \(\mathrm{x}=\pm 1\) m
2. \(\mathrm{x}=\pm 2\) m
3. \(\mathrm{x}=\pm 3\) m
4. \(\mathrm{x}=\pm 4\) m

Subtopic:  Elastic Potential Energy |
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NEET 2023 - Target Batch - Aryan Raj Singh
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NEET 2023 - Target Batch - Aryan Raj Singh