The potential energy function for a particle executing linear simple harmonic motion is given U(x)=1/2kx^(2), where k is the force constant of the oscillator. For k=0.5Nm^(-1), the graph of U(x) on Y-axis x is shown figure. The particle has total energy 1J. Moving this potential it turns back when it reaches x=+-_________m

The potential energy function for a particle executing linear simple h

1.	The potential energy function for a particle executing linear simple harmonic motion is given U(x)=1/2kx^(2), where k is the force constant of the oscillator. For k=0.5Nm^(-1), the graph of U(x) on Y-axis x is shown figure. The particle has total energy 1J. Moving this potential it turns back when it reaches x=+-_________m
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