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Explanation
1. Potential Energy in Simple Harmonic Motion
In a simple harmonic oscillator (e.g., a mass on a spring), the restoring force is given by
$F = -kx$, where $k$ is the force constant and $x$ is the displacement from the equilibrium position.
The corresponding potential energy function $U(x)$ of such a system is
$U(x) = \frac{1}{2} k x^2$.
2. Behavior at Equilibrium and Extreme Positions
Equilibrium Position (O): Here, $x = 0$. Substituting into $U(x) = \frac{1}{2} k x^2$,
we get $U(0) = 0$. This implies the potential energy is minimum (zero) at the equilibrium position.
Extreme Positions ($x_1$ and $x_2$): At these extreme displacements, $|x|$ is maximum. Consequently,
$U = \frac{1}{2} k x^2$ is largest at these points. Thus, the potential energy is maximum at the extremes.
3. Graphical Representation
The plot of $U(x) = \frac{1}{2} k x^2$ versus $x$ is a parabola opening upwards, having its minimum at $x = 0$
(the mean position). At $x = \pm x_1$ or $\pm x_2$ (the turning points of the motion), the potential energy reaches
its maximum value.
4. Correct Answer
The graph showing a parabolic curve with its minimum at $O$ (zero potential energy) and maximum at the extreme
positions corresponds to the option:
5. Conclusion
For a particle executing SHM, the potential energy is minimum (zero) at the equilibrium position and
increases symmetrically on either side, forming a parabolic shape. Hence, the correct graph matches a parabola
centered at $x=0$ with maximum values at $x = x_1$ and $x = x_2$.
