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(C) The correct answer is $C$.In the given potential energy curve $V(x)$,the particle is confined between the turning points $r_1$ and $r_2$ where the

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periodic but not a simple harmonic

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(C) The correct answer is $C$.In the given potential energy curve $V(x)$,the particle is confined between the turning points $r_1$ and $r_2$ where the total energy $E$ equals the potential energy $V(x)$.Since the particle is confined between two points and the potential energy is finite,the particle will oscillate back and forth between $r_1$ and $r_2$,making the motion periodic.However,for a motion to be simple harmonic,the potential energy must be symmetric about the equilibrium position $r_0$ and follow the form $U(x) = \frac{1}{2} k (x - r_0)^2$. As seen in the figure,the potential energy curve $V(x)$ is not symmetric about the equilibrium position $r_0$ (it is asymmetric). Therefore,the motion is periodic but not simple harmonic.

Consider a one-dimensional potential $V(x)$ as shown in the figure below. $A$ classical particle of mass $m$ moves under its influence and has total energy $E$ as shown below. The motion is

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