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(A) The potential energy $V(x)$ is given by a parabolic curve,which implies $V(x) = \frac{1}{2}kx^2$. This represents a simple harmonic oscillator whe

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(A) The potential energy $V(x)$ is given by a parabolic curve,which implies $V(x) = \frac{1}{2}kx^2$. This represents a simple harmonic oscillator where the restoring force is $F = -\frac{dV}{dx} = -kx$. Since the particle is released from rest at some positive displacement $x = A$ at $t = 0$,its motion is described by the equation $x(t) = A \cos(\omega t)$. At $t = 0$,$x = A$,which is the positive extreme position. As time increases,the particle moves towards the mean position $(x = 0)$ and then to the negative extreme position $(x = -A)$. This behavior is represented by a cosine graph starting from a positive maximum value at $t = 0$.

$A$ particle of mass $m$ is released from rest and follows a parabolic path as shown. Assuming that the displacement of the mass from the origin is small,which graph correctly depicts the position of the particle as a function of time $?$

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