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

(N/A) Total energy of the particle,$E = 1 \; J$.
Force constant,$k = 0.5 \; N m^{-1}$.
According to the law of conservation of mechanical energy,the total energy $E$ is the sum of kinetic energy $K$ and potential energy $V(x)$:
$E = K + V(x)$
$E = K + \frac{1}{2} k x^{2}$
At the turning points,the particle momentarily comes to rest,meaning its velocity is zero,and consequently,its kinetic energy $K$ is zero.
Therefore,at the turning points,the total energy is entirely potential energy:
$E = V(x)$
$1 = \frac{1}{2} k x^{2}$
Substituting the given values:
$1 = \frac{1}{2} \times 0.5 \times x^{2}$
$1 = 0.25 \times x^{2}$
$x^{2} = \frac{1}{0.25} = 4$
$x = \pm 2 \; m$.
Thus,the particle must 'turn back' when it reaches $x = \pm 2 \; m$.