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 $k=0.5 \mathrm{~N} m^{-1}$, the graph of $\mathrm{V}(\mathbf{x})$ versus $\mathrm{x}$ is shown in Figure. Show that a particle of total energy $1 \mathrm{~J}$ moving under this potential must ‘turn back’ when it reaches $x=\pm 2 \mathbf{m}$.

Solution:

Energy of the particle will be, $\mathrm{E}=1 \mathrm{~J}$

$\mathrm{K}=0.5 \mathrm{~N} \mathrm{~m}^{-1}$

$\mathrm{K} . \mathrm{E}=\frac{1}{2} \mathrm{mv}^{2}$

Based on law of conservation of energy, we can write,

$E=V+K$

$1=\frac{1}{2} \mathrm{kx}^{2}+\frac{1}{2} \mathrm{mv}^{2}$

When it turns back, velocity becomes zero, so

$1=\frac{1}{2} \mathrm{kx}^{2}$

$\frac{1}{2} \times 0.5 x^{2}=1$

$x^{2}=4$

$x=\pm 2$

Thus, the particle turns back on reaching $\mathrm{x}=\pm 2 \mathrm{~m}$.