The potential energy function for a particle executing linear
Question.
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 $k=0.5 \mathrm{~N} \mathrm{~m}^{-1}$, the graph of $V(x)$ versus $x$ is shown in Fig. 6.12. Show that a particle of total energy $1 \mathrm{~J}$ moving under this potential must ‘turn back’ when it reaches $x=\pm 2 \mathrm{~m}$.

solution:

Total energy of the particle, E = 1 J

Force constant, $k=0.5 \mathrm{~N} \mathrm{~m}^{-1}$

Kinetic energy of the particle, $\mathrm{K}=\frac{1}{2} m v^{2}$

According to the conservation law:

$E=V+K$

$1=\frac{1}{2} k x^{2}+\frac{1}{2} m v^{2}$

At the moment of ‘turn back’, velocity (and hence $K$ ) becomes zero.

$\therefore 1=\frac{1}{2} k x^{2}$

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

$x^{2}=4$

$x=\pm 2$

Hence, the particle turns back when it reaches $x=\pm 2 \mathrm{~m}$.
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