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 Figure. Show that a particle of total energy $1 \;J$ moving under this potential must 'turn back" when it reaches $x=\pm 2 m$
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 Figure. Show that a particle of total energy $1 \;J$ moving under this potential must 'turn back" when it reaches $x=\pm 2 m$
Total energy of the particle, $E=1 J$
Force constant, $k=0.5 N m ^{-1}$
Kinetic energy of the particle, $K =\frac{1}{2} m v^{2}$
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 $x=\pm 2 m$
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