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 $x=\pm 2 \mathbf{m}$.

Home » The potential energy function for a particle executing linear simple harmonic motion is given by , where is the force constant of the oscillator. For , the graph of versus is shown in Figure. Show that a particle of total energy moving under this potential must ‘turn back’ when it reaches .

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 $x=\pm 2 \mathbf{m}$ .

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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 $x=\pm 2 \mathbf{m}$ .