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The potential energy function for a particle executing simple harmonic motion is given by \({V}{\left({x}\right)}=\frac{{{1}}}{{{2}}}{k}{x}^{{{2}}}\), where k is the force constant of the oscillatore. For \({k}=\frac{{{1}}}{{{2}}}{N}{m}^{{-{1}}}\), show that a particle of total energy 1 joule moving under this potential must turn back when it reaches \({x}=\pm{2}{m}.\)
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The potential energy function for a particle executing simple harmonic motion is given by \({V}{\left({x}\right)}=\frac{{{1}}}{{{2}}}{k}{x}^{{{2}}}\), where k is the force constant of the oscillatore. For \({k}=\frac{{{1}}}{{{2}}}{N}{m}^{{-{1}}}\), show that a particle of total energy 1 joule moving under this potential must turn back when it reaches \({x}=\pm{2}{m}.\)

Answer

At any instant, the total energy of an oscillator is the sum of K.E. and P.E.
i.e. \({E}={K}.{E}.+{P}.{E}.=\frac{{{1}}}{{{2}}}{m}{u}^{{{2}}}+\frac{{{1}}}{{{2}}}{k}{x}^{{{2}}}\)
The particle turns back at the instant, when its velocity becomes zero i.e. \({u}={0}\) .
\(\therefore{E}={0}+\frac{{{1}}}{{{2}}}{k}{x}^{{{2}}}\)
As \({E}={1}\) joule and \({k}=\frac{{{1}}}{{{2}}}{N}/{m}\) \(\therefore{1}=\frac{{{1}}}{{{2}}}\times\frac{{{1}}}{{{2}}}{x}^{{{2}}}\) or \({x}^{{{2}}}={4},{x}=\pm{2}{m}\)
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