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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 $${x}=\pm{2}{m}.$$  Speed
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02:06 ## QuestionPhysicsClass 11

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

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 $${x}^{{{2}}}={4},{x}=\pm{2}{m}$$          