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A particle of mass m is executing oscillations about the origin on the x-axis with amplitude A. Its potential energy is given as $U (x) = α x^{4}$ where $α$ is a positive constant. The x-coordinate of the particle where its potential energy is one third the corresponding kinetic energy is

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\pm \frac{A}{\sqrt{3}}

\pm \frac{A}{2}

\pm \frac{A}{\sqrt{2}}

\pm \frac{A}{3}

answer is B.

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Detailed Solution

Potential energy

= a x^{4}

Potential energy, $u = \frac{1}{3}$ kinetic energy

$k = 3 u$

$k = 3 a x^{4}$

At maximum amplitude A

Total energy, $E = {aA}^{4}$

$E = u + k$

${aA}^{4} = a x^{4} + 3 a x^{4}$

$x^{4} = \frac{A^{4}}{4}$

$x^{2} = \pm \frac{A^{2}}{2}$

$x = \pm \sqrt{\frac{A^{2}}{2}}$

$x = \pm \frac{A}{\sqrt{2}}$

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