Q. A particle of mass m is executing oscillations about the origin on the $x$-axis with amplitude A. Its PE is given as $U_{(x)} = \alpha x^{4},\ $where $\alpha$ is positive constant. The$x$- coordinate of mass where potential energy is one-third of the KE of particle, is
A) $\pm \frac{A}{\sqrt{3}}$
B) $\pm \frac{A}{\sqrt{2}}$
C) $\pm \frac{A}{3}$
D) $\pm \frac{A}{2}$
Explanation
(b) Energy of oscillation, $E = \alpha A^{4}$
Kinetic energy of mass at $x = x$ is
$K = E - U = \alpha(A^{4} - x^{4})$
As $K = 3U$
$\alpha\left( A^{4} - x^{4} \right) = 3\alpha x^{4}$ or $x = \pm \frac{A}{\sqrt{2}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.