Question

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

• A. $\pm \frac{A}{2}$
• B. $\pm \frac{A}{\sqrt{2}}$
• C. $\pm \frac{A}{3}$
• D. $\pm \frac{A}{\sqrt{3}}$

Answer 1

Answer: (B)

The energy of oscillation is $\text{E}=\alpha {\text{A}}^4$
$\therefore$ the kinetic energy of mass at x = x is
$\text{K}=\text{E}-\text{U}=\alpha \left({\left(\text{A}\right)}^4-{\left(\text{x}\right)}^4\right)$
Given, K = 3U
$\therefore \alpha \left({\left(\text{A}\right)}^4-{\left(\text{x}\right)}^4\right)=3\alpha {\left(\text{x}\right)}^4$
Or $\text{x}=\pm \frac{\text{A}}{\sqrt{2}}$