Q. When the displacement is half the amplitude, the ratio of potential energy to the total energy is
A) $\frac{1}{2}$
B) $\frac{1}{4}$
C) 1
D) $\frac{1}{8}$
Explanation
(b) $\frac{U}{E} = \frac{\frac{1}{2}m\omega^{2}y^{2}}{\frac{1}{2}m\omega^{2}a^{2}} = \frac{y^{2}}{a^{2}} = \frac{\left( \frac{a}{2} \right)^{2}}{a} = \frac{1}{4}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.