Q. The total energy of a particle executing SHM is 80 J. What is the potential energy when the particle is at a distance of 3/4 of amplitude from the mean position?
A) 60 J
B) 10 J
C) 40 J
D) 45 J
Explanation
(d) $\frac{1}{2}m\omega^{2}r^{2} = 80\ J;$
PE$= \frac{1}{2}m\omega^{2}y^{2} = \frac{1}{2}m\omega^{2} \times \left( \frac{3}{4}r \right)^{2}$
$= \frac{9}{16}\left( \frac{1}{2}m\omega^{2}r^{2} \right) = \frac{9}{16} \times 80 = 45\ J$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.