Q. A particle starts SHM form the mean position. Its amplitude is $a$ and total energy E. At one instant its kinetic energy is 3 E/4 its displacement at this instant is
A) $y = a/\sqrt{2}$
B) $y = \frac{a}{2}$
C) $y = \frac{a}{\sqrt{3/2}}$
D) y = a
Explanation
(b) Total energy, $E = \frac{1}{2}m\omega^{2}a^{2};$
KE$= \frac{3E}{4} - \frac{1}{2}m\omega^{2}(a^{2} - y^{2})$
So $\frac{3}{4} = \frac{a^{2} - y^{2}}{a^{2}}\ \ \ or\ \ \ y^{2} = a^{2}/4\ \ \ \ or\ \ y = a/2$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.