Q. The amplitude of a particle executing SHM is made three-fourth keeping its time period constant. Its total energy will be
A) $\frac{E}{2}$
B) $\frac{3}{4}E$
C) $\frac{9}{16}E$
D) None of these
Explanation
(c) $E = \frac{1}{2}m\omega^{2}a^{2} \Rightarrow \frac{E^{'}}{E} = \frac{a^{'2}}{a^{2}} \Rightarrow \frac{E^{'}}{E} = \frac{\left( \frac{3}{4}a \right)^{2}}{a^{2}}\ \left( \because a^{'} = \frac{3}{4}a \right)$
$\Rightarrow E^{'} = \frac{9}{16}E$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.