MCQ
The total energy of the body executing $S.H.M.$ is $E$. Then the kinetic energy when the displacement is half of the amplitude, is
- A$\frac{E}{2}$
- B$\frac{E}{4}$
- ✓$\frac{{3E}}{4}$
- D$\frac{{\sqrt 3 }}{4}E$
Potential energy $U = \frac{1}{2}m{\omega ^2}({a^2} - {y^2}) $
$= E - \frac{1}{2}m{\omega ^2}{y^2}$
When $y = \frac{a}{2}$
$⇒$ $U = E - \frac{1}{2}m{\omega ^2}\left( {\frac{{{a^2}}}{4}} \right)$
$= E - \frac{E}{4} = \frac{{3E}}{4}$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.