Question

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. $E/2$
• B. $E/4$
• C. $3E/4$
• D. $\sqrt{3}E/4$

Answer 1

Answer: (C)

$3E/4$

Answer 2

Kinetic energy $= \frac{1}{2}m\omega^{2}(a^{2} - y^{2}) = \frac{1}{2}m\omega^{2}\left( a^{2} - \frac{a^{2}}{4} \right)$

$= \frac{3}{4}\left( \frac{1}{2}m\omega^{2}a^{2} \right)$=$\frac{3E}{4}$ [As $y = \frac{a}{2}$]