Question

The total energy of the body executing simple harmonic motion (SHM) is $E$. Then the kinetic energy when the displacement is half of the amplitude is

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

Answer 1

Answer: (C)

Total energy in SHM $E=\frac{1}{2} m \omega^{2} a^{2}$,
(where $a =$ amplitude)
Kinetic energy $K=\frac{1}{2} m \omega^{2}\left(a^{2}-y^{2}\right)$
$=E-\frac{1}{2} m \omega^{2} y^{2}$
when $y=\frac{a}{2}$
$ \Rightarrow K=E-\frac{1}{2} m \omega^{2}\left(\frac{a^{2}}{4}\right)$
$=E-\frac{E}{4} K=\frac{3 E}{4}$