Question

At a displacement from the equilibrium position, that is one-half the amplitude of oscillation, what fraction of the total energy of the oscillator is kinetic energy?

• A. $ \frac{1}{2} $
• B. $ \frac{1}{4} $
• C. $ \frac{1}{\sqrt{2}} $
• D. $ \frac{3}{4} $

Answer 1

Answer: (D)

In case of SHM $K E=\frac{1}{2} m \omega^{2}\left(a^{2}-x^{2}\right)$
and at $x=a / 2$
$K E=\frac{1}{2} m \omega^{2}\left[a^{2}-(a / 2)^{2}\right]$
$=\frac{3}{4}\left[\frac{1}{2} m \omega^{2} a^{2}\right]$
Total energy $=\frac{1}{2} m \omega^{2} a^{2}$
$\therefore $ Fraction of total energy $=\frac{K E}{\text { Total energy }}$
$=\frac{\frac{3}{4}\left[\frac{1}{2} m \omega^{2} a^{2}\right]}{\frac{1}{2} m \omega a^{2}}=\frac{3}{4}$