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