Question

If a particle cover the displacement half of amplitude; then what fraction of the total energy of a simple harmonic oscillator is kinetic energy

• A. $2/7$
• B. $3/4$
• C. $5/7$
• D. $7/3$

Answer 1

Answer: (B)

$3/4$

Answer 2

Kinetic energy, $K=\frac{1}{2}m\omega^{2}\left(a^{2}-y^{2}\right)$ or $\frac{1}{2}m\left(a^{2}-\frac{a^{2}}{4}\right)\omega^{2}\quad\left(\because y=\frac{a}{2}\right)$ $=\frac{1}{2}m\omega^{2}\,a^{2}\left(\frac{3}{4}\right)$ Total energy, $E=\frac{1}{2}m\omega^{2}a^{2}$ $\therefore \frac{K}{E}=\frac{\frac{1}{2}m\omega^{2}\,a^{2}\left(\frac{3}{4}\right)}{\frac{1}{2}m\omega^{2}\,a^{2}}$ $\Rightarrow \frac{K}{E}=\frac{3}{4}$