Question

The ratio of total energy of a harmonic oscillator, to its average potential energy with respect to position will be

• A. 1:2
• B. 3:1
• C. 1:3
• D. 2:11

Answer 1

Answer: (B)

3:1

Average potential energy, $U\left(x\right)=\frac{1}{2x_m}{\int }_{-x_m}^{x_m}\frac{1}{2}k{x}^2=\frac{1}{2x_m}\frac{1}{3}k{x_m}^3=\frac{1}{6}k{x_m}^2$

Total Energy $=\frac{1}{2}m{v}^2+\frac{1}{2}{kx}^2=\frac{1}{2}{k{x}_m}^2$