When the displacement of a linear harmonic oscillator is `1//3` of its amplitude, the ratio of its total energy to its potential energy is

Options

A. 3
B. 4
C. 9
D. `(1)/(9)`

Answer

C

Explanation

`x=(1)/(3)A, ("Total Energy")/("Potential Energy")=((1)/(2)KA^(2))/((1)/(2)Kx^(2))`
`=(A^(2))/(x^(2))=(A^(2))/(A^(2)//9)=9`