A particle is vibrating in a simple harmonic motion with an amplitude of 4 cm . At what displacement from the equilibrium position, is its energy half potential and half kinetic

Options

A. 1 cm
B. `sqrt(2) cm`
C. 3 cm
D. `2sqrt(2) cm`

Answer

D

Explanation

`K.E.=P.E.`
`(1)/(2)momega^(2)(A^(2)-x^(2))=(1)/(2)momega^(2)x^(2)`
`A^(2)-x^(2)=x^(2) therefore A^(2)=2x^(2)`
`therefore x^(2)=(A^(2))/(2) therefore x=(a)/(sqrt(2))=(4)/(sqrt(2))=2sqrt(2)cm`