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. 2 cm
C. `sqrt(2) cm`
D. `2sqrt(2) cm`

Answer

D

Explanation

The energy is half potential and half kinetic.
Thus its `K.E.=(1)/(2)("Total energy ")=(1)/(2)[(1)/(2)m omega^(2)A^(2)]`
`therefore (1)/(2)m omega^(2)x^(2)=(1)/(2)[(1)/(2) m omega^(2)A^(2)]`
`x^(2)=(A^(2))/(2) therefore x=(A)/(sqrt(2))=(4)/(sqrt(2))=2sqrt(2)cm`