If the KE of a particle performing a SHM of amplitude A is `(3)/(4)` of its total energy, then the value of its displacement is

Options

A. `x=+-(A)/(2)`
B. `x =+-(A)/(4)`
C. `x=+-(sqrt(3)A)/(2)`
D. `x=+-(A)/(sqrt(2))`

Answer

A

Explanation

`(1)/(2)m omega^(2)(A^(2)-x^(2))=(3)/(4)((1)/(2)m omega^(2)A^(2))`
`therefore (A^(2)-x^(2))=(3)/(4)A^(2) therefore x^(2)=(A^(2))/(4) therefore x= +-(A)/(2)`