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

Answer

C

Explanation

`K.E.=(1)/(2)mv^(2)=(1)/(2)m omega^(2)(A^(2)-x^(2))`
`therefore (1)/(2)m omega^(2) (A^(2)-x^(2))=(1)/(4)[(1)/(2)m omega^(2)A^(2)]`
` therefore A^(2)-x^(2)=(A^(2))/(4) therefore x=(sqrt(3))/(2)A`