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

Answer

A

Explanation

`K.E.=(3)/(4)T.E.`
`(1)/(2)momega^(2)(A^(2)-x^(2))=(3)/(4)xx(1)/(2)momega^(2)A^(2)`
`A^(2)-x^(2)=(3)/(4)A^(2)`
`A^(2)-(3)/(4)A^(2)=x^(2)`
`(A^(2))/(4)=x^(2)`
`therefore x=(A)/(2)`