Amplitude of SHM particle is 4 cm. At what distance from mean positon will the potential energy and kinetic energy be equal?

Explanation

`omega =K`
`(1)/(2) kx^(2) = (1)/(2)k(A^(2)-x^(2))`
`therefore x^(2) = A^(2) -x^(2)`
`therefore 2x^(2) = A^(2)`
`therefore x= pm (A)/(sqrt(2)) = pm (4)/(sqrt(2))`
`therefore x = pm 2sqrt(2) cm`.