Amplitude of a SHO is `sqrt(5) cm`. At what displacement from the mean position the ratio of kinetic energy to potential energy is 4?

Explanation

`implies ("Kinetic energy")/("Potential energy")= 4`
Kinetic energy = 4 potential energy
`therefore (1)/(2)k(A^(2)-x^(2))= 4(1)/(2)kx^(2)`
`therefore A^(2)- x^(2) = 5x^(2)`
`therefore A^(2)= 5x^(2)`
`therefore x = pm (A)/(sqrt(5))= pm (sqrt(5))/(sqrt(5))= 1 cm`.