A particle executes simple harmonic motion with an amplitude of 10 cm. At what distance from the mean position are the kinetic and potential energies equal?

Answer

B, C

Explanation

`r=10cm `
Because K.E. =P.E.
`=so, (1/2)momega^2(r^2-y^2)=(1/2)momega^2y^2`
`r^2-y^2=y^2`
`2y^2=r^2`
`rarr y=r/sqrt2=10/sqrt2 =5sqrt2` from the mean position