The amplitude of a particle executing SHM about `O` is `10 cm`. Then

Options

A. when the kinetic energy is `0.64` times of its max. kinetic energy its displacement is `6 cm` from `O`
B. when the displacement is `5 cm` from O its kinetic energy is `0.75` times its maximum kinetic energy
C. Its total energy of SHM at any point is equal to its maximum kinetic energy
D. Its speed is half the maximum speed when its displacement is half the maximum displacement

Answer

A, B, C

Explanation

`y = 10sin (omegat + phi)`
Maximum KE `= 1/2 momega^(2)A^(2)`
`rArr 64/100 xx 1/2 momega^(2)A^(2) = 1/2momega^(2) (A^(2) - x^(2))`
`rArr 64A^(2) = (A^() - x^(2)) 100 rArr x = 0.6 A`
`rArr 1/2momega^(2) (A^(2) - x^(2)) = 1/2momega^(2) A^(2) xx x`
`rArr A^(2) - 0.25 A^(2) = A^(2)x`
`rArr x = 0.75` mean `75%` of energy