A body is performing SHM, then its

Options

A. average total energy per cycle is equal to its maximum kinetic energy.
B. average kinetic energy per cycle is equal to half of its maximum kinetic energy.
C. mean velocity over a complete cycle is equal to `(2)/(pi)` times of its maximum velocity.
D. root mean square velocity is `(1)/(sqrt(2))` time of the maximum velocity.

Answer

A, B, D

Explanation

Total energy, `E_(T) = (1)/(2)momega^(2)r^(2)`
`[K.E.]_(max) = (1)/(2)momega^(2)r^(2)`
`[K.E.]_(min) = 0`
`therefore` Average K.E. per cycle
` = ((1)/(2)momega^(2)r^(2) + 0)/(2)`
` = (1)/(2)((1)/(2)momega^(2)r^(2)) = (1)/(2)E_(T)`
And `u_(max) = romega`
`v_(min) = 0`
Mean velocity ` = (romega + 0)/(2) = (1)/(2)romega = (1)/(2)v_(max)`
`v_(rms) = sqrt((0 + r^(2)omega^(2))/(2)) = (romega)/(sqrt(2)) = (1)/(sqrt(2)) = v_(max)`