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))` times of its maximum velocity.

Answer

A, B, D

Explanation

Total energy (mechanical energy) of a body performing simple harmonic motion
`E= (1)/(2) ka^(2)`
where A= amplitude and maximum kinetic energy obtained at mean position due to maximum kinetic energy.
Maximum kinetic energy,
`K_("max")= (1)/(2)m(v_("max"))^(2)`
`=(1)/(2)m A^(2) omega^(2)`
`=(1)/(2) kA^(2) " "[therefore m omega^(2) = k]`
Option A is correct.
Average kinetic energy of a body performing SHM at any time
`lt K gt = (0+ K_("max"))/(2)`
`= ((1)/(2)KA^(2))/(2)= (1)/(4) KA^(2)`
and half value of maximum kinetic energy `= (K_("max"))/(2)`
`=((1)/(2)KA^(2))/(2)`
`=(1)/(4)KA^(2)`
hence, option B is correct.
Average velocity in one cycle
`lt v gt = (0+v_("max"))/(2)= v(A omega^(2))/(2)`
Maximum velocity, `v_("max")= A omega`
`therefore (2)/(pi)" times "v_("max")= (A omega xx pi)/(2)`
hence, option C is incorrect.
Root mean square speed `v_("max")= (v_("max"))/(sqrt(2))` and hence, option D is correct.