The avergae kinetic energy of particle of mass `m` undergoing `S.H.M.` with angular frequency `omega` and amplitude `A,` over half of one time period is

Options

A. `(1)/(2) m omega^(2) A^(2)`
B. `(1)/(4) m omega^(2)A^(2)`
C. `m omega^(2)A^(2)`
D. `2 m omega^(2)A^(2)`

Answer

B

Explanation

average kinetic energy of a particle is
`K.E._(avg) = (1)/(2)m omega^(2) A^(2) (int_(0)^(T//2)cos^(2)omegat.dt)/(int_(0)^(T//2)dt)`
by solving we get `K.E_(avg) = (1)/(4)m omega^(2) A^(2)`