A particle of mass `m` executes `SHM` with amplitude 'a' and frequency 'v'. The average kinetic energy during motion from the position of equilibrium to the end is:

Options

A. `pi^(2)ma^(2)epsi^(2)`
B. `(1)/(4)ma^(2)epsi^(2)`
C. `4pi^(2)ma^(2)epsi^(2)`
D. `2pi^(2)ma^(2)epsi^(2)`

Answer

A

Explanation

`K_(av)=(int_(0)^(T//4)(1)/(4)m[aomega cos(omegat)]^(2)dt)/(int_(0)^(T//4)dt)=(ma^(2)omega^(2))/(2(T)/(4))int_(0)^(T//4) cos^(2)(omegat)dt=(2ma^(2)omega^(2))/(T).(T)/(8)=(1)/(4)ma^(2)omega^(2)`
`=(1)/(4)ma^(2)(2piepsi)^(2) , =pi^(ma^(2)epsi^(2)`.