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:
(1) `2pi^(2) ma^(2) v^(2)` (2) `pi^(2)ma^(2) v^(2)`
(3) `(1)/(2) ma^(2) v^(2)` (4) `4pi^(2) ma^(2) v^(2)`

Options

A. `1/4m a^(2)v^(2)`
B. `4pi^(2)m a^(2)v^(2)`
C. `2pi^(2)m a^(2)v^(2)`
D. `pi^(2)m a^(2)v^(2)`

Answer

D

Explanation

For a particle to excute simple harmonic motion its displacement at any time t is given by `x(t)= a sin omega t` where a =amlitude, `omega`= angular frequency
velocity of a particle v= `dx/dt= a cos omega t`
kinectic energy is `k =1/2 m v^(2)` `implies k=1/2m a^(2) omega^(2) cos^(2) omega t`
Average kinectic energy, =`1/2m omega^(2)a^(2) lt cos^(2) omega tgt`
`=1/2 m omega^(2)a^(2)(1/2)` `[therefore ltcos^(2)thetage1/2]` =`1/2m a^(2)(2piv)^(2)` `[therefore omega=2piv] implies