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

Options

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

Answer

b

Explanation

K.E. of a particle =`1/2mv^2`
For S.H.M `y=a sin omega cos omegat`
`:. K.E.=1/2ma^2omega^2cos^2omegat`
Max.`(K.E.)_("average")=1/4(ma^2)(2piv)^2=pi^2ma^2v^2`