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 equilinrium to the end is.

Options

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

Answer

B

Explanation

(b) KEY CONCEPT: The instantaneous kinetic energy of a particle executing S.H.M. is given by.
`K=1/2ma^(2) omega^(2) sin^(2)omegat`
:. Average K.E. `=ltKgelt1/2momega^(2)a^(2) sin^(2)omegatgt`
`=1/2momega^(2)a^(2)ltsin^(2)omegatgt`
`=1/2momega^(2)a^(2)(1/2) (:. ltsin^(2)thetage1/2)`
`=1/2momega^(2)a^(2)=1/2ma^(2)(2piv)^(2) (:. omega=2piv)`
`=1/2momega^(2)a^(2) =1/2ma^(2) (2piv)^2) (:. omegapiv)`
or, `ltKgtpi^(2)ma^(2)v^(2)`.