A particle is executing linear `SHM`. The average kinetic energy and avearge potential energy over a period of oscillation are `K_(av)` and `U_(av)` Then

Options

A. `K_(av) = (U_(av))/(2)`
B. `U_(av) = (K_(av))/(2)`
C. `K_(av) = U_(av)`
D. `U_(av) = (K_(av))/(3)`

Answer

C

Explanation

`K = (1)/(2) mv ^(2) = (1)/(2)mA^(2)omega^(2)cos^(2) omega t`
Average value of `cos^(2) omega t` is `(1)/(2)` over one cycle
`K_(av) = (1)/(4)m omega^(2)A^(2)`....(i)
`U = (1)/(2)m omega^(2)x^(2) = (1)/(2) m omega^(2)A^(2)sin^(2) omegat`
Average value of `sin^(2) omega t` is `(1)/(2)` over one cycle
`:. U_(av) = (1)/(4)m omega^(2)A^(2)`...(ii)
From Eqs (i) and (ii), we get `K_(av) = U_(av)`