A body is performing SHM, then its

Options

A. average total enregy per cycle is equal to its maximum kinetic energy
B. average kinetic energy per cycle is equal to half of its maximum kinetic energy
C. mean velocity over a complete cycle is equal to `(2)/(pi)` times of maximum velocity
D. root mean square velocity is `(1)/(sqrt2)` times of its maximum velocity

Explanation

Let the equation of a SHM is represented as `x=asinomegat`
Assume mass of the body is m.
(a) Total mechanical energy of the body at any time t is
`E=(1)/(2)momega^(2)a^(2)`……(i)
Kinetic energy at any instant is
`K=(1)/(2)mv^(2)=(1)/(2)m[(dx)/(dt)]^(2)` `[:'v=(dx)/(dt)]`
`=(1)/(2)momega^(2)a^(2)cos^(2)omegat` `[:'x=asinomegat]`
`rArrK_(max)=(1)/(2)momega^(2)a^(2)=E` `[:'"For "k_(max),cosomegat=1].....(ii)`
(b) KE at any instant t is
`K=(1)/(2)momega^(2)a^(2)cos^(2)omegat`
`(K_(av))"for a cycle"=(1)/(2)momega^(2)a^(2)[(cos^(2)omegat)_(av)]"for a cycle"`
`=(1)/(2)momega^(2)a^(2)[(0+1)/(2)]`
`=(1)/(4)momega^(2)a^(2)=*(K_(max))/(2)` [form Eq. (ii)]
(c) Velocity = `v=(dx)/(dt)=acosomegat`
`v_(mean)=(v_(max)+v_(min))/(2)`
`=(aomega+(-aomega))/(2)=0` [For a complete cycle]
`v_(max)nev_(mean)`
(d) `v_(rms)=sqrt((v_(1)^(2)+v_(2)^(2))/(2))=sqrt((0+a^(2)omega^(2))/(2))=(aomega)/sqrt(2)`
`rArrV_(rms)=V_(max)/(sqrt2)`