A particle of mass `m = 1 kg` excutes `SHM` about mean position `O` with angular frequency `omega = 1.0 rad//s` and total energy `2J`. `x` is positive if measured towards right from `O`. At `t = 0`, particle is at `O` and movel towards right.

Options

A. `{:(,P,Q,R,S),((A),1,2,4,3):}`
B. `{:(,P,Q,R,S),((B),2,1,3,4):}`
C. `{:(,P,Q,R,S),((C),4,1,3,2):}`
D. `{:(,P,Q,R,S),((D),4,2,3,1):}`

Explanation

`KE_(max)=1/2 mv_(max)^(2)=TE`
`rArr v_(max)=(sqrt(2xx2)/1)=2 m//s`
amplitude `A=(v_(max))/(omega)=2m`.
`x=A sin omega t=2 sin t`
`v=2 cos t=sqrt(4-x^(2))`
`(P) v=sqrt(2)m//s rArr xpmsqrt(2) m`.
`(Q) KE=1/2 mv^(2)rArr 1+1/2xx1xxv^(2) rArr v=sqrt(2) m//s`
`:. x=pmsqrt(2) m`.
(R) at `t=pi//6s, x=2 sin pi//6 =1,`.
(S) `KE=3/2 rArr 1.5 =1/2 xxmv^(2)`
`rArr v=sqrt(3) rArr x=pm1m`.