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. speed of particle is `sqrt(2)m//s` at `x = +- sqrt(2)m`
B. Kinetic energy of the particle is `1J` at `x = +- sqrt(2)m`
C. At `t = pi//6` sec. particle is at `x = -1 m`
D. Kinetic energy is `1.5 J` at `x = +- 1m`

Answer

A, B, D

Explanation

`KE_(max)=(1)/(2)mv_(max)^(2)=TE`
`rArr v_(max)=sqrt((2xx2)/(1))=2 m//s`
amplitude `A = (v_(max))/(omega) = 2 m`
`x = A sin omega t = 2 sin t`
`v =2 cos t = sqrt(4 - x^(2))`
(A) `v = sqrt(2) m//s rArr x = pm sqrt(2) m`
(B) `KE = (1)/(2) mv_(2) rArr 1 = (1)/(2) xx 1 xx v_(2)`
`rArr v = sqrt(2) m//s`
`:. x = pm sqrt(2) m`
(C) at `t = pi//6 s, x =2 sin pi//6 = 1m`
(D) `KE = (3)/(2) rArr 1.5 =(1)/(2) xx mv_(2)`
`rArr v = sqrt(3) rArr x = pm 1 m`.