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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. 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` |
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Answer» Correct Answer - A::B::D `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`. |
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