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InterviewSolution
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A body of mass `0.10kg` is attached to vertical massless spring with force constant `4.0 xx 10^(3)N//m`. The body is displaced `10.0cm` from its equilibrium position and released. How much time elapses as the body moves from a point `8.0 cm` on one side of the equilibrium position to a point `6.0cm` on the same side of the equilibrium position ? |
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Answer» Correct Answer - A::C::D `omega = sqrt ((k)/(m)) = sqrt ((4 xx 10^(3))/(0.1)) = 200 rad//s` `X = 10 sin (200 t)` `6 = 10 sin 200 t_(1)` `:. 200 t_(1) = sin^(-1)((3)/(5)) = 0.646 rad` `:. t_(1) = 3.23 ms` `8 = 10 sin 200t_(2)` or `200 t_(2) = sin^(-1)((4)/(5)) = 0.925 rad` `:. t_(2) = 4.62 ms` `:. Delta t = 1.4 ms = 1.4 xx 10^(-3)s`. |
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