InterviewSolution
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InterviewSolution
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The masses in figure slide on a frictionless table. `m_(1)` but not `m_(2)`, is fastened to the spring. If now `m_(1)` and `m_(2)` are pushed to the left , so that the spring is compressed a distance `d`, what will be the amplitude of the oscillation of `m_(1)` after the spring system is released ? |
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Answer» Correct Answer - A::B::D While returning to equilibrium position, `(1)/(2)kd^(2) = (1)/(2)(m_(1) + m_(2))v^(2)` `:. v = (sqrt((k_(1))/(m_(1) + m_(2))))d` Now, after mean position `m_(2)` is detached from `m_(1)` and keeps on moving with this constant velocity `v` towards right. Block `m_(1)` starts SHM with spring and this `v` becomes its maximum velocity at mean position. `:. v = omega A` `:. (sqrt((k)/(m_(1) + m_(2))))d = (sqrt((k)/(m_(1))))A` `:. A= (sqrt((m_(1))/(m_(1) +m_(2))))d`. |
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