InterviewSolution
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InterviewSolution
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Two blocks of masses `3kg` block is attached to a spring with a force constant, `k k = 900N//ma` which is compressed `2m` initially from its equilibrium position. When `3kg` mass is released, it strikes the `6kg` mass and the two stick togther in an inelastic collision. The velocities of a particle executing `S.H.M.` are `30cm//s` and `16 cm//s` when its displacements are `8cm` and `15cm` from the equilibrium position. then its amplitude of oscillation in cm is:A. `25`B. `21`C. `17`D. `13` |
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Answer» Correct Answer - C `v^(2) = omega^(2) (A^(2)-x^(2))` For `x = 8cm, 30^(2) = omega^(2) (A^(2) - 8^(2)) …………..(1)` For `x = 15 cm, 16^(2) = omega^(2) (A^(2) -15^(2)) …………(2)` Solving (1) and (2), `16^(2)A^(2) - 16^(2) (8)^(2) = 30^(2)A^(2) - (30)^(2) - (15)^(2)` `A^(2) (30^(2)-16^(2)) = (30)^(2) (15)^(2) - (16)^(2) (8)^(2) rArr A = 17` |
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