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For the arrangement shown in figure, the spring is inilially compressed by 3cm. When the spring is released the block collides with the wall and rebounds to compress the spring again. If the time starts at the instant when spring is released, find the minimum time after which the block becomes stationary. |
Answer» Solution :If collision is elastic then In the case of spring - mass system, since the time period is INDEPENDENT of the amplitude of oscillation THEREFORE.  `T = (T_(0))/(2pi) [PI + 2 sin^(-1) ((1)/(3))]` `T = sqrt((m)/(k)) [pi + 2sin^(-1) ((1)/(3))]` Case- II if collision is inelasticFrom t = 0 to before collsion let time is `t_(1)` `t_(1) = (T_(0))/(4) + (T_(0))/(2pi) sin^(-1) ((1)/(3))` After collision the amplitude will changed to `sqrt5` therefore if the time is `t_(2)` before coming rest instantaneously then `t_(2) = (T_(0))/(4) + (T_(0))/(2pi) sin^(-1) ((1)/(sqrt5))` Total time taken `T = t_(1) + t_(2)` `T = sqrt((m)/(k)) [pi + sin^(-1) ((1)/(3)) + sin^(-1) ((1)/(sqrt5))]` |
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