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Block `A` in the figure is released from the rest when the extension in the spring is `x_(0)`. The maximum downward displacement of the block will be `:` A. `(Mg)/(2k)-x_(0)`B. `(Mg)/(2k)+x_(0)`C. `(2Mg)/(k)-x_(0)`D. `(2Mg)/(k)+x_(0)` |
Answer» Correct Answer - A For maximum displacement `Mg(x) = 1/2k(2x)^(2) rArr x = (Mg)/(2k) rArr x = (mg)/(2k) - x_(0)` |
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