InterviewSolution
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InterviewSolution
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A mass `m` is suspended separately by two different spring of spring constant `K_(1)` and `k_(2)` given the time period `t_(1)` and`t_(2)` respectively if the same mass `m` is shown in the figure then time period `t` is given by the relation A. `t = t_(1) + t_(2)`B. `t =( t_(1)t_(2))/(t_(1) + t_(2))`C. `t^(2) = t_(1)^(2) + t_(2)^(2)`D. `t^(-2) = t_(1)^(-2) + t_(2)^(-2)` |
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Answer» Correct Answer - D `t_(1) = 2pi sqrt((m)/(k_(1)))` and `t_(2) = 2pi sqrt((m)/(k_(2)))` Equivalant constant for shown combination is , `(K_(1) +K_(2)` so time period `t` is given by `t= 2pi sqrt((m)/(K_(1) + K_(2)))` By solving these equations we get `T^(2) = t_(1)^(2) + t_(2)^(2)` |
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