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The point of suspension lambda of a simple pendulum with normal time perid T_(1) is moving upward according to equation y = kt^(2) where k =1 m//s^(2). If new time period is T^(2) the ratio T_(1)^(2)/T_(2)^(2) will be |
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Answer» `2//3` `(dy)/(DT) = k.2t RARR (d^(2)y)/(dt^(2)) = 2k` GIVEN, `k = 1 m//s^(2)` The point of suspension of the pendulum is moving upwards with an ACCELERATION of `2 m//s^(2)`. This is the case where the pseudo acceleration is acting downwards. g. (effective acceleration due to gravity) `=g +2 = 12 m//s^(2)` `T_(1) = 2pi sqrt(l/g), T_(2) =2pi sqrt(l/g.), THEREFORE T_(1)^(2)/T_(2)^(2) =g^(.)/g = 12/10 = 6/5` |
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