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A simple pendulum has time period (T_1). The point of suspension is now moved upward according to the relation `y = K t^2, (K = 1 m//s^2)` where (y) is the vertical displacement. The time period now becomes (T_2). The ratio of `(T_1^2)/(T_2^2)` is `(g = 10 m//s^2)`.A. `5//6`B. `6//5`C. 1D. `4//5` |
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Answer» Correct Answer - B `y = kt^(2)` `:. (dy)/(dt) = 2kt` `rArr (d^(2)y)/(dt^(2)) = 2k = 2 m//s^(2)`…(i) `(because k = 1 m//s^(2), given)` we know that `T = 2 pi sqrt((l)/(g))` `:. (T_(1)^(2))/(T_(2)^(2)) =(g_(2))/(g_(1)) rArr (T_(1)^(2))/(T_(2)^(2)) = (12)/(10) =(6)/(5)` `[because g_(1) =10 m//s^(2), g_(2) =g+2=12 m//s^(2)]`. |
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