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Find the time period of another pendulum of length equal to half of it at the same place. |
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Answer» Solution :The PERIOD of SIMPLE PENDULUM is given by `T= 2pisqrt((l)/(g)) ….(1) :. G= 4pi^(2)(l)/(T^(2))` i.e., `(l)/(T_(1)^(2))= (l_(2))/(T_(2)^(2)) implies (1.0)/(2^(2))= (0.50)/(T_(2)^(2))` or `T_(2)^(2)= (0.50)/(1.01) xx 4= 2 ` `:. T_(2)= sqrt(2)= 1.414s` |
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