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If the length of second pendulum becomes (1)/(3) what will be its periodic time? |
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Answer» Solution :`implies (2)/(SQRT(3)) s` Time period of second PENDULUM `T_(1) = 2s` In EQUATION `T= 2pi sqrt((l)/(g)), 2pi, g` are constant `therefore T propto sqrt(l)` `therefore (T_2)/(T_1)= sqrt((l_2)/(l_1)) = sqrt((l)/(3XX l_(1)))= (1)/(sqrt(3))""[therefore l_(2)= (l_1)/(3)]` `therefore T_(2) = (T_1)/(sqrt(3))= (2)/(sqrt(3))s`. |
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