Two pendulums of lengths 100 cm and 121 cm suspended side by side. A suspended bobs suspended at its end stretched and leave together. After how many minimum oscillations of big pendulum, both are in same phase?

Two pendulums of lengths 100 cm and 121 cm suspended side by side. A s

1.	Two pendulums of lengths 100 cm and 121 cm suspended side by side. A suspended bobs suspended at its end stretched and leave together. After how many minimum oscillations of big pendulum, both are in same phase?
Answer» `11` `10` `21` `20` Solution :Suppose `T_(1)" and "T_(2)` are the PERIODIC time of two pendulums `T_(1)= 2pi sqrt((100)/(g))" and "T_(2)= 2pi sqrt((121)/(g))""[therefore T_(1) LT T_(2)," because "l_(1) lt l_(2)]` Suppose at t=0, both oscillate at same time but their oscillations are the same due to different period. Suppose pendulum of long string MAKES n oscillations and pendulum of short string makes `(n+1)` oscillations. `therefore (n+1)T_(1) = nT_(2)` `therefore (n+1)xx 2pi sqrt((100)/(g))= n xx 2pi sqrt((121)/(g))` `therefore (n+1)10= 11n` `therefore 10n +10 = 11n` `therefore n = 10`.