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?

Options

A. `11`
B. `10`
C. `21`
D. `20`

Answer

B

Explanation

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`.