The length of a seconds pendulum at a place is 1.02m. Find the time period of another pendulum of length 0.51m at the same place.

The length of a seconds pendulum at a place is 1.02m. Find the time pe

1.	The length of a seconds pendulum at a place is 1.02m. Find the time period of another pendulum of length 0.51m at the same place.
Answer» Solution :The TIME period of a SIMPLE pendulum is `T=2pisqrt((l)/(g))`………….(1) i.e., `(l)/(T^(2))= (g)/(4pi^(2))= "constant"` at a given place ……………(2) If `T_(1)` and `T_(2)` are the time periods of the pendulum for length `l_(1)` and `l_(2)`. From Eq(2) we cab WRITE `(l_(1))/(T_(1)^(2))= (l_(2))/(T_(2)^(2))`..............(2) Here `T_(1)=2s` when `l_(1)= 1.02m` when `l_(2)= 0.5 m, T_(2)=? ` Using Eq. (3)`(1.02)/(2^(2))= (0.51)/(T_(2)^(2))` or `T_(2)^(2)= (0.51)/(1.02) xx 4 =2 ` `:. T_(2)= sqrt(2) = 1.414 s` When the lenght increases the time period of the pendulum increases. So the clock loses time or RUNS slow.