The period of a simple pendulum is found to increases by 50% when the length the pendulum is increases by 0.6m. Calculate the initial length and the initial period of oscillation at a place where `g=9.8 m//s^(2)`.

Options

A. 0.16 m
B. 0.32 m
C. 0.48 m
D. 0.60 m

Answer

C

Explanation

`T alpha sqrt(l)`
`(T_(1))/(T_(2)) = sqrt((l_(P))/(l_(p) + 0.6))`
`rArr (4)/(9) = (l)/(l + 0.6)`
`4l + 0.24 = 4l`
5l = 0.24
l = 0.48