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)`.

Explanation

a) Increases in length of pendulum `=0.6m:`
Increases in time period of pendulum `=50%=1.5T`
Let original length of pendulum =1
Original time period `=T, g=9.8m//s^(2).`
For 1st case `9.8=4pi^(2)(l)/(T^(2)) to l,`
For 2nd case `l_(1)=(l+0.6),T_(1)=1.5T`
`9xx8=3pi^(2)(l_(1))/(T_(l1)^(2)) to (2),`
Divide eq. (2) with eq. (1)
`1=(l_(1))/(l)(T^(2))/(T_(1)^(2)) rArr (l_(1))/(l)=2.25 rArr l_(1) =2.25l`
But `l_(1)=l+0.6:`
`therefore l+0.6=2.25l rArr 0.6=1.25l`
`therefore` Length of pendulum `l=(0.6)/(1.25)=0.48m`
b) Time period `T=2pisqrt(l/g)`
`=2xx3.142sqrt((0.48)/(9.8))=6.284xx=0.2213=1.391sec.`