The time `t` of a complete oscillation of a simple pendulum of length `l` is given by the equation `T=2pisqrt(1/g)` where `g` is constant. What is the percentage error in `T` when `l` is increased by 1%?

Explanation

Percentage error in `l` is `(Delta l)/l xx100=0.01`
Given,
`t=2pisqrt(l/g)`
apply log on both sides,
`log (t)=log(2pisqrt(l/g))`
`=>log (t)=log(2pi)+log(sqrt(l/g))`
`=>log (t)=log(2pi)+log(l/g)^(1/2)`
`=>log(2pi)+1/2(log l - log g )`
taking differentiate on both sides we get,
`=>1/t Delta t=0+1/2*1/lxxDelta l-(0)`
`=>(Delta t)/t=1/2xx(Delta l)/l`
multiplying both sides with `100`
`=>(Delta t)/txx100=1/2xx(Delta l)/lxx100`
`=1/2 xx0.01`
`=0.5xx0.01`
`=0.005`
`therefore` percentage error in`t=0.005`