A simple pendulum has a time period `T_(0)` at north pole. Find time period at equator. Accunt for earth's rotation only.

Explanation

`T_(0)=2pisqrt(L/(g_("pole")))=2pisqrt(L/g)`
`T=2pisqrt(L/(g_("equator")))=2pisqrt(L/(g-omega^(2)R_(e)))`
`T/(T_(0))=sqrt(g/(g-omega^(2)R_(e)))=(1/(1-(omega^(2)R_(e))/s))^(1//2)=(1-(omega^(2)R_(e))/g)^(-1//2)`
`=(1+(omega^(2)R_(e))/(2g))implies T=T_(0)(1-(omega^(2)R_(e))/g)`
`omega=(2pi)/(24hr),R_(e)=6400 km, g=9.8 m//s^(2)`