A simple pendu,um has a time eriod exactly 2 s when used i a laboratory at north pole. What wil be the time period if the same pendulum is used in a labroator at equator? Ccount for the earth's rotation only. Take `g=(GM)/R^2=9.8ms^-2 ` and radius of earth =6400 km

Explanation

Consider the pendulum in its mean position ast the north pole. As the pole is one the axis of rotation, thke bob is in equilibrium. Hence in the mean positin the tension T is balanced by earth's attraction. Thus, `T=(GMm)/R^2=mg`. The tiem period t is
`t=2pi sqrt(l/(T/m))2pisqrt(l/g)`............i
At equator the lab and the pendulum rotate with the earth angulr velocity `omega=(2piradian)/(24 hour) `in a circle of radius equal to 6400 km. Using Newton's second law, `(GMm)/R^2-T'=momega^2R or, T'=mg(g-omega^2R)`
where T' is the tension in the string.
the time period will be
`t'=2pi sqrt(l/((T'/m)))=2pi sqrt(l/(g-omega^2R))`......ii
By i and ii
`t'/t=sqrt(v/(g-omega^2-R))=(1-(omega^2R)/g)^(-1/2)`
or `t'=t(1+omega^2R)/(2g)`.
Putting the value t'=2.004 seconds`.