A simple pendulum has a time period exactly 2s when used in a laboratory at north pole. What will be the time period if the same pendulum is used in a laboratory at equator? Account for the earth's rotation only. Take g Gmr2 = 9.8 m/s2 and radius of earth = 6400 km.
2.004 seconds
Consider the pendulum in its mean position at the north pole. As the pole is on the axis of rotation, the bob is in equilibrium. Hence in the mean position, the tension T is balanced by earth's attraction. Thus, T =g Gmr2 = mg. The time period t is
t = π √1Tm = 2 π lg ..(i)
At equator , the lab and the pendulum rotate with earth at angular velocity ω =2πradian24hour in a cirlce of raduis equal to 6400 km. Using Newton's second law,
Gmr2 - T′ = mω2R or, T' = mg−ω2R)
Where T' is the tension in the string
The time perioud will be
t' = 2 π √1Tm = 2 π √g−ωR2 ...(ii)
By (i) and (ii)
t′t = √gg−gω2R = (1−ω2Rg)−12
or t ' = t (1 + ω2Rg)
Putting the values, t ' = 2.004 seconds.