Question

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 $\frac{Gm}{r^2}$ = 9.8 m/s² and radius of earth = 6400 km.

• A. 0.004 seconds
• B. 4.004 seconds
• C. 2 seconds
• D. 2.004 seconds

Answer 1

The correct option is D

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 $\frac{Gm}{r^2}$ = mg. The time period t is

t = $\pi$ $\sqrt{\frac{1}{\frac{T}{m}}}$ = 2 $\pi$ $\frac{l}{g}$ ..(i)

At equator , the lab and the pendulum rotate with earth at angular velocity $\omega$ =$\frac{2\pi radian}{24hour}$ in a cirlce of raduis equal to 6400 km. Using Newton's second law,

$\frac{Gm}{r^2}$ - $T^{{}^{\prime }}$ = m${\omega }^{2}R$ or, T' = m$g-{\omega }^{2}R)$

Where T' is the tension in the string

The time perioud will be

t' = 2 $\pi$ $\sqrt{\frac{1}{\frac{T}{m}}}$ = 2 $\pi$ $\sqrt{g-{\omega R}^2}$ ...(ii)

By (i) and (ii)

$\frac{t^{\prime }}{t}$ = $\sqrt{\frac{g}{g-g{\omega }^{2}R}}$ = ${(1-\frac{{\omega }^{2}R}{g})}^{\frac{-1}{2}}$

or t ' = t (1 + $\frac{{\omega }^{2}R}{g}$)

Putting the values, t ' = 2.004 seconds.