What happens to the time period of a simple pendulum when it is taken from equator to the pole?

Answer

Verified

Hint:Use the formula for time period of a simple pendulum and find the relation between the time period and acceleration due to gravity. Then use the fact that the value of g is inversely proportional to the square of the distance (r) from the centre to the surface of earth.

Formula used:

$T=2\pi \sqrt{\dfrac{l}{g}}$

Complete step by step answer:

A pendulum clock consists of a pendulum. Time is calculated by the oscillations of the pendulum in the clock. Time period of the oscillations of the pendulum is known, which helps to calculate the time.The time period of oscillation of a pendulum depends on the length (l) of the string to which the bob is attached. It also depends on the acceleration to gravity (g) due to the gravitational force acting on the bob.The time period of a pendulum is given as $T=2\pi \sqrt{\dfrac{l}{g}}$.From the above equation for time period (T), we understand that the time period of a pendulum is directly proportional to the square root of its length but inversely to the acceleration due to gravity.i.e. $T\propto \dfrac{1}{\sqrt{g}}$ …. (i)In the given case, since the same clock is taken from the equator to pole, the length of the pendulum remains the same.However, the value of g changes from place to place on earth.The value of g is inversely proportional to the square of the distance (r) from the centre to the surface of earth.i.e. $g\propto \dfrac{1}{{{r}^{2}}}$The value of r decreases as we go from the equator to the poles. As a result, the value of g increases as we go from equator to the pole.And from (i) we know that if the value of g increases, the value of T decreases.This means that the time period of the pendulum is less at the poles than that of at the equator. Since the time period has reduced, the pendulum will undergo more number of oscillations than that of at the equator. As a result, the clock at the pole will show more time than the correct time at the equator.Therefore, we can say that when the pendulum is taken to a pole, it will gain time.

Hence, the correct option is A.

Note:The formula for the time period given above is only applicable when the pendulum is placed in air and the air resistance is ignored.

When another external force (including gravity) acts on the bob of the pendulum, the formula for the time period changes to $T=2\pi \sqrt{\dfrac{l}{{{g}_{eff}}}}$ .Here, ${{g}_{eff}}$ is the effect acceleration due to the gravitational force and the other force. In other words, it is the acceleration due to the net external force.Note that tension is an internal force.