On an artificial satellite, orbiting the earth, net gravity is zero. Detailed solution time period of simple pendulum t = 2 π l g ' in artificial satellite g ′ = 0 ∴ t = infinite. Hence time period of simple pendulum is infinite.
As g = 0, so t = ∞. The correct option is ∞. To determine the time period of a simple pendulum on a freely moving artificial satellite, we can follow these steps:
Now for a body of mass m on the satellite the gravitational force due to earth is r2gm m towards the centre of the earth. Artificial satellites don't possess any natural g. The time period of a simple pendulum on a freely moving artificial satellite is (a) zero (b) 2 sec (c) 3 sec (d) infinite Identify the condition in a satellite:
The time period of a simple pendulum on a freely moving artificial satellite is infinite (d), as the pendulum cannot oscillate in the absence of gravity. On an artificial satellite orbiting the earth the acceleration is given by r2gm towards the centre of the earth. The value of g is zero in orbiting satellite: The time period of a simple pendulum is defined as the time taken by the pendulum to finish one complete oscillation.
Understand the formula for time period: When a point mass is attached to an inextensible string and suspended from fixed support then it is called a simple pendulum. On a freely moving artificial satellite, the effective gravity g becomes zero because the satellite is in free fall. In artificial satellite, effective value of g is zero.
The time period of a simple pendulum is given by the formula t =2π gl where l is the length of the pendulum and g is the acceleration due to gravity. In an experiment to determine the gravitational acceleration g of a place with the help of a simple pendulum, the measured time period squared is plotted against the string length of the pendulum in the figure.