Question #1acc9

1 Answer
May 14, 2016

Answer:

None of the answers are applicable.
#T# would be very large but #!=oo#, in the case discussed below.

Explanation:

We know that for small angle of deviation #theta# from the mean position the time period #T# of a pendulum is given as
#T=2pisqrt(L/g)#
where #L# is the length of the pendulum and #g# is the local gravity.
When the pendulum is situated in a satellite, we need to find local #g# to calculate the time period.

Assuming that the satellite has left the earth's gravitational field and is cruising in a gravity free space where influence of the nearest celestial body can be ignored.

As such only force of gravitational attraction would be between the bob of the pendulum and the satellite. If left hanging in space of satellite and calculate, we will find that it is a very small force, as such local gravity #->0#
#=>T->oo#