Question

A seconds pendulum is taken to a place where acceleration due to the gravity falls to one-fourth. How is the time period of the pendulum affected, if at all? Give reasons. What will be its new time period?

Answer 1

Time period of 'a' is inversely proportional to the square root of acceleration due to gravity.

i.e `T ∝ 1/sqrt("g")`

Now, if the acceleration due to gravity falls to one-fourth, the time period will be doubled.

Let the new time period be T' and let g' be the acceleration due to gravity.

Then,

`T/(T') = sqrt("g'"/"g")`

or , `T/(T') = sqrt((1/4"g")/"g")`

or , `T/(T') = 1/2`

or, T' = 2T

or , T = 2 × 2 = 4s