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

As we know,

`T = 2pisqrt(l/g)`

We observe that time period is inversely related to the square root of gravity's acceleration.

As a result, as 'g' decreases by one-fourth, the time period increases.

When the acceleration due to gravity is reduced by one-fourth, we see that

`T = 2pisqrt((l/g)/4)`

`T = 2pisqrt((4l)/g)`

`T = 2 xx 2pisqrt(l/g)` 

As a result, we can conclude that when gravity's acceleration is lowered to one-fourth, the time period of a basic pendulum doubles.

As, the given pendulum is a seconds' pendulum so T = 2s

∴ New T = 2 x 2 = 4s