How is the time period of a simple pendulum affected, if at all, in the following situations:

(a) The length is made four times,

(b) The acceleration due to gravity is reduced to one-fourth.

As we know that,

$T = 2 π \sqrt{\dfrac{l}{g}}$

where,

T = Time period

l = effective length of the pendulum

g = acceleration due to gravity.

(a) In the case when length is made four times, let time period be T₁, we see that —

$T$1 = 2 π \sqrt{\dfrac{4l}{g}} \\[0.5em] T1= 2 \times 2 π \sqrt{\dfrac{l}{g}} \\[0.5em] T_1 = 2 \times T \\[0.5em]

Hence, we can say that when the length is made four times, time period of a simple pendulum is doubled.

(b) In the case, when acceleration due to gravity is reduced to one fourth, let time period be T₁, we see that —

$T$1 = 2 π \sqrt{\dfrac{l}{\dfrac{g}{4}}} \\[0.5em] T1 = 2 π \sqrt{\dfrac{4l}{g}} \\[0.5em] T1 = 2 \times 2 π \sqrt{\dfrac{l}{g}} \\[0.5em] T1 = 2 \times T \\[0.5em]

Hence, we can say that when acceleration due to gravity is reduced to one fourth, time period of a simple pendulum is doubled.