Measurements and Experimentation — Question 4

As we know that,

$\text T = 2 \text π \sqrt{\dfrac{\text l}{\text g}}$

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 —

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

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 —

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

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