Measurements and Experimentation — Question 5

As we know that,

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

Time period is directly proportional to the square root of the length of the pendulum.

In the case when length is 1m,

$\text T_1 = 2 \text π \sqrt{\dfrac{1}{\text g}} \\[0.5em]$

and

In the case when length is 9m,

$\text T_2 = 2 \text π \sqrt{\dfrac{9}{\text g}} \\[0.5em]$

So, comparison of T₁ and T₂ gives —

$\text T_1 : \text T_2 = 2 \text π \sqrt{\dfrac{1}{\text g}} : 2 \text π \sqrt{\dfrac{9}{\text g}} \\[0.5em] \text T_1 : \text T_2 = \sqrt{\dfrac{1}{9}} \\[0.5em] \Rightarrow \text T_1 : \text T_2 = \dfrac{1}{3} \\[0.5em]$

Hence, T₁ : T₂ = 1 : 3