# Question #4c8b0

##### 1 Answer
Mar 8, 2017

The Period becomes $\sqrt{2}$ times the original one.

#### Explanation:

Let, $T , l \mathmr{and} g$ be the Period, Length and Gravitational

Acceleration of a Simple Pendulum.

$\therefore T = 2 \pi \sqrt{\frac{l}{g}} \ldots \ldots \ldots \ldots \ldots \ldots \ldots . . \left(1\right) .$

Suppose that, the period is $T '$ when, the length $l ' = 2 l .$

Then, by $\left(1\right) , T ' = 2 \pi \sqrt{\frac{l '}{g}} = 2 \pi \sqrt{\frac{2 l}{g}} \ldots \ldots \ldots \ldots . . \left(2\right) .$

$\therefore \frac{T '}{T} = \frac{2 \pi \sqrt{\frac{2 l}{g}}}{2 \pi \sqrt{\frac{l}{g}}}$.

$\Rightarrow \frac{T '}{T} = \sqrt{\frac{2 l}{l}} = \sqrt{2} , i . e . ,$

$T ' = \left(\sqrt{2}\right) T ,$ meaning that the Period becomes $\sqrt{2}$ times the

original one.