# When the length of simple pendulum is dubuled ; find the ratio of new frequency to old frequency?

Aug 5, 2018

The answer is $= \frac{1}{\sqrt{2}}$

#### Explanation:

The period $T$ of a simple pendulum is

$T = 2 \pi \sqrt{\frac{l}{g}}$

Where $l$ is the length of the pendulum

And $g$ is the acceleration due to gravity

The frequency is

$f = \frac{1}{T} = \frac{1}{2 \pi} \sqrt{\frac{g}{l}}$

Squaring both sides

${f}^{2} = \frac{1}{4 {\pi}^{2}} \cdot \frac{g}{l}$

Suppose l_1=2l

Then,

${f}_{1}^{2} = \frac{1}{4 {\pi}^{2}} \cdot \frac{g}{2 l}$

Then,

${f}_{1}^{2} / \left({f}^{2}\right) = \frac{\frac{1}{4 {\pi}^{2}} \cdot \frac{g}{2 l}}{\frac{1}{4 {\pi}^{2}} \cdot \frac{g}{l}} = \frac{1}{2}$

${f}_{1} / f = \frac{1}{\sqrt{2}}$