Question

What is the period, in seconds, of a simple pendulum of length 2 meters? Use the gravitational constant g = 9.8 m/s^2 and round your answer to two decimal places.

Answer 1

Answer:

The period, in seconds to two decimal places is, 2.84 sec

Step-by-step explanation:

Using formula:

$T = 2 \pi \sqrt{\frac{l}{g}}$             ......[1]

where

T represents the Time period in second

l represents the length of the pendulum/

As per the statement:

Use the gravitational constant g = 9.8 m/s^2 , length of the simple pendulum(l) = 2 meters and $\pi = 3.14$

Substitute the given values we have;

$T = 2\cdot 3.14 \cdot \sqrt{\frac{2}{9.8}}$

⇒$T = 6.28 \cdot \sqrt{\frac{1}{4.9}}=6.28 \cdot \frac{1}{2.21359436212}= \frac{6.28}{2.21359436212} = 2.83701481512$

Therefore, the period, in seconds to two decimal places is, 2.84 sec