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.

Question

asked Jul 1, 2018 170k views

2 Answers

Poli · Answer 1 · 2018-07-05T08:55:32+0000

Answer:

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

Explanation:

Using formula:

$T = 2 \pi \sqrt{(l)/(g)}$ ......[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;

$T = 2\cdot 3.14 \cdot \sqrt{(2)/(9.8)}$

⇒
$T = 6.28 \cdot \sqrt{(1)/(4.9)}=6.28 \cdot (1)/(2.21359436212)= (6.28)/(2.21359436212) = 2.83701481512$

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