What is the period, in seconds, of a simple pendulum of length 3 meters? Use the gravitational constant g = 9.8 m/s2 and round your answer to two decimal places.

The answer is 3.47638 (AP3X)

Rounded it is 3.48

the period, in seconds = 3.47 s

Step-by-step explanation:

Given:

length of pendulum, l = 3 m

gravitational constant, g=9.8 m/s^2

Period of pendulum, T=?

Applying formula for period of pendulum:

T=2π√ (l/g)

Putting values we get:

T=2π√ (3/9.8)

T=3.47 s!