Question

"In 1851 Jean Bernard Leon Foucault demonstrated the rotation of the earth using a pendulum 11.0 m long, which was set up in the Paris Observatory. How long would it have taken for Foucault's pendulum to make one complete swing back to its starting point if g = 9.81 m/s2 at the observatory?"

Answer 1

It will take 6.656 seconds

Step-by-step explanation:

Period of oscillation is given by:

T = 2 × pi sqrt(L/g)

Where L is length of pendulum

g = acceleration due to gravity

T = 2 × 3.142 sqrt(11.0/9.8)

T= 6.284× sqrt(1.122)

T=6.656seconds

Answer 2

Answer:3.34seconds

Step-by-step explanation:

The relationship between period of oscillation Tand length of pendulum, given a small amount of amplitude displacement is given as

T=π×√l/g

π or pi=3 142

T=period is the time taken for the pedulum from a point through one complete swing back to that point.

l=length of pendulum=11m

g=acceleration due to gravity=9.81m/s2

T=3.142×√11/9.81=3.33s