T^2 = 4pi^2(L/g) = 3.4^2
39.48(2.87/g) = 11.56
113.3/g = 11.56
11.56g = 113.3
g = 9.80 m/s^2
What is the free-fall acceleration in a location
where the period of a 2.87 m long pendulum
is 3.4 s?
Answer in units of m/s
2
1 answer
1 answer