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
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
To find the free-fall acceleration in a location, we can use the formula for the period of a pendulum:
T = 2π √(L / g)
Where T is the period of the pendulum, L is the length of the pendulum, and g is the acceleration due to gravity.
In this case, we are given the period of the pendulum as 3.4 seconds and the length of the pendulum as 2.87 meters. We can rearrange the formula to solve for g:
g = (4π² L) / T²
Plugging in the values we have:
g = (4π² * 2.87) / (3.4)²
Now we can calculate the value of g.