What is the free-fall acceleration in a location
where the period of a 0.594 m long pendulum
is 1.55 s s?
Answer in units of m/s
period=2PI*sqrt(l/g)
square both sides
T^2=4PI^2 (l/g0
g= 4 PI^2 length/period^2
To calculate the free-fall acceleration in a specific location, we can use the equation for the period of a pendulum:
T = 2π √(L/g)
where T is the period, L is the length of the pendulum, and g is the free-fall acceleration.
In this case, we are given the period T = 1.55 s and the length of the pendulum L = 0.594 m. We can rearrange the equation to solve for g:
g = (4π² L) / T²
Now we can substitute the given values into the equation:
g = (4π² * 0.594 m) / (1.55 s)²
Let's calculate the value.