A "seconds pendulum" is one that moves through its equilibrium position once each second. (The period of the pendulum is precisely 2.000 s.) The length of a seconds pendulum is 0.9925 m at Tokyo and 0.9948 m at Cambridge, England. What is the ratio of the free-fall accelerations at these two locations?
To find the ratio of the free-fall accelerations at Tokyo and Cambridge, we need to use the formula for the period of a simple pendulum:
T = 2π * √(L/g)
where T is the period, L is the length of the pendulum, and g is the acceleration due to gravity.
Given that the period of the seconds pendulum is precisely 2.000 s, we can rearrange the formula to solve for g:
g = 4π² * (L/T)²
First, let's calculate the acceleration due to gravity at Tokyo:
Length of the seconds pendulum at Tokyo, L1 = 0.9925 m
Period of the pendulum, T = 2.000 s
g1 = 4π² * (0.9925/2.000)²
Next, let's calculate the acceleration due to gravity at Cambridge:
Length of the seconds pendulum at Cambridge, L2 = 0.9948 m
g2 = 4π² * (0.9948/2.000)²
Finally, we can find the ratio of the free-fall accelerations by dividing g2 by g1:
Ratio of the free-fall accelerations = g2 / g1
Just plug in the values and calculate the ratio to find the answer.