What is the ratio of the free-fall accelerations at Japan and England? -

A quot;seconds pendulumquot; is one that moves through its equilibrium position once each second. (The period of the pendulum is precisely 2 sec.) The length of a seconds pendulum is 0.9927 m at Tokyo, Japan, and 0.9942 m at Cambridge, England. What is the ratio of the free-fall accelerations at these two locations?

T = 2 pi sqrt.(l/g) or g = pi^2 * l g(Tokyo) / g (Japan) = pi^2 * l(Tokyo) / p^2 * l(Japan) = l(Tokyo) / l(Japan) = 0.9927 / 0.9942 = 3307 / 3314 Hope this helps. your_guide123@yahoo.com

Quite simple, really. The period of a simple pendulum is T = 2π√(l/g) Since T and π are constants, we can re-write the above expression in terms of l and g, thus l/g = k where k is just another constant to replace (T/2π)2 So there you have it, a simple ratio l/g which must remain constant. 0.9927 / 0.9942 = 0.9985 is the Tokyo:Cambridge ratio.