Question

A pendulum that moves through its equilibrium position once every 1.000 s is sometimes called a "seconds pendulum."

Answer 1

I would solve this problem by starting from the equation for a pendulum's period:

T ≈ 2π√( L / g )

Rearranging for g, and substituting the length in Cambridge:

g ≈ ( 4π² * L ) / T²
. . = ( 4π * 0.9942 m ) / ( 1.000 s + 1.000 s )²
. . = 9.812 m/s²

And for Tokyo:

g ≈ ( 4π² * L ) / T²
. . = ( 4π * 0.9927 m ) / ( 1.000 s + 1.000 s )²
. . = 9.798 m/s²