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