Question

A pendulum with a period of 2.00000s in one location (g=9.80ms^-2 ) is moved to a new location where the period is now 1.99796s. What is the acceleration due to gravity at its new location

Answer 1

$$T_1=2\pi\sqrt{l/g_1}\\ T_2=2\pi\sqrt{l/g_2}$$

$$g_2=g_1\frac{T_1^2}{T_2^2}=9.80*\frac{2.00000^2}{1.99796^2}=9.82\:\rm m/s^2$$